Sleeping Beauty Paradox [Archive] - Two Plus Two Newer Archives

bigmonkey

06-08-2007, 10:10 AM

Siegmund:[ QUOTE ]
I'll be interested in seeing what good argument, if any, there is for 1/3... sure haven't heard one yet and have trouble imagining one yet.

[/ QUOTE ]

See PairTheBoard's link to Elga's paper, where his argument for 1/3 begins properly in the second section, although all these Sleeping Beauty papers are pretty short and accessible. It's easier to sympathise with 1/3 when you grant that the scientists may as well flip the coin after already waking Beauty up on Monday.

The weird process is how Elga's argument goes something like:

P(H) is obviously 1/2. (intuition)
*Do some calculations*
OK she should believe P(H) is 1/3.

David Lewis replies to Elga and uses this method.

Being rational she should believe P(H) is 1/2.
As far as Lewis is concerned this is the end of the story.
But if you apply Elga's method to Lewis' answer you get the following strange conclusion:
Upon waking on Monday if Beauty were told it was Monday she should believe the probability of a future coin landing heads is 2/3.

Both of their explanations seem to contradict themselves. I think the conclusion is that Elga reasons from Beauty's pre-credence (1/2) to her betting odds (1/3), and Lewis reasons from her post-credence (1/2) to her pre-credence (1/2) and in the process reveals her betting odds for a different game on Monday (2/3)

I'm not quite sure how to interpret what that game is that could make her believe or bet on heads with credence 2/3.

From Elga we know that if she's told it's Monday she should have credence 1/2 in H, because it hasn't been flipped yet. If she's told it's Tuesday then she should have credence 0 in heads. So waking up with no information she knows there's some chance it's Monday and some chance it's Tuesday, therefore some chance P(H)=1/2 and some chance P(H)=0. This is why it looks like her overall value for P(H) < 1/2 and can be worked out as 1/3. Lewis responds by giving her an abnormal credence that H is true given it's Monday, to counterbalance the 1/3 claim.

I think the explnation might be: she knows there's some chance P(H)=1/2 and some chance P(H)=0. Interpreted this means there's some chance she is being offered a fair bet and some chance she is being offered a guaranteed-to-lose bet. She therefore demands abnormally good odds on the bet in case it is a guaranteed-to-lose bet.

Every time I re-read Elga and Lewis they seem pretty convincing, which suggests my mind is a bit fickle.

f97tosc:[ QUOTE ]
I can't say I am familiar with the reflection principle, but I suspect that the "resolution" will have to do with the fact during certain experimental outcomes, we supress Sleeping Beauty's ability to state (or even form) her belief.

[/ QUOTE ]

The reflection principle is that:

Px(A|Px+y[A]= c)=c

Where Px and Px+y are credence functions at time x and time x+y. So the principle is that given that at some later time your credence for A is c then your credence for A now should be C. If you believed now that tomorrow you would receive some important information about soemthing that will make you have credence c in A, then you should now already have credence c in A. I think it could be argued that Beauty isn't completely rational because she believes she will be given an amnesiac, and therefore cannot be in a rational cognitive state, but then people argue that she knows exactly what memories she will forget, and also the amnesiac is only given to her after going back to sleep to forget her awakening, so when she wakes up she is supposed to be fully rational.

PairTheBoard: [ QUOTE ]
Solution in White:

[/ QUOTE ]

I did originally think Beauty was using a principle of indifference over whether it's Monday or Tuesday where she shouldn't be.
Then in the re-computation you feed in the probabilities of heads and tails being 1/2 and effectively get them straight back out again, which is what Lewis does. But he also has the weird consequence that upon being told it is Monday Beauty believes a future coin has 2/3 chance of landing heads.
And then you get the Elga paper involved arguing for 1/3.

My current conclusion is that there are credences and betting odds, and these aren't always the same. However, it's quite hard to locate where Beauty is using credences and where she is using betting odds. The long-term argument whereby she knows that 2/3 of her total awakenings will have been preceded by a tails really seems like a credence and not a betting odds.Perhaps the reason this seems like a paradox is because as Bayesian thinkers we find it hard to tell the difference between credences and betting odds, and get punished in examples like this.

KipBond: [ QUOTE ]
The reason 1/3 is now valid is because it can be implied that the question is actually: "what is the probability that the coin landed heads and then we woke you and asked you this question?"

[/ QUOTE ]

They could ask her before the experiment "what is the probability of heads and that we will ask you again later?" As she knows for a fact that they will ask her again later she should still say 1/2 at this point. Is your point that the supposed paradox is really a counter-example and refutation of the Reflection Principle?

KipBond

06-08-2007, 10:29 AM

[ QUOTE ]
KipBond: [ QUOTE ]
The reason 1/3 is now valid is because it can be implied that the question is actually: "what is the probability that the coin landed heads and then we woke you and asked you this question?"

[/ QUOTE ]

They could ask her before the experiment "what is the probability of heads and that we will ask you again later?" As she knows for a fact that they will ask her again later she should still say 1/2 at this point.

[/ QUOTE ]

Once you combine the "waking" event with the "coin flipping" event, you get a separate event: a "flipping+waking" event. Since she will be woken up twice as often after a "tails" outcome of the flipping event, the odds are 1/3 she will have a "waking+heads" event, and 2/3 she will have a "waking+tails" event. It doesn't matter if she's asked this before or after being woken. The probability that the coin will be heads is 1/2. The probability that the coin will be heads and she will be woken is 1/3.

After 1,000,000 coin flips, there will be about 500,000 heads+waking events, and 1,000,000 tails+waking events.

bigmonkey

06-08-2007, 12:18 PM

I don't see how you are offering a solution that doesn't just concede that Beauty violates Reflection.

The scientists ask her before, "What is the probability that between t1 and t2 a heads will land and you will awake?"

When she wakes up they ask her "What is the probability that between t1 and t2 a heads landed and you woke up?"

She knows before that she will wake up, and she can already know exactly what the experience of waking up is like, just to make sure she gets no new information.

If you answer 1/2 to the first question and 1/3 to the second question, it sounds like they only wake her up 2/3 of the time after a heads lands, but we know that is false, so I don't see why this is not a contradiction.

KipBond

06-08-2007, 02:35 PM

[ QUOTE ]
I don't see how you are offering a solution that doesn't just concede that Beauty violates Reflection.

[/ QUOTE ]

Well, she should be able to answer question in 2 different ways, while clarifying the ambiguity.

[ QUOTE ]
The scientists ask her before, "What is the probability that between t1 and t2 a heads will land and you will awake?"

[/ QUOTE ]

What is t1 & t2?

It depends on the exact question being asked as to whether the answer is 1/2 or 1/3.

1/2 of the coin flips will be heads.
1/3 of the flip-wakings will be heads.

It's ambiguous as to what question is being asked, so it needs to be phrased precisely. If the question doesn't differentiate between the "coin flips" and the "flip-wakings", then it can be answered 2 ways.

bigmonkey

06-08-2007, 03:09 PM

t1 and t2 were just two moments in time I made up and didn't define. It's not really important, and should be something like t1 as when Beauty fell asleep and t2 when she awoke.

[ QUOTE ]
1/2 of the coin flips will be heads.
1/3 of the flip-wakings will be heads.

[/ QUOTE ]

That does seem like the intuitive thing to say before she falls asleep. When she wakes up they seem to be the same thing.

Are you saying that when she is woken up and asked what credence she has that she just experienced a heads-awakening she answers 1/3? And if asked if the coin landed heads she answers 1/2?

But if a heads occurs then she will experience a heads-awakening, and if she experiences a heads-awakening, then a heads occurred. They are equivalent, so P(H) should equal P(heads-awakening)

Siegmund

06-08-2007, 03:19 PM

[ QUOTE ]
But if a heads occurs then she will experience a heads-awakening, and if she experiences a heads-awakening, then a heads occurred. They are equivalent, so P(H) should equal P(heads-awakening)

[/ QUOTE ]

Well, there's your explanation. No paradox at all; your "they are equivalent, P(H) should equal P(heads-awakening)" is just plain wrong. P(H) is (heads)/(coinflip outcomes) while P(heads-awakening) is (heads-awakenings)/(all awakenings). Unless EVERY coin flip results in the same number of awakenings, P(H) and P(heads-awakening) are not equivalent.

It amazes me so much has been written about this.

KipBond

06-08-2007, 04:57 PM

Here's a simpler scenario:

Same setup w/ the amnesiac Beauty -- only this time if the coin is heads, I don't wake her up at all. I flip the coin again. If it's tails, I wake her up & ask what the P(H) is (*). What's her answer?

(*) The way I phrase the question is the ambiguous part. What is the P the coin will be heads when you flipped it? Or what is the P that the coin you last flipped before waking me (**) is heads?

(**) Note that this "waking" (& the whole scenario) is actually additional information, so it's wrong to say Beauty doesn't have different information to change her mind regarding the P(H).

PairTheBoard

06-08-2007, 08:12 PM

[ QUOTE ]
I did originally think Beauty was using a principle of indifference over whether it's Monday or Tuesday where she shouldn't be.
Then in the re-computation you feed in the probabilities of heads and tails being 1/2 and effectively get them straight back out again, which is what Lewis does. But he also has the weird consequence that upon being told it is Monday Beauty believes a future coin has 2/3 chance of landing heads.

[/ QUOTE ]

I don't see how my "valid?" model gives the conclusion Lewis arrives at. The tricky thing my Model does is that it takes the Two Full Awakinings produced by a Tails, and turns them into Two Half Awakinings in the Model. If she looks at that Model it seems to point to the incorrect conclusion that If she is awaken and told it is Monday, she concludes she is twice as likely to have gotten there by Heads than by Tails. That's just false. Especially when you consider the point you raised that we could just as well wait till she goes back to sleep on Monday to flip the coin. Somethings wrong there but I'm not sure what it is. I don't think it's a clear Model.

I've been thinking about looking at an analogous model like the following:

We flip a coin and put one or two Bags in an empty Basket. The bags are indistingishable from the outside. On the inside are two kinds of prizes. H-prizes and T-prizes. If Heads comes up we put Bag with an H-prize in the basket. If Tails comes up we put Two Bags with T-prizes in the basket. We put a little mark on the Two T-prizes. We mark one "Monday" and the other "Tuesday".

After putting the prize or prizes on the Basket, one will be drawn at random and handed to the Beauty. She can't see how many prizes are in the basket.

Now before we flip the coin we ask the Beauty the probability it will land heads. She answers 50%. Also, before we flip the coin we ask her questions about what happens when she recieves a Bag from the basket. We ask her, "When you recieve a Bag from the basket, what is the probability the Bag contains an H-prize?" She will rationally answer 1/3. But this is no puzzle to us, is it? We're talking about two different Events so it's not suprising they have two different probabilities. One Event is "Coin Lands Heads". The other event is "H-Prize in random Bag from Basket".

Now, is there anything we might ask her after She actually recieves a Bag from the Basket where her answer would puzzle us. If she is asked the above question she will still answer 1/3 but it's no puzzle because that's still the probability of a different Event than the Event "Coin Lands Heads".

ok. Let's think. Now she is in the Position of being handed a Bag from the Basket. She knows one of two things has happened. Either a Heads landed or a Tails landed. She knows she can find out which outcome happened by opening the bag and looking inside. She thinks to herself 1/3 of the time I open the bag I will find out that Heads landed, and 2/3 of the time I open the Bag I will find out Tails landed. Now she thinks,

"Does that mean my credence for believing Heads landed is 1/3?"

If she repeats the experiment and puts herself in that position numerous times she will open the bag and discover Heads has landed 33% of the times.

There's the Paradox. The number of times she puts herself in that position is the same number of times the Coin has been flipped. She discovers Heads has landed 33% of the times she is put in that position.

Does she "Discover heads has landed" if and only if "Heads has Landed"? Yes. There is a 1-1 correspondence between her "Discovery of an H-Prize" and the "Coin having landed Heads". There is a 1-1 correspondence between her "Bag Openings" and "Coin Flips". And she "Discovers H-Prize" 1/3 of the "Bag Opening" times. It looks like she must conclude that the "Coin Flips" have been landing Heads 1/3 of "Coin Flip" times.

Are any of these 1-1 correspondences false? Are any of my terms in quotes ambiguous? Doesn't the logic undeniably follow?

No. All the statements in bold are False. Half of the "Bag Opening" times she will discover an H-prize and 50% of the "Bag Opening" times she will discover a T-prize. That's if the experiment is exactly repeated over and over.

So there is no Paradox. At least not so far. We should maybe ask at this point, "Are our "Bags" truly analogous to Beauty's "Awakenings?". I'm not seeing anything different. I suppose we should track Elga's Logic through our Bags Model to see where it goes. We have T-Prizes labeled "Monday" and "Tuesday" if he needs to refer to that.

Let's see if I can do that. He says,

[ QUOTE ]
Elga -
I will argue first that P(T1) = P(T2), and then that P(H1) = P(T1).

[/ QUOTE ]

For us, T1 is a Monday labeled T-Prize Bag. T-2 is a Tuesday labeled T-Prize Bag. And H1 is an H-Bag Prize, which we could label Monday if he likes. So when Beauty looks at an unopened Bag will she think the bag is equally likely to contain a Monday T-Prize as a Tuesday T-Prize? Yes. She knows there is a probability of 25% that the bag contains either of those. Now, does she think it is equally likely the bag contains an H-Prize as a Monday T-Prize? NO! She does NOT think that. The First has probability 50% while the second has probability 25%. So here, OUR Beauty is more logical than Elga's. So where does Elga's beauty get her logic to say they are equally likely. Let's look,

[ QUOTE ]
Elga -
Now: if (upon awakening) you were to learn that it is Monday, that would amount to your learning that you are in either H1 or T1. Your credence that you are in H1 would then be your credence that a fair coin, soon to be tossed, will land Heads. It is irrelevant that you will be awakened on the following day if and only if the coin lands Tails - in this circumstance, your credence that the coin will land Heads ought to be 1/2. But your credence that the coin will land Heads (after learning that it is Monday) ought to be the same as the conditional credence P(H1 | H1 or T1). So P(H1| H1 or T1) = 1/2, and hence P(H1) = P(T1).

[/ QUOTE ]

Wait a minute now. If she "learns that it is Monday"??? That's not an unopened Bag anymore. That's going to produce Conditional Probabilites. A perfectly rational Beauty will not equate conditional probablities to those were the condition is left out. That would not be logical. And are they equally likely? Is it true that,

P(H-prize|moday prize) = P(T-Prize|monday prize)

Well, there is a 75% chance the Bag will contain a Monday prize. 50% is for an H-prize and 25% for a Monday T-prize. So those conditional probabilities are not equal. The first is 2/3 and the second is 1/3. Aha. I'm back to my original Model and this is the point where I objected to it. For the Sleeping Beauty, Heads produces just as many Mondays for her as Tails. Our Bag Beauty sees it different. Why? Is this another statement to put in Bold? Why are they seeing the Monday Condition differently? Should we adjust our Bag Model to better fit the Sleeping Model?

ok. Here's the difference. In the Sleeping Model, it has been claimed that it would make no difference if we waited until after Beauty had gone back to sleep on Monday to flip the coin. The Coin really only decides if she gets an Awakening Tuesday. She gets an Awakening Monday whether we flip a coin or not. The only reason we talk about flipping the coin before Monday is so we can distinguish between a Monday Heads Awakening and a Monday Tails Awakening. So how can we use our Bag Model to describe that situation. In our Bag Model Beauty is only getting one Bag each trip. In the Sleeping Model, Beauty may be getting one Awakening or Two. We could adust our Model so that Beauty is handed the Whole Covered Basket. She then either gets One H-Prize or Two T-Prizes. I don't think this changes any of my statements above. Just replace phrases where "she gets a T-Prize" with ones where "she gets 2 T-Prizes". Her unopened Bag position is replaced by an unopened Basket position.

The trouble with that adjustment is that in our new Basket Model she gets both T-Prize Bags at the same time. They can remain unopened, but once she gets Two Bags she knows they have T-prizes in them. So she needs to get the Two T-Prize Bags one at a time. The trouble is, when she gets the second one she knows she has already gotten the first one. In the Sleeping Model the Unconditional State is an Amnesia Awakening which she gets Two of if the coin is Tails. With the Unconditional State of an Unopened Bag in our old Bag Model she can't get both T-prizes. With the Unconditional State of the Unopened Basket in our new Basket Model she can't get the two T-prize bags one at a time. If we give them to her one at a time and she gets a T-prize she knows there's another T-prize coming. That's a Conditional State for us.

ok. Let's try this. We stay with the new Basket Model. We intend to give her both T-prize Bags before we're finished with it. We give her an unopened Bag, one at a time. But instead of putting the Monday Label on the Prizes we put the Monday Label on the outside of the Unopened Bag. And if there are Two Bags in the Basket we give her the one labeled Monday First. Lets call this our Sequential Basket Model. Notice we have the Unconditional State of the Unopened Basket. We also have the State of an Unopened Monday Bag, which is Unconditional. Unfortunately it doesn't solve the problem of the Tuesday Bag being Conditional for us but Unconditional for the Amnesia Sleeping Model.

I need to take a break. How can we adjust our Sequential Basket Model to make the Unopened Tuesday Bag Unconditional And Sequential to the Unopened Monday Bag?

PairTheBoard

soon2bepro

06-08-2007, 09:08 PM

What do you mean what is the solution?

A priori, the chance of the coin landing heads is 1/2. A posteriori, it's 1/3. (assuming whenever she wakes up during the experiment she will be asked this)

Put another way, if she had to bet $9 even money on the coin up front, it'd be a $0 EV to bet on heads or tails; but if she will be asked whenever she wakes up, it'd be a +$6 EV to bet on tails and $-6 EV to bet on heads.

I don't see any paradox. There's new information so the answer changes.

jason1990

06-08-2007, 09:33 PM

Here is some formal mathematics. I will let you decide if it applies to the problem.

Step 1. I modify the problem. Fix p, 0 < p < 1. Each time they go to wake her up, they will only wake her up with probability p. With probability 1 - p, they let her sleep.

This experiment can be modeled by a probability space on which we have constructed three independent random variables C, M, and T. The coin toss is C, it is 0 (heads) or 1 (tails) with equal probability. The variables M and T are uniform on [0,1]. We wake her up on Monday if and only if M < p. In the case C = 1, we wake her up on Tuesday if and only if T < p. Even more explicitly, the probability space is {0,1}x[0,1]x[0,1] with the uniform product measure, and C, M, and T are the coordinate projection functions.

Okay, now that the probability space has been constructed, and we know that it is really there, I will henceforth ignore it and just compute things normally. Let

H = the event a heads is flipped,
T = the event a tails is flipped, and
W = the event she wakes up at least once during the experiment.

When we wake her up, the only thing she really knows is that W occurred. So what we are interested in is the probability of H given W. By Bayes theorem,

P(H|W) = P(H)P(W|H)/[P(H)P(W|H) + P(T)P(W|T)]
= 0.5p/[0.5p + 0.5(1 - (1 - p)^2)]
= p/(p + 1 - (1 - p)^2).

Step 2. Take the limit as p goes to 1. Answer: 1/2.

Step 3. Wonder what happened. Can someone else come up with an event other than W which makes sense and which gives the answer 1/3? If not, then can someone come up with a different (formal) probability space that makes sense and gives the answer 1/3?

jason1990

06-08-2007, 09:58 PM

Well, I thought a little more. I originally introduced the random awakenings so that she really would learn something when she woke up. I found that easier to think about. But it was not strictly necessary for the mathematics. We could treat the case p = 1 directly, as AWoodside already observed. In that case,

P(H|W) = P(H and W)/P(W) = P(H)/1 = 1/2.

I am going to go out on a limb and firmly go with 1/2. The only objection that gives me pause is what f97tosc and others have mentioned. If she bets even money on heads every time she wakes up, she will lose.

I would explain this "paradox" as follows. The probability of heads is indeed 1/2. But she loses because she is "tricked" by the amnesia into betting two times when she is losing, and only once when she is winning.

It would be the same if you and I bet even money on a fair coin. If I could trick you into betting $2 whenever you guessed wrong, but only $1 when you guessed right, then I would be a big winner. The coin would still be fair; it is the wagering system that would be rigged.

jogger08152

06-08-2007, 10:15 PM

[ QUOTE ]
OK I should've phrased the question a bit better. Before she sleeps it's obvious that the rational answer is 1/2, no argument there. But after she wakes up it seems there are two equally valid arguments, one saying 1/2, one saying 1/3. You've made the valid argument that probabilities don't change when no new evidence has been received, but the most disconcerting part of the paradox is that the answer of 1/3 also seems correct. It looks like it should be a different way of working out the same answer but it gives us a different answer instead, leading to paradox.

Is your position that, the answer is definitely 1/2, therefore it cannot be 1/3, therefore the reasoning for 1/3 is somehow wrong. If that's your position then how is the reasoning for 1/3 invalid?

[/ QUOTE ]
The reasoning for 1/3 assumes that waking up gives the Princess useful information about which outcome was more likely; however, the scenario is designed such that the waking/sleeping process does not provide her with relevant data (day, and/or whether or not she has awakened previously in the week), and therefore the probability remains unchanged from her initial assumption of 1/2.

PairTheBoard

06-08-2007, 11:29 PM

[ QUOTE ]
Step 3. Wonder what happened. Can someone else come up with an event other than W which makes sense and which gives the answer 1/3? If not, then can someone come up with a different (formal) probability space that makes sense and gives the answer 1/3?

[/ QUOTE ]

I thought at first that your Model when p=1 is the same as my Unopened Bag Model. But I'm not sure it is. I think I reached your same conclusion. But I don't think that's enough. In order to resolve the paradox I believe we have to look at Elga's argument and see how it applies in our Model.

[ QUOTE ]
Elga -
I will argue first that P(T1) = P(T2), and then that P(H1) = P(T1).

[/ QUOTE ]

His P(T1)=P(T2) agrees with my Unopened Bag Model. His argument goes,

[ QUOTE ]
If (upon first awakening) you were to learn that the toss outcome is Tails, that would amount to your learning that you are in either T1 or T2. Since being in T1 is subjectively just like being in T2, and since exactly the same propositions are true whether you are in T1 or T2, even a highly restricted principle of indifference yields that you ought then to have equal credence in each. But your credence that you are in T1, after learning that the toss outcome is Tails, ought to be the same as the conditional credence P(T1| T1 or T2), and likewise for T2. So P(T1| T1 or T2) = P(T2 | T1 or T2 ), and hence P(T1) = P(T2).

[/ QUOTE ]

So he has this perspective of a State of Awakening. His Model is conditioned throughout on this State of Awakening. In my Bag Model, I Modeled that by Her being presented with an Unopened Bag. It looks like you Model that by the Condition, (M1 that condition approaches Certainty. So it comes to have null effect when conditioned on. I'm not sure what T1 and T2 mean in your model though. Even under no conditions, or when p=1. When p=1 I don't see how your model can distingish between an "Awakening" on Monday and an "Awakening" on Tuesday. But looking at the Sleeping Beauty from the outside we certainly can distinguish between the two. So why can't we model these distinct "Awakenings"? We can see them happening after all.

I found even more Model difficulties with his second equation, P(H1)=P(T1). I noticed it was actually a statement about the conditional probalities, P(H1|Moday) and P(T1|Moday). So his chain really looks more like,

P(H1)=P(H1|Moday)=P(T1|Monday)=P(T1)=P(T2)

So I figured I'd just show how my model gives the same Conditional equation,

P(H1|Monday)=P(T1|Monday)

and conclude his fallacy was in equating conditional probablities with unconditional ones. The problem was that my model did not agree with his for that conditional equation. Then when I looked at the actual problem settup I saw that the Conditional Equation was clearly true. The Condition of being told it's Monday is a Null condition with respect to the Coin Toss. That's why we could just as well wait until after She has gone back to sleep Monday Night to flip the Coin.

What I finally realized is that Her Amnesia does a strange thing to her perspective. It turns Conditional Statements into Unconditional Statements. I'm not sure how to apply this concept to your Model because I can't see how your Model can identify States of Awakening on Monday and Tuesday. When I modified my unopened Bag model to an unopened Basket of Bags, I had to give her two unopened bags sequentially when Tails flipped. This made the first unopened Bag a good model for her State of Awakening not knowing it's Monday. But when She received the second unopened Bag it could not Model her State of Awakening not knowing it's Tuesday.

Here's Elga's argument for why his P(H1)=P(T1)

Elga -
------------------
I will argue first that P(T1) = P(T2), and then that P(H1) = P(T1).

Now: if (upon awakening) you were to learn that it is Monday, that would amount to your learning that you are in either H1 or T1. Your credence that you are in H1 would then be your credence that a fair coin, soon to be tossed, will land Heads. It is irrelevant that you will be awakened on the following day if and only if the coin lands Tails - in this circumstance, your credence that the coin will land Heads ought to be 1/2. But your credence that the coin will land Heads (after learning that it is Monday) ought to be the same as the conditional credence P(H1 | H1 or T1). So P(H1| H1 or T1) = 1/2, and hence P(H1) = P(T1).
---------------------------------

PairTheBoard

jason1990

06-08-2007, 11:56 PM

The clock is ticking on me, and I am in fact packing right now, so I must be brief. I confess to having read only part of your post so far, and I have no idea who Elga is.

[ QUOTE ]
If (upon first awakening) you were to learn that the toss outcome is Tails, that would amount to your learning that you are in either T1 or T2. Since being in T1 is subjectively just like being in T2, and since exactly the same propositions are true whether you are in T1 or T2, even a highly restricted principle of indifference yields that you ought then to have equal credence in each.

[/ QUOTE ]
Fine. Beauty is a "perfect Bayesian". She is free to make any assumptions she wants, indifference principle or otherwise. But a perfect Bayesian does not introduce assumptions in the middle of an experiment. A perfect Bayesian uses "priors". That is, the assumptions are made prior to the experiment.

I am guessing the argument is something like this. There are three states, H1, T1, T2. Beauty wakes up. She has no information about what state this is. So she assumes it is equally likely to be any state. Hence, the probability of heads is 1/3. That is perfectly valid Bayesian reasoning -- provided the "experiment" begins when she wakes up. But you cannot compare that conclusion to her earlier belief that the probability of heads was 1/2, because that earlier belief is not part of this experiment.

Bayesians are supposed to be consistent. Their beliefs are supposed to obey the laws of probability. The only way in which to check that consistency is to put all the beliefs on one probability space. A "perfect" Bayesian does not start with one probability space, and then suddenly in the middle of the experiment throw it away and make a new one.

Once the assumptions are in place and the experiment begins, the Bayesian is bound by the formal rules of probability. And the formal rules of probability say that you condition on events or information (sigma-algebras). You do not take "no information", convert it into assumptions, and then condition on that. The place for converting "no information" into assumptions is in the beginning, when you choose your prior and build your probability space.

bigmonkey

06-09-2007, 04:51 PM

Thanks for all the detailed replies. I've been pretty busy the past couple of days so I'm just re-reading now. It may take me a while to absorb the arguments being used here...

bigmonkey

06-09-2007, 08:55 PM

Me: [ QUOTE ]
But if a heads occurs then she will experience a heads-awakening, and if she experiences a heads-awakening, then a heads occurred. They are equivalent, so P(H) should equal P(heads-awakening)

[/ QUOTE ]

To this Siegmund responds that:

[ QUOTE ]
Well, there's your explanation. No paradox at all; your "they are equivalent, P(H) should equal P(heads-awakening)" is just plain wrong. P(H) is (heads)/(coinflip outcomes) while P(heads-awakening) is (heads-awakenings)/(all awakenings). Unless EVERY coin flip results in the same number of awakenings, P(H) and P(heads-awakening) are not equivalent.

[/ QUOTE ]

This seems to be an issue in a few posts. I am arguing that because they are equivalent there is a paradox and you seem to be arguing that because there is no paradox they cannot be equivalent. But I think their equivalence is intuitively obvious. That a heads entails a heads-awakening (true) and a heads-awakening entails a heads (true), is proof that they entail each other and are therefore equivalent, in terms of probability.

KipBond: Your example is ambiguous as you point out. Are you saying that the original example has a similar ambiguity to yours? I think there may be a double-ambiguity anyway as I'm not sure whether in your example only a maximum of two coins are tossed or whether potentially infinite coins are tossed until a tails flips. If the former then 1/4 of the time she never wakes up so I'll assume you didn't mean that. If the latter then she knows the very last coin was definitely tails, so the question must be about the first coin. I don't see any reason why she doesn't say 1/2. I don't see how this is a simpler scenario because she only ever wakes up once doesn't she?

PairTheBoard: I think your post is an attempt to show that the P(H) and the P( heads-awakening) are not equivalent, as I asserted previously and again at the start of this post.

OK I've read your analogy once quickly and am doing a concentrated read through it.

[ QUOTE ]
We ask her, "When you recieve a Bag from the basket, what is the probability the Bag contains an H-prize?" She will rationally answer 1/3. But this is no puzzle to us, is it? We're talking about two different Events so it's not suprising they have two different probabilities. One Event is "Coin Lands Heads". The other event is "H-Prize in random Bag from Basket".

[/ QUOTE ]

I agree that they are different events. But my position is that their probabailities must be equal. Whenever there is a heads the Bag contains an H-prize, and vice versa. P(H|H-prize) = 1 and P(H-prize|H) = 1, so I'm fairly certain that makes P(H) = P(H-prize), although they are still seperate events.

[ QUOTE ]
No. All the statements in bold are False. Half of the "Bag Opening" times she will discover an H-prize and 50% of the "Bag Opening" times she will discover a T-prize. That's if the experiment is exactly repeated over and over.

So there is no Paradox. At least not so far. We should maybe ask at this point, "Are our "Bags" truly analogous to Beauty's "Awakenings?". I'm not seeing anything different.

[/ QUOTE ]

I think I'm seeing something importantly different. In your example when tails has flipped, does Beauty have to receive both T-prizes at two seperate times? You seem to be representing it where if heads then she takes the single H-prize in the single Bag from the basket, and if tails she either takes or is given a random Bag which contains either the T-prize marked "Monday" or the T-prize marked "Tuesday", and then that is the end. To be truly analogous if tails then she takes out one T-prize and collects the other one later when it will technically already be decided (but she won't know) that she is getting a T-prize. Is how I've said your analogy should be, how it is and I misinterpreted it? I think in the revised analogy all the bold lines in your post are true.

[ QUOTE ]
In our Bag Model Beauty is only getting one Bag each trip. In the Sleeping Model, Beauty may be getting one Awakening or Two.

[/ QUOTE ]

OK so my interpretation of your analogy was correct.

[ QUOTE ]
We could adust our Model so that Beauty is handed the Whole Covered Basket. She then either gets One H-Prize or Two T-Prizes. I don't think this changes any of my statements above.

[/ QUOTE ]

OK I can see you are bringing your analogy closer to the Sleeping example, so I needn't have said some of that stuff earlier.

[ QUOTE ]
With the Unconditional State of an Unopened Bag in our old Bag Model she can't get both T-prizes. With the Unconditional State of the Unopened Basket in our new Basket Model she can't get the two T-prize bags one at a time. If we give them to her one at a time and she gets a T-prize she knows there's another T-prize coming. That's a Conditional State for us.

[/ QUOTE ]

I'm not quite sure what you mean by unconditional states and conditional states. I think earlier in your argument you were offended by Elga's suggestion that Beauty is told it is Monday and is then asked whether it is Monday-with-a-heads or Monday-with-a-tails. Let me get it.

You said: [ QUOTE ]
Wait a minute now. If she "learns that it is Monday"??? That's not an unopened Bag anymore. That's going to produce Conditional Probabilites. A perfectly rational Beauty will not equate conditional probablities to those were the condition is left out.

[/ QUOTE ]

I don't see where the problem is with Beauty having conditional probabilities for events. In fact if she is a Bayesian she will not understand what is meant by an unconditional probability since all probabilities are just a relation between hypothesis and evidence, P(H|e).

For the analogy to be correct (as far as I can see), if tails then Beauty has to receive one gift at a time, the Monday gift first then the Tuesday gift. There's no problem whereby if she receives the first she knows she will receive the second, because she is given the amnesiac to forget ever receiving a gift. In the original example we can tell Beauty exactly what day it is and what the coin landed on, as long as we get her credence before we tell her, and then give her an amnesiac so she forgets that information the next day. (I wonder if there's any point in telling her. Does it give her peace of mind? She loses it as soon as she is put back to sleep so maybe there is no point.) To follow Elga's argument in your analogy, we don't need to put the label of "Monday" or "Tuesday" outside the bag, she just needs to conditionalize on the unseen gift in the bag being a "Monday" gift. I think that if she knows she's receiving a "Monday" gift then, like Sleeping Beauty can infer that the coin hasn't even been flipped yet, so Elga's P(H1)=P(T1)=P(T2) holds up. I think if your analogy is applied correctly it's only superficially different to the original Sleeping Beauty version.

From your later post to jason:

[ QUOTE ]
I found even more Model difficulties with his second equation, P(H1)=P(T1). I noticed it was actually a statement about the conditional probalities, P(H1|Moday) and P(T1|Moday). So his chain really looks more like,

P(H1)=P(H1|Moday)=P(T1|Monday)=P(T1)=P(T2)

[/ QUOTE ]

He says P(H1)=P(T1)=P(T2)=1/3, but P(H1|Monday)=P(T1|Monday)=1/2.

[ QUOTE ]
What I finally realized is that Her Amnesia does a strange thing to her perspective. It turns Conditional Statements into Unconditional Statements.

[/ QUOTE ]

Could you elaborate on this? The amnesia just means that she behaves exactly the same both days (if tails). It means that she can give her "strategy" on the probabilities of ehads and tails before the experiment starts, because as far as she is concerned there is only one action/belief to have after falling asleep. It might just be that she does it twice.

soon2bepro: [ QUOTE ]
A priori, the chance of the coin landing heads is 1/2. A posteriori, it's 1/3. (assuming whenever she wakes up during the experiment she will be asked this)...I don't see any paradox. There's new information so the answer changes.

[/ QUOTE ]

What new evidence is she receiving then? She knows the exact conditions for the experiment before it starts, and she can infer then that later she will believe 1/3 whereas now she believes 1/2. The experiment never really needs to even be done, since she's so rational she already knows exactly what she will believe later given whatever evidence.

jason1990,

[ QUOTE ]
P(H|W) = P(H)P(W|H)/[P(H)P(W|H) + P(T)P(W|T)]

[/ QUOTE ]

Is that really derived from Bayes Theorem? The answer on the right side is "1 + something"! I agree with your conclusion about the paradox though - ish. I think 1/2 is her credence and 1/3 her betting odds, but it does seem intuitive that Elga's argument also reveals her credence. I jut think the way Beauty conditionalizes is how we, as regular people, work out our credences, not betting odds. So maybe the paradox says something about how we ought to work out our credences.

[ QUOTE ]
Once the assumptions are in place and the experiment begins, the Bayesian is bound by the formal rules of probability. And the formal rules of probability say that you condition on events or information (sigma-algebras). You do not take "no information", convert it into assumptions, and then condition on that. The place for converting "no information" into assumptions is in the beginning, when you choose your prior and build your probability space.

[/ QUOTE ]

Beauty isn't making assumptions. Being a Bayesian she has a belief set, which is simply the set of all her credences for every possible logical statement. In Elga's example she isn't suddenly assuming that it is Monday, i.e. assuming the belief "It is Monday" and putting that in her belief set. She already has a belief "If it is Monday then P(H)=P(T)= 1/2" She has loads of beliefs of the form, for example, "I believe to degree 0.6 that if Hilary Clinton wins the Democrat vote she will be the next president", which becomes P(becomes president|wins democrat vote) = 0.6. So she isn't making any assumptions or being told that it is Monday.

jogger08152,

[ QUOTE ]
The reasoning for 1/3 assumes that waking up gives the Princess useful information about which outcome was more likely; however, the scenario is designed such that the waking/sleeping process does not provide her with relevant data (day, and/or whether or not she has awakened previously in the week), and therefore the probability remains unchanged from her initial assumption of 1/2.

[/ QUOTE ]

I don't follow how the reasoning for 1/3 assumes she is given useful information. She's just conditionalizing on what is possible, and could've made the sme calculation before she slept.

uberfish22

06-09-2007, 09:22 PM

The easiest way I can think to resolve this paradox is to look at the following matrix:

Heads Tails
Mon X X
Tue - X

Sleeping Beauty gets woken up during the experiment. She reasons that any of the possibilities marked X in the matrix are equally likely, so she will correctly answer that the probability the coin is currently heads up given that she's awake is 1/3.

The fallacy is to assume that waking her up during the experiment doesn't give her any information. It does, because it enables her to rule out the possibility (Tuesday+Heads). This information is not available to her before or after the experiment, so she will answer 1/2 then.

SNOWBALL

06-09-2007, 09:29 PM

Waking up absolutely IS new information.

I'll make it simpler. The lakers are playing the knicks. Bookmakers have the odds at dead even. You have an annoying neighbor who is betting the knicks. There is a 50% probability that he will come over and sulk if he loses, and a 100% probability that he will come over and brag if he wins.

Before the game, you are asked what the probability is that the knicks will win. You answer 50%.
After your neighbor knocks on your door, the probability is 2/3rds.

KipBond

06-09-2007, 10:27 PM

[ QUOTE ]
This information is not available to her before or after the experiment, so she will answer 1/2 then.

[/ QUOTE ]

That's not true. If it's available to you right now, without being woken during the experiment, then it's available to her. She can answer just as well as you can before, during, and after the experiment.

KipBond

06-09-2007, 10:38 PM

[ QUOTE ]
KipBond: Your example is ambiguous as you point out. Are you saying that the original example has a similar ambiguity to yours? I think there may be a double-ambiguity anyway as I'm not sure whether in your example only a maximum of two coins are tossed or whether potentially infinite coins are tossed until a tails flips. If the former then 1/4 of the time she never wakes up so I'll assume you didn't mean that. If the latter then she knows the very last coin was definitely tails, so the question must be about the first coin. I don't see any reason why she doesn't say 1/2. I don't see how this is a simpler scenario because she only ever wakes up once doesn't she?

[/ QUOTE ]

Sorry, let me reconfigure the new experiment to remove that confusion:

Same setup w/ amnesiac Beauty -- only this time, if the coin is heads, I end the experiment without asking her P(H). If it's tails, I wake her up & ask what the P(H) is. What is her answer?

We can repeat this experiment with as many trials as we want. The P(H) is 1/2. But, when I wake Beauty and ask her what the P(H) is, she can answer 2 ways: the "general" P(H) is 1/2 (the coin has a 50% chance of being heads or tails; or the "specific" P(H) for this last coin flip that I'm now asking her about (*), which is 0% (since I only ask her if it is tails).

(*) In the original Beauty scenario, I'm asking her TWICE about the same outcome IF the result was tails, and only ONCE about the outcome if it was heads.

PairTheBoard

06-09-2007, 11:50 PM

I agree with you that the models I described did not do what we need them to do. My goal was to describe a physical model that would simulate Sleeping Beauty having an Awakening. I thought I could do that by having her reciece a Bag or Basket. I've been working on another one today that uses Twin Bag Beauties. A Memory-Beauty who simulates what Sleeping Beauty would experience if she didn't get Amnesia. And an Amnesia Beauty who simulates a generic awakining where she doesn't know if it's monday, tueday, heads, or tails.

I did simulate such an Amnesia Beauty but the P(tails|monday) according to her bag did not simulate the experience Sleeping Beauty would have if she was awakened and told it was Monday. I'm beooming doubtful it can be done with the kind of physical model I had in mind. We need some kind of rigorous model by which to interpret Elga's statements. He uses a Credence function. That should work like a probability or else "credence" is not the right word. If it works like a probability we should be able to describe the proability model rigorously. I'm still not seeing it though.

PairTheBoard

Siegmund

06-10-2007, 12:48 AM

[ QUOTE ]
This seems to be an issue in a few posts. I am arguing that because they are equivalent there is a paradox and you seem to be arguing that because there is no paradox they cannot be equivalent. But I think their equivalence is intuitively obvious.

[/ QUOTE ]

And, I daresay, only people labouring under that misapprehension think there is a paradox here. I am merely pointing-out the fact, only intuitively obvious to some of us, that they are non-equivalent /images/graemlins/smile.gif

[ QUOTE ]

That a heads entails a heads-awakening (true) and a heads-awakening entails a heads (true), is proof that they entail each other and are therefore equivalent, in terms of probability.

[/ QUOTE ]

The point is that a heads leads to one heads-awakening but a tails leads to two tails-awakenings. If you want to say that a heads-awakening and a heads entail each other, in the sense that one happening implies the other must have happened / must be about to happen, fine - but that's no reason for you to believe that if the coin comes up heads half the time, half of the awakenings must be heads-awakenings.

Suppose you could perform this experiment 100 times in a row. Heads will come up about 50 times, and tails will come up about 50 times; the sleeper will be awakened about 150 times, 50 of them after heads and 100 of them after tails. Frequentist or Bayesian, drugged or sober, there's nothing remotely paradoxical, or even interesting, about the fact that a randomly selected flip has a 1/2 chance of being heads and a randomly selected awakening has a 1/3 chance of being a heads-awakening.

Again, I can't believe I've wasted my time reading and replying to this thread. It's the old battle of wits with an unarmed man.

PairTheBoard

06-10-2007, 03:52 AM

I think I've been making this way too complicated. I think it's easy to see once you look at statements being made in the 1/3 argument as conditional probabilities.

bigmonkey's original simple argument was this,
[ QUOTE ]
bigmonkey -

A brief argument for 1/3: she knows that the chance of heads is 1/2, and heads results in one awakening. The chance of tails is 1/2 and tails results in two awakenings. Therefore over the time period of the experiment she expects to wake up 1.5 times, once a tails-awakening and half a time a heads-awakening, on average. Therefore when asked how likely was it that the coin landed heads she will answer 1/3, and tails 2/3, conditionalizing on the evidence that she is awake

[/ QUOTE ]

First I'd like to compare that to Elga's argument here,
Elga's Sleeping Beauty Argument (http://www.princeton.edu/~adame/papers/sleeping/sleeping.html)
First Elga describes H1,T1,T2. I think these correspond to bigmonkey's three (.5) types of awakenings.

[ QUOTE ]
Elga -
Suppose that the first waking happens on Monday, and that the second waking (if there is one) happens on Tuesday. Then when you wake up, you're certain that you're in one of three `predicaments':

H1 HEADS and it is Monday.
T1 TAILS and it is Monday.
T2 TAILS and it is Tuesday.

[/ QUOTE ]

He introduces a "Credence" function P and describes his Goal,
[ QUOTE ]
Elga -
Let P be the credence function you ought to have upon first awakening. Upon first awakening, you are certain of the following: you are in predicament H1 if and only if the outcome of the coin toss is Heads. Therefore, calculating P(H1) is sufficient to solve the Sleeping Beauty problem. I will argue first that P(T1) = P(T2), and then that P(H1) = P(T1).

[/ QUOTE ]

First he shows P(T1)=P(T2)

[ QUOTE ]

If (upon first awakening) you were to learn that the toss outcome is Tails, that would amount to your learning that you are in either T1 or T2. Since being in T1 is subjectively just like being in T2, and since exactly the same propositions are true whether you are in T1 or T2, even a highly restricted principle of indifference yields that you ought then to have equal credence in each. But your credence that you are in T1, after learning that the toss outcome is Tails, ought to be the same as the conditional credence P(T1| T1 or T2), and likewise for T2. So P(T1| T1 or T2) = P(T2 | T1 or T2 ), and hence P(T1) = P(T2).

[/ QUOTE ]

So what is he doing there. A "Credence Function" should work like a probability function. So P is defined by the credence-probability values it gives to Events. He is trying to Deduce what P(coin landed heads) ought to be according to what we see it must be for other events, and for conditional credence/probablities. Notice we have a confusion of notation here. His Credence P(Heads) may turn out to be different from our normal P(Heads). Yet we are using P for both. So I am going to change his P to a Q.

Notation: Denote Elga's Credence function by Q

Now for his argument that Q(T1)=Q(T2). He is conditoning on the coin having landed on Heads. He is also assuming by the indiffence princple that if we wake Beauty and tell her the coin has landed on tails we are equally likely to do it on Monday as on Tuesday. So what Elga is really saying in his equation is this,

Q(T1)=Q(T2) means
Q(monday|tails)=Q(tuesday|tails)=50%

That's all his "Q(T1)=Q(T2)" means. It means that if Beauty is told the coin has landed on tails, she will give equal Credence to the day being Monday or Tuesday. Elga has identifed a conditional Credence-probability.

Now look at his argument for Q(H1)=Q(T1)
[ QUOTE ]
if (upon awakening) you were to learn that it is Monday, that would amount to your learning that you are in either H1 or T1. Your credence that you are in H1 would then be your credence that a fair coin, soon to be tossed, will land Heads. It is irrelevant that you will be awakened on the following day if and only if the coin lands Tails - in this circumstance, your credence that the coin will land Heads ought to be 1/2. But your credence that the coin will land Heads (after learning that it is Monday) ought to be the same as the conditional credence P(H1 | H1 or T1). So P(H1| H1 or T1) = 1/2, and hence P(H1) = P(T1).

[/ QUOTE ]

Once again he is conditioning. This time he is conditioning on Beauty being told it is Monday. If she is told it is Monday she should again give equal Credence-probability to the coin having been Heads or Tails. In fact she should give the value of that Credence-probability to be 50%. So all Elga is saying with his Q(H1)=Q(T1) is

Q(H1)=Q(T1) means
Q(heads|monday)=Q(tails|monday)=50%

In the first equation he has identified Q(T1) with Q(monday|tails). In the second equation he has identified Q(T1) with Q(tails|monday). I'm not sure that it makes sense to identify Q(T1) with both those things, whatever T1 really means, but at least it's consistent. Because all the above values for those conditional Credence-probabilites are 50%.

He finally claims that Q(H1) is indeed the thing we were looking to find out about Q. That is Q(H1)=Q(coin having landed on heads). I think that's reasonable according to his identification of Q(H1) with Q(heads|monday). But what that ends up implying for our rational Beauty is that,

Q(heads)=Q(H1)=Q(heads|monday)=Q(tails|monday)=Q(T 1)=Q(monday|tails)=Q(tuesday|tails)=Q(T2)=50%

That does not disagree with our P(heads)=50%. What it does imply is that H1,T1,T2 are not mutually exclusive Credence-probability Events[/b]. Either that or Q is not a true Credence-probability function. Or Q(H1),Q(T1) and Q(T2) should not have been identified with those conditional Credence-probabilites!

Let's look again at how they were defined,

[ QUOTE ]
Suppose that the first waking happens on Monday, and that the second waking (if there is one) happens on Tuesday. Then when you wake up, you're certain that you're in one of three `predicaments':

H1 HEADS and it is Monday.
T1 TAILS and it is Monday.
T2 TAILS and it is Tuesday.

[/ QUOTE ]

Those certainly look like mutually exclusive Credence Events for the NONINFORMED Awakened Beauty. Assuming Q is a legitimate Credence-probability function we must conclude that Q(H1),Q(T1), and Q(T2) should not have been identified with the conditional Credence-probabilities as they were.

Elga said,

Q(T1) = Q(monday|tails). Why? Why should we think the NONINFORMED awakened Beauty should have the same Credence that it's monday and tails was flipped as she would have if she was told tails was flipped and asked her Credence that it is monday? Here's what Q(T1) should really be by definition.

Q(T1)== Q(tails and monday) = 25%.

That's certainly not the same as,

Q(monday|tails)=50%

nor is it

Q(tails|monday)=50%

That is the fallacy in Elga's logic.

Similiarly for Q(T2),

Q(T2)==Q(tails and tuesday)

It also is not as Elga claims equal to

Q(tuesday|tails).

Elga is ok with his Q(H1),

Q(heads)=Q(H1)=Q(heads|monday)=50%

So Elga's Credence-probability function does have H1,T1,T2 to be mutually exclusive events. And they are exactly what we thought they should be to begin with.

Q(H)=50%
Q(T1)=25%
Q(T2)=25%

That makes perfect sense. An UNINFORMED awakened beauty should give 50% Credence the the possiblity that heads flipped and it is monday, 25% credence to the possibility that tails flipped and this just happens to be her monday tails awakening, and 25% Credence to the possibility that tails flipped and this just happens to be her second tails awakening.

A rational Beauty will not get confused by asking her how her Credences would change if she were given more information. Her Q(H1),Q(T1),Q(T2) must remain based on having the No Information that she actually has.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

Going back to bigmonkey's original argument
and the one most people are familiar with.

[ QUOTE ]
A brief argument for 1/3: she knows that the chance of heads is 1/2, and heads results in one awakening. The chance of tails is 1/2 and tails results in two awakenings. Therefore over the time period of the experiment she expects to wake up 1.5 times, once a tails-awakening and half a time a heads-awakening, on average. Therefore when asked how likely was it that the coin landed heads she will answer 1/3, and tails 2/3, conditionalizing on the evidence that she is awake

[/ QUOTE ]

So on a trip through one of these weeks, Beauty averages 1.5 awakenings. One of these is due to heads and two are due to tails. Should she believe the awakening she is now experiencing is due to heads or due to tails and it's being monday or due to tails and it's being tuesday. She has three choices.

H=(heads so it must be monday)
T1(tails and it happens to be monday)
T2(tails and it happens to be tuesday)

Normally she would give these Q Credence values of,

Q(H)=50%
Q(T1)=25%
Q(T2)=25%

But bigmonkey says she expects 1.5 awakenings. Half awakenings are hard to conceptualize. Suppose Beauty went through two of these weeklong trips. Suppose luck evens out right away and on the first trip she gets a heads and on the second trip she gets tails. Then she has 1 awakening the first week on monday. She has 2 awakenings the second week on monday and tuesday. Shouldn't Beauty believe she is equally likely to be having a Week1 trip as a Week2 trip? There is a 50% chance she is having a Week1 trip and this is her Monday awakening. There is a 50% chance she is having a Week2 trip and this is either her Monday or Tuesday Awakening.

She knows that if she takes two Week long trips she will get on average 3 awakenings, one due to heads and two due to tails. But should she think the three of them are equally likely to be the one she is experiencing right now? No. She is more likely to be exeriencing the one due to heads! She knows there is a 50% chance she is experiencing That one. There is only a 25% chance she is experiencing the one due to tails on monday. And only a 25% chance she is experiencing the one due to tails on tuesday. Just because there are three types of awakenings doesn't mean they are equally likely to be the one she is experiencing right now.

The logical fallacy is once again another example of missappying Sklansky's Principle.

PairTheBoard

PairTheBoard

06-10-2007, 06:47 AM

What do you do when your Credence Function is not giving you the right betting odds?

There's one final objection I must resolve for Beauty. A Credence function is suppose to give Credence-probabilities which tell you the odds you should get to make bets. The objection is that the Credence function I gave Beauty does not do this. Normally that would mean the Credence function is incorrect and should be changed so that it does provide correct betting odds. I concluded the Credence Function Q that Beauty must rationally have is,

Q(heads so must be monday) = 50%
Q(tails and happens to be monday) = 25%
Q(tails and happens to be tuesday) = 25%

These are the Credence values Beauty should have at any awakening. As the three types of awakenings are indistinguishable to her she must logically have the same Credence function when she experiences any of them. Now suppose Beauty uses her credence function at each awakening to bet $1 at even money that a heads was flipped. If she did that she would lose money wouldn't she?

You can see this more easily if you look at the Week1,Week2 scenario. Suppose she takes two Week trips like this, Week1 a heads week and Week2 a tails week. Then in the two weeks she will make 3 bets of $1 at even money that it's a heads week. Over the two weeks she will lose $1 that way. It's a -EV bet for her. Therefore her Credence function is wrong. If it is changed to (1/3,1/3,1/3) for the three type of awakenings she will demand 2-1 odds to bet on heads and come out ok over the two weeks. Thus we should change her Credence function so that it does its job giving her the right odds. Right? Wrong.

This is an example of a principle I've pointed out at work in the Two Envelope Problem as well. It is the principle of the outcome of the bet dictating the amount you must bet. When this happens you don't change your estimate of the probability of the outcome. You just refuse to bet on the wrong side or you demand odds different from what your Credence Function tells you.

Consider the following example of this principle at work. I have a standard deck of playing cards. On the back of all the black cards I write the number 2. On the back of all the red cards I write the number 1. Now I come out of my room and greet you at the table. I tell you we will shuffle the deck and you can cut it. I tell you that you may then bet that the top card on the deck is a red card. I will give you even money. You have dollar chips in front of you. Even money agrees with your Credence Function for the chance the top card will be red. You figure it's 0 EV so you don't mind gambling. You would not gamble at -EV but 0 EV is ok with you. So you agree.

However, now I tell you that you must bet the amount of money written on the back of the cards. You become suspicious. You ask me about those numbers. You are a friend so I decide to be honest with you. I tell you I have written 2 on the back of all the black cards and 1 on the back of all the red cards. So you must bet $2 on red at even money when it's a black card. And $1 on red at even money when it's a red card. Do you still like the bet? Of course not.

You are being forced by the outcome of the bet to bet twice as much when that outcome is a loser for you. The outcomes of black cards are forcing you to bet $2 on your Red choice. The outcomes of red cards are forcing you to bet the minimum $1 on your winning red choice. Taking even money on all those bets is a losing proposition for you.

Now here's the point. It is your Credence function for the random card being red that tells you even money is ok. Even money is not ok for these bets. But is it because your Credence function is wrong? No. Your Credence function is sill correct. Before your cut the deck you properly believe there is a 50% chance you will cut a red card. You keep your Credence function for red cards. You just don't take the bet when the amount of the bet is being forced on you by the outcome of the bet.

That is exactly what is happening with Beauty if she uses her correct Credence function to take even money betting on heads. When she is in the losing outcome of the tails Week2 she is being forced to make twice as many losing bets on heads than she gets to make when she is in the winning outcome of the heads Week1. She doesn't change her Credence function. She just demands different odds than what it tells her to bet on heads.

The betting objection has been answered.

PairTheBoard

uberfish22

06-10-2007, 08:00 AM

[ QUOTE ]
[ QUOTE ]
This information is not available to her before or after the experiment, so she will answer 1/2 then.

[/ QUOTE ]

That's not true. If it's available to you right now, without being woken during the experiment, then it's available to her. She can answer just as well as you can before, during, and after the experiment.

[/ QUOTE ]

More specifically, the extra information that's only available to her during the time of the experiment and not from an outside perspective can be stated as "I am inside the experiment, and awake."

I suspect there's some confusion going on between the two different questions "what is the a chance in a vacuum of the coin having landed heads" and "what is the chance given my current knowledge that the coin landed heads".

KipBond

06-10-2007, 09:27 AM

[ QUOTE ]
Q(heads so must be monday) = 50%
Q(tails and happens to be monday) = 25%
Q(tails and happens to be tuesday) = 25%

[/ QUOTE ]

I'm pretty sure this is wrong. (I've read your posts twice, but I may be misunderstanding you?)

This would be right IF after a tail flip I randomly chose a day to wake her (either Monday OR Tuesday). Not both. (I'm assuming the "happens to be monday" means you are waking her on Monday; otherwise it doesn't correspond to the problem.)

P(Coin is Heads) = 50%
P(Coin is Tails) = 50%

P(Coin is Heads & I then wake her on Monday) = 50%
P(Coin is Tails & I then wake her on Monday) = 50%
P(Coin is Tails & I then wake her on Tuesday) = 50%

If you want to combine these events to make a total "event space" that totals 100%, then each "flip-waking" event has a 1/3 chance of happening.

If your model denies the following fact, it HAS to be wrong:

After 1,000,000 coin flips:
~500,000 will be heads
~500,000 will be tails

and we will wake Beauty ~1,500,000 times:
~500,000 will be heads-wakings (coin was heads & we woke her up on Monday)
~1,000,000 will be tails-wakings (coin was tails & we woke her up on Monday & Tuesday)

f97tosc

06-10-2007, 10:50 AM

[ QUOTE ]

When we wake her up, the only thing she really knows is that W occurred.

[/ QUOTE ]

I think this is false. Obviously, she will know that W has occured - but it is not the only thing. Since she cannot tell the difference between waking up the first and second time, we have also introduced a bias in favor of the path that she will wake up more than once.

Suppose we do the following experiment instead. First we flip the coin. Then if it is T, we put two black tokens in an urn. If its W, we put one black and one white token in the urn.

Now, let S take one token and guess if it is H or T. Obviously, it is black, she should give the probability one third to H.

Also note, that this doesn't change if we first remove one at random (=she took one before but does not remember what it was).

Also, one ideosyncracy of the problem is that she never actually gets to observe the white token, because she is kept asleep. I would guess that that is why the reflection principle is violated (although I don't really know what it says!), however, this makes no difference for her when she takes a black token and tries to infer the flip of the coin.

KipBond

06-10-2007, 10:58 AM

[ QUOTE ]
She knows that if she takes two Week long trips she will get on average 3 awakenings, one due to heads and two due to tails. But should she think the three of them are equally likely to be the one she is experiencing right now? No. She is more likely to be exeriencing the one due to heads! She knows there is a 50% chance she is experiencing That one. There is only a 25% chance she is experiencing the one due to tails on monday. And only a 25% chance she is experiencing the one due to tails on tuesday. Just because there are three types of awakenings doesn't mean they are equally likely to be the one she is experiencing right now.

[/ QUOTE ]

This can't be right. So, if we tell her it is Monday, then there is a 2/3 chance the coin was heads? I don't see how. It has to be 1/2:

P(H|monday) = 50%
P(H|tuesday) = 0%
P(T|monday) = 50%
P(T|tuesday) = 100%

P(monday|H) = 100%
P(tuesday|H) = 0%
P(monday|T) = 50%
P(tuesday|T) = 50%

If any of those are not right, I'd like do know why. It seems intuitive to me.

Now:

P(monday|waking) = 2/3
P(tuesday|waking) = 1/3

I take it this is where the disagreement/misunderstanding is. Because if you agree with those, then P(H|waking) & P(T|waking) are pretty simple to compute.

Deorum

06-10-2007, 03:04 PM

I did not read all of the posts, but I do not see a paradox.

To the question of, "What is the probability that you were awakened by a tails-flip?" the answer is 2/3. This is the equivalent of asking, "What day is it?" If she answers Monday, she will be correct 2/3 of the time, if she answers Tuesday, she will be correct 1/3 of the time. To illustrate this, imagine that we run the experiment 100 times (and for the sake of simplicity, let's fix the flip number at 50 heads and 50 tails). 50 times she will wake on Monday because of heads. 50 times she will wake on Monday because of tails. 50 times she will wake on Tuesday because of tails. If she answers Monday, she will get 100 correct and 50 wrong. If she answers Tuesday, she will get 50 correct and 100 wrong.

But this is not the same as saying that there was a 2/3 chance that the coin came up tails. The number of times she awakens due to a flip has nothing to do with whether or not it came up heads or tails. Rather, it has to do with the precedent set at the beginning of the experiment that she will wake twice when the coin was flipped tails and once when the coin was flipped heads.

She knows that she will wake twice as often on a tails flip than she will on a heads flip. She knows that her awakening is 1/3 likely due to a heads flip, and 2/3 likely due to a tails flip. But the awakenings that occur on Monday from a tails flip and the awakenings that occur on Tuesday from a tails flip are not derived from independent flips. Each Monday-Tuesday pair is due to one flip. In other words, 1 heads-Monday = 1 tails-Monday + 1 tails-Tuesday.

She knows all of this from the beginning. Since she knows that each tails flip results in her waking twice as often as a heads flip, she knows that there were only half as many "tails flips" as "tails awakenings". She also knows that the amount of "heads flips" and "heads awakenings" are equivelant. Therefore, the amount of "tails flips" and "heads flips" are equivalent, because:

.5(2/3) = 1/3

Since the amount of "heads flips" and "tails flips" are equivalent, she will state that the probability of the coin being flipped heads or tails is 50% for each.

Q.E.D.

Quick Proof:

She knows:

1 tails flip = two awakenings

1 heads flip = one awakening

Through substitution, 1 tail flip = 2 heads flips

She will have 2/3 awakenings due to tails, and 1/3 due to heads (on average - remember, she knows that).

Since each "tails awakening" is only worth half of a "heads awakening" in terms of flips, and there are twice as many "tails awakenings" than "heads awakenings" she will come to the conclusion that the coin comes up 50% heads and 50% tails.

PairTheBoard

06-10-2007, 05:43 PM

First, let me simply restate my full argument, then respond to your points. Beauty has an information based Credence-probability function P before she enters the experiment. When she finds herself Awakened she considers if she should adopt a new information based Credence-probability function Q. She asks herself, "Do I have any new information?". She decides No. The information she had for forming P included the fact that she would now be in this situation. So she concludes that her Q should remain P. She let's Q=P and starts to think what that means.

She decides the probability the coin landed heads is still 50%. Just because she is now experiencing one of the three kinds of awakenings she knew beforehand she would be experiencing doesn't add any new information for how the coin might have landed. So Q(heads)=50%. That means she must conclude there is a 50% chance that this awakening is a monday heads awakening. Similiarly she reasons there is a 25% chance that this awakening is a monday tails awakening or tuesday tails awakening.

But she is now confused. That means she should bet $1 at even money that heads landed. But, she thinks, if she were to do that everytime she awakens in this experiment she will lose $1 per experiment. She wonders, should she change her Credence function Q so that it will give her the odds she needs to break even on these bets on heads?

Then she realizes something. When heads comes up she is only allowed to bet once. When tails comes up she is forced to bet twice. She thinks, that's not a legitimate bet. That's like offering me to bet even money on heads but telling me I must bet $1 when heads comes up and $2 when tails comes up. I don't change my estimate for the probability that heads will come up. I just don't take the bet. So Beauty stays with her Prior Credence function and just refuses the unfair bet on heads. She knows, that for this experiment there is a 50% chance the coin came up heads and if it did this must be monday.
=======================

Now to your points.

[ QUOTE ]

P(Coin is Heads & I then wake her on Monday) = 50%
P(Coin is Tails & I then wake her on Monday) = 50%
P(Coin is Tails & I then wake her on Tuesday) = 50%

If you want to combine these events to make a total "event space" that totals 100%, then each "flip-waking" event has a 1/3 chance of happening.

[/ QUOTE ]

You have 3 Events there. The first is just the Event that the Coin lands heads. Both the second and third are the same event. They are the Event that the coin lands tails.

Beauty keeps those probabilities. They tell her that this experiment has a 50% chance of Heads. So if heads flipped this must be monday. If tails flipped she applies indifference to conclude it's equally likely to be monday or tuesday. Thus the Beauty's Event, "I am awake", has properly divided probabities to add to 100%.

[ QUOTE ]
After 1,000,000 coin flips:
~500,000 will be heads
~500,000 will be tails

and we will wake Beauty ~1,500,000 times:
~500,000 will be heads-wakings (coin was heads & we woke her up on Monday)
~1,000,000 will be tails-wakings (coin was tails & we woke her up on Monday & Tuesday)

[/ QUOTE ]

Beauty is well aware of this fact. That's why she won't use her correct Credence of 50% that Heads landed this experiment to bet on heads at even money. She can see that the situation you describe above implies that the Outcome of the Bet will be Forcing Beauty to bet twice in losing situations and only allows her to bet once in winning situations.

You do not change your Probability for an outcome because there is a betting situation where the outcome of the bet dictates to you how much you will bet.

PairTheBoard

KipBond

06-10-2007, 06:40 PM

[ QUOTE ]
"What is the probability that you were awakened by a tails-flip?" the answer is 2/3.
...
Since the amount of "heads flips" and "tails flips" are equivalent, she will state that the probability of the coin being flipped heads or tails is 50% for each.

[/ QUOTE ]

So, after being awakened, what is her answer to the question: "What is the probability that the coin was heads?"

The question is ambiguous. You give answers to both interpretations above.

PairTheBoard

06-10-2007, 06:54 PM

Consider the following analogy. You have a computer with a good algorithm for processing information to produce a Credence-probability for Events defined by that information. You input into the computer the following information. You are going to flip a fair coin. The computer must output a Probability that the coin has landed heads. But you also tell it that if the coin lands Heads the computer will be outputing its computed probability for one day. If it lands Tails the computer will be outputing its computed probability for one year. But the computer will not know which of these conditions it is in because with Heads, its processor will be sped up by a factor of about 365. It will have no way of distingishing between the normal and sped up processing speed.

A signal will be given to the computer telling it to begin giving its output for the probability the coin landed heads. We will also ask it to give it's Credence-probability for What State it is In. The computer must output it's Belief probability for its being in a Sped up Processing State or a Normal Processing State. We tell it we will measure the number of real days it holds the correct Belief probability for the State it is in.

You also tell the computer that we will be taking its outputs and each real day we will bet 1 computer measuring unit on heads at the odds determined by its output for the probability that heads has landed. But we don't tell it how that should affect its output of the probability that heads has landed. We don't tell it to optimize anything with respect to those bets we are making. We just tell it to take all this information and compute the best Credence-probability that the coin landed heads, and its Belief probability for its State. Do this when we say Begin.

You have input that information. You then flip a coin, set the computer's processor speed accordingly and ask it to Begin giving outputs for the probability the coin landed heads. Let's suppose it outputs a 50% probability the coin landed on heads.

Now, are we going to criticize the computer because it believes there is a 50% chance the coin landed heads for only a day when the coin did land heads, and it believes the 50% for a year when the coin landed tails?

Are we going to criticize the computer because if we bet $1 on heads at odds determined by the computer's output of 50%, and we make 1 bet a day for every day the computer is giving that output we will lose money?

No. The computer is correct to give the output of 50%. The fact that it believes 50% longer for tails than for heads does not make it an incorrect belief.

PairTheBoard

KipBond

06-10-2007, 07:18 PM

[ QUOTE ]
She decides the probability the coin landed heads is still 50%. Just because she is now experiencing one of the three kinds of awakenings she knew beforehand she would be experiencing doesn't add any new information for how the coin might have landed.

[/ QUOTE ]

I disagree. See my "new experiment" above where I only wake her when the coin is tails. Waking her DOES give her more information about the previous outcome of the coin she's being asked about.

[ QUOTE ]
the Outcome of the Bet will be Forcing Beauty to bet twice in losing situations and only allows her to bet once in winning situations.

You do not change your Probability for an outcome because there is a betting situation where the outcome of the bet dictates to you how much you will bet.

[/ QUOTE ]

That's about what I said in my 1st & 2nd posts in this thread(*). I still agree. However --- I now think that because the question being asked Beauty is ambiguous, one of the ways she can answer the question allows her to use the new information she has when she is woken up.

(*) I then brought up why the question in this thread was not as ambiguous as the question posed in the website link you provided, and why "1/3" can be a valid answer to the more ambiguous wording of the question posed to Beauty.

KipBond

06-10-2007, 07:45 PM

PTB: Please tell me where you disagree with the following:

P(H|monday) = 50%
- (Probability the coin that was flipped prior to Beauty's waking is Heads, given that today is Monday)
P(H|tuesday) = 0%
P(T|monday) = 50%
P(T|tuesday) = 100%

P(monday|H) = 100%
- (Probability that today is Monday, given that the coin we flipped prior to waking Beauty is Heads)
P(tuesday|H) = 0%
P(monday|T) = 50%
P(tuesday|T) = 50%

P(monday|waking) = 2/3
- (Probability that today is Monday, given that we are waking Beauty up in accordance with the outlined experiment parameters)
P(tuesday|waking) = 1/3
P(wednesday|waking) = 0

P(waking|monday) = 100%
- (Probability that we are waking Beauty up in accordance with the outlined experiment parameters, given that today is Monday)
P(waking|tuesday) = 50%
P(waking|wednesday) = 0%

KipBond

06-10-2007, 08:18 PM

Sorry for posting so much... but I remembered that there was something in your first reply that I disagreed with:

[ QUOTE ]
So having awoken she knows there is a 75% chance it is Monday, and a 25% chance it is Tuesday.

[/ QUOTE ]

There are 3 flip+waking events, all equally likely:
1) Heads+Monday+Waking
2) Tails+Monday+Waking
3) Tails+Tuesday+Waking

If you ran this experiment 1,000,000 times there would be:
500,000 Heads+Monday+Wakings
500,000 Tails+Monday+Wakings
500,000 Tails+Tuesday+Wakings

So, when she wakes up, she knows that there is a 2/3 probability that it is Monday, and a 1/3 probability that it is Tuesday.

[ QUOTE ]
2/3 of the time that it's Monday will be because of Head. 1/3 of the time that it's Monday will be because of Tails. And all the time that it's Tuesday will be because of Tails.

[/ QUOTE ]

50% of the time that it's Monday will be because of Heads. 50% of the time that it's Monday will be because of Tails. 100% of the time that it's Tuesday will be because of Tails.

bigmonkey

06-10-2007, 08:22 PM

Sorry for the delay. I'm going to try to find time (about 4 hours?) to read and respond to these tomorrow. Feel free to continue the debate though...

Deorum

06-10-2007, 11:37 PM

[ QUOTE ]
[ QUOTE ]
"What is the probability that you were awakened by a tails-flip?" the answer is 2/3.
...
Since the amount of "heads flips" and "tails flips" are equivalent, she will state that the probability of the coin being flipped heads or tails is 50% for each.

[/ QUOTE ]

So, after being awakened, what is her answer to the question: "What is the probability that the coin was heads?"

The question is ambiguous. You give answers to both interpretations above.

[/ QUOTE ]

There is no ambiguity in the question. The answer to the question, "What is the probability that the coin was heads?" is 1/2. The answer to the question, "What is the probability that you were awakened this time by a 'tails flip'?" is 2/3.

These questions are not the same. The first refers to the probability of the state of the coin, while the second refers to the probability that she was awakened by a specific flip. The probability of the state of the coin and the probability that she was awakened by a specific flip are not the same.

Since different types of flips result in different numbers of awakenings, but flipping the coin results in an equal number of heads and tails, it follows that the probabilities for each question are different. However, neither question is ambiguous; both have definitive answers.

Edit: Check out the quick proof in my post above, it is pretty simplistic and will be the thought process which she would use to determine that the probability of either flip is 1/2. All of what is written above my quick proof is simply a more in depth explanation to the quick proof process.

KipBond

06-11-2007, 12:55 AM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
"What is the probability that you were awakened by a tails-flip?" the answer is 2/3.
...
Since the amount of "heads flips" and "tails flips" are equivalent, she will state that the probability of the coin being flipped heads or tails is 50% for each.

[/ QUOTE ]

So, after being awakened, what is her answer to the question: "What is the probability that the coin was heads?"

The question is ambiguous. You give answers to both interpretations above.

[/ QUOTE ]

There is no ambiguity in the question. The answer to the question, "What is the probability that the coin was heads?" is 1/2. The answer to the question, "What is the probability that you were awakened this time by a 'tails flip'?" is 2/3.

[/ QUOTE ]

That doesn't make sense. I assume you would say the answer to "What is the probability that you were awakened this time by a 'heads flip'"? is 1/3? When I say the coin was heads that can mean the coin was heads when I flipped it, or the coin was heads this time prior to me waking you up. They are different.

[ QUOTE ]
These questions are not the same. The first refers to the probability of the state of the coin, while the second refers to the probability that she was awakened by a specific flip. The probability of the state of the coin and the probability that she was awakened by a specific flip are not the same.

[/ QUOTE ]

Yeah, I think I agree with you here. The questions are not the same, but the original question can be interpreted two different ways, resulting in 2 different answers. Thus, the ambiguity.

[ QUOTE ]
Since different types of flips result in different numbers of awakenings, but flipping the coin results in an equal number of heads and tails, it follows that the probabilities for each question are different. However, neither question is ambiguous; both have definitive answers.

[/ QUOTE ]

It's ambiguous as to which question is being asked. Ambiguity is obviously subjective.

PairTheBoard

06-11-2007, 01:24 AM

Suppose you Flip and put 1 black ball in the bag if heads is flipped. And you put 100 White balls in the bag if tails is flipped. Now pretend you are a ball in the bag. You can't tell your color nor how many balls are in the bag. You are asked, what is the probability you are a black ball?

Shouldn't you say there is a 50% probability you are a black ball? What does that probability mean? It means that if we repeat this experiment you will find yourself being a black ball in half the experiments. You will find yourself being a white ball in half the experiments. And it the white balls are numbered 1-100 you will find yourself being White Ball #100 in .5% of the experiments.

Beauty's awakenings are exactly the same as the balls in the bag.

PairTheBoard

PairTheBoard

06-11-2007, 05:28 AM

Here's the thing. When you give a probability you are giving a frequency wrt "number of times". Our probability "number of times" here is "number of times" the experiment is run. The 1/3 people are changing the probaility to "number of times" the awakening happens. Do you want the frequency per Experiment, or do you want the frequency per Awakening?

When Beauty is asked, what's the chance this is Monday, does she want to be right per experiment or per Awakening?

Consider my 1 Black Heads ball or 100 White Tails balls in the bag example. The white balls are numbered 1-100. You are a ball in the bag. When Tails, you will be White Ball-1 on day 1, White Ball-2 on day 2, ..., White Ball-100 on day 100. When Heads you will be the Black ball on day 1 after which you will be removed. You will be asked this question on day 1. When Tails you will be asked this question on each of days 2-100.

You are a ball. What is the probability you are a Black Ball? If you answer 50% you will be right per experiment. ie. In half of the experiments, when you are asked that question you will be a black ball.

If you answer 1/101 you will be right per Ball experience. In 1 out of 101 ball experiences you will be a Black Ball.

When Beauty is asked the probability she is having a Heads-Awakening does she want to be right per experiment or per Awakening experince?

If she answers 1/2 she will be right per experiment. In half the experiments, when she is asked that question she will be in a Heads-Awakening.

If she answers 1/3 she will be right per Awakening experience. In 1/3 of her awakening experiences she will be in a Heads-Awakening.

So what should Beauty's probability measure be. The one she started out with, ie. per Experiment. Or a new one she creates for some reason which is per Awakening. Why should she adopt a new probability measure? If she insists on using this new probability measure of her creation why doesn't she just give both probabilites to be clear?

Isn't the full answer just the one Deorum is saying she should give?

She can simply say, the probability I am in a Heads Awakening is 1/2 per Experiment and 1/3 per Awakening.

PairTheBoard

Deorum

06-11-2007, 09:17 AM

Kipbond,

Ah, okay I think I understand now what you mean by ambiguity. I think all three of us are pretty much in agreement. Let's recap:

If she is asked, "what is the probability that you were awakened by a 'heads flip' this awakening", her answer will be "1/3" because the total number of heads-awakenings over an average trial period will be 1/3. Likewise, tails-flips awakenings constitute 2/3 of the total awakenings. Are we all in agreement about this?

If she is asked, "what is the probability that the coin came up heads when we flipped it" her answer will be "1/2" because she knows that she is twice as likely to have been awakened at any given point by a tails-flip, as well as knowing that the parameters of the experiment are such that she is awakened twice as often when the coin comes up tails, and thus will conclude that the coin came up tails half as often as she is awakened by tails-flips thereby reducing that number by half. The result of this reduction is that the amount of heads-flips and tails-flips are equal, giving a probability of 1/2 to each. Are we in agreement about this?

So as long as we can agree with these two parameters, there is no paradox, right? It is simply unclear as to which question the OP is asking (although I would argue that the question as posed, "What is the probability that the coin came up heads", infers that they mean the absolute state of the coin, regardless of which caused her to wake, such that the answer is 1/2 - but okay, semantics). It appears that the only problem is in the clarity of the question stated by the OP, not what is actually going on mathematically, correct?

KipBond

06-11-2007, 09:54 AM

[ QUOTE ]
Here's the thing. When you give a probability you are giving a frequency wrt "number of times". Our probability "number of times" here is "number of times" the experiment is run. The 1/3 people are changing the probaility to "number of times" the awakening happens. Do you want the frequency per Experiment, or do you want the frequency per Awakening?
...
So what should Beauty's probability measure be. The one she started out with, ie. per Experiment. Or a new one she creates for some reason which is per Awakening. Why should she adopt a new probability measure? If she insists on using this new probability measure of her creation why doesn't she just give both probabilites to be clear?

[/ QUOTE ]

That's pretty much what I've been saying.(*) But, it doesn't require a "changing" of her probability measure after she wakes up. She's perfectly logical, so given the experiment parameters, she can answer both ways before, during, and after the experiment. Just like we can, right now.

Since there really are two different events (coin-flips, and coin-flip-wakings), it's not clear which one should be the probability space in regards to the question being asked. Answer both ways to remove any ambiguity.

(*) Although, the "number of times" doesn't refer to the "number of times" the experiment is run vs. the "number of times" she is awakened. It refers to the "number of times" the coin is tossed, vs. the "number of times" she is awakened. Both events are in the same experiment; it just so happens that the coin is tossed the same # of times the experiment is run, while the awakenings occur 1.5x more frequently. This could be different with different experiment parameters.

KipBond

06-11-2007, 10:07 AM

[ QUOTE ]
If she is asked, "what is the probability that you were awakened by a 'heads flip' this awakening", her answer will be "1/3" because the total number of heads-awakenings over an average trial period will be 1/3. Likewise, tails-flips awakenings constitute 2/3 of the total awakenings. Are we all in agreement about this?

[/ QUOTE ]

Agreed.

[ QUOTE ]
If she is asked, "what is the probability that the coin came up heads when we flipped it" her answer will be "1/2" because she knows that she is twice as likely to have been awakened at any given point by a tails-flip, as well as knowing that the parameters of the experiment are such that she is awakened twice as often when the coin comes up tails, and thus will conclude that the coin came up tails half as often as she is awakened by tails-flips thereby reducing that number by half. The result of this reduction is that the amount of heads-flips and tails-flips are equal, giving a probability of 1/2 to each. Are we in agreement about this?

[/ QUOTE ]

Agreed. Although if she is being asked about the outcome of the coin flip event, then she can just disregard the entire experiment and answer as she normally would: 1/2.

[ QUOTE ]
So as long as we can agree with these two parameters, there is no paradox, right? It is simply unclear as to which question the OP is asking (although I would argue that the question as posed, "What is the probability that the coin came up heads", infers that they mean the absolute state of the coin, regardless of which caused her to wake, such that the answer is 1/2 - but okay, semantics). It appears that the only problem is in the clarity of the question stated by the OP, not what is actually going on mathematically, correct?

[/ QUOTE ]

Agreed. In my third reply in this thread (http://forumserver.twoplustwo.com/showthreaded.php?Cat=0&Number=10710957), after I had read the link PTB gave in which an argument for the answer being 1/3 was given, I wrote:

[ QUOTE ]
I think the problem, as stated, was ambiguous -- or rather, not as ambiguous as is required for the "paradox". This part here:

[ QUOTE ]
On Sunday evening we ask Beauty what she thinks the probability is of the coin landing heads. Unsurprisingly she says 1/2. Whenever she wakes up she will say 1/3.

[/ QUOTE ]

The probability of the coin landing heads??? 1/2. The probability of it having landed heads?? 1/3 is a valid answer.

The reason 1/3 is now valid is because it can be implied that the question is actually: "what is the probability that the coin landed heads and then we woke you and asked you this question?"

[/ QUOTE ]

In that article (""), the question to Beauty is asked as: "When you are first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?"

The key word is "is". It's ambiguous as to what probability space Beauty should be using. The coin-flip space, or the coin-flip-awaking space.

So, I think we all agree. No paradox, for sure. /images/graemlins/smile.gif

f97tosc

06-11-2007, 10:11 AM

One more explanation here, in favor of 1/3.

First let's disregard the coin, and simply wake her up Monday and Tuesday, and she loses memory in between. And she knows this. Then when she wakes up, she will assign the probability 0.5 to it being Monday or Tuesday.

Suppose now that we independently of this throw the fair coin once (she will still wake up both days regardless of outcome). Now clearly when she wakes up she will assign the probability 0.25 to each of the 4 combinations in (M,T)x(H,T).

Now suppose further that after we wake her up, for the specific case Tuesday and Heads, we tell her, "go back to sleep" but for any other combination we ask her "what is the probability for heads?", and that she knows this.

Hopefully, we can all agree that for this version of the problem, she will arrive at 1/3, by Bayes' theorem, when she gets the question.

And the only difference between this problem, and the stated problem, is that in the stated problem, she never gets to wake up on Tuesday. But I don't see how that should change her information when she is woken up. It may throw us off a bit since it distorts the meaning of an 'observation' (did she get to 'observe' what happened the second day?).

Intuitively speaking, when woken up, she learns that she was not told "go back to sleep" in the modified problem, or that she "is not asleep" in the original problem.

KipBond

06-11-2007, 10:21 AM

[ QUOTE ]
Beauty's awakenings are exactly the same as the balls in the bag.

[/ QUOTE ]

I disagree. Beauty is being woken twice and asked P(H) after a tails event, thus giving her more information about the P(H). In your ball experiment, the question was only asked once for both outcomes; no new information.

If I only asked you the P(H) if the result was tails, it's easy to see how my asking you the P(H) is in fact more information to determine P(H). (Assuming the event space is understood to be P(H & I'm asking you this question)).

jason1990

06-11-2007, 12:28 PM

I thought I would be away, and yet I am still here. The internet is like the moon. It follows you wherever you go. Anyway, my ability to read and post here may be a bit sporadic in the next couple of weeks as I am traveling. So if my replies are delayed or nonexistent, please do not be offended.

[ QUOTE ]
jason1990,

[ QUOTE ]
P(H|W) = P(H)P(W|H)/[P(H)P(W|H) + P(T)P(W|T)]

[/ QUOTE ]

Is that really derived from Bayes Theorem? The answer on the right side is "1 + something"!

[/ QUOTE ]
Actually, this is Bayes theorem itself, at least the form that I typically see presented in the textbooks I teach from. The right side is of the form a/(a + b). It is a true statement.

[ QUOTE ]
I agree with your conclusion about the paradox though - ish. I think 1/2 is her credence and 1/3 her betting odds

[/ QUOTE ]
Frankly, I do not think she has well-defined "betting odds" here. Her betting odds would depend on how often she is forced to bet. The experimenters can freely manipulate that and she will never know because of the amnesia. A person's betting odds are supposed to be tuned to the precise amount that protects the person from being exploited. There is no way to protect herself from exploitation here, no matter what odds she chooses. I think betting odds are not the right concept for this problem.

[ QUOTE ]
[ QUOTE ]
Once the assumptions are in place and the experiment begins, the Bayesian is bound by the formal rules of probability. And the formal rules of probability say that you condition on events or information (sigma-algebras). You do not take "no information", convert it into assumptions, and then condition on that. The place for converting "no information" into assumptions is in the beginning, when you choose your prior and build your probability space.

[/ QUOTE ]

Beauty isn't making assumptions.

[/ QUOTE ]
My reply was hasty and my language informal. I chose my language because I thought PairTheBoard would understand it, based on our previous conversations. Here is a less informal, but still non-rigorous, explanation of what I meant. When I said Elga was making assumptions, I meant he was defining primitive credences. SB's belief set is nothing more than an assignment of probabilities to the events in some probability space. If we assign a few of these (which I am informally calling the primitive ones), then the rest will be forced on us by the laws of probability and the requirement that SB's beliefs are consistent. To show that 1/2 before the experiment and 1/3 during are consistent, we must build a probability space that encodes both of these probabilities. Elga does not do that. Elga (implicitly) builds a probability space that encodes only the 1/3 probability. His sample space (apparently) is {H1,T1,T2}. He then gives an argument for why it is reasonable to assign 1/3 to each outcome in this space. But he cannot formulate the 1/2 before probability in this space. So his argument is incomplete. It does not show that these two credences are consistent. To show they are consistent, we must build a probability space that models the entire experiment from beginning to end. We assign the primitive credences when we build that model. Elga builds a probability space that only models the middle of the experiment.

jason1990

06-11-2007, 12:41 PM

[ QUOTE ]
Suppose we do the following experiment instead. First we flip the coin. Then if it is T, we put two black tokens in an urn. If its [H], we put one black and one white token in the urn.

Now, let S take one token and guess if it is H or T. Obviously, [if] it is black, she should give the probability one third to H.

[/ QUOTE ]
I do not believe that this is a fair translation of the problem. In your urn model, seeing a black token would be genuinely new information, precisely because of the fact that she knew it was possible to see a white token.

I think a fairer translation would be this. Flip the coin and fill your urn exactly as you describe. But instead of letting S take a token, the experimenter looks inside the bag, takes out a black token, and shows it to her. In that case, the answer is still 1/2.

PairTheBoard

06-11-2007, 12:56 PM

[ QUOTE ]
One more explanation here, in favor of 1/3.

First let's disregard the coin, and simply wake her up Monday and Tuesday, and she loses memory in between. And she knows this. Then when she wakes up, she will assign the probability 0.5 to it being Monday or Tuesday.

Suppose now that we independently of this throw the fair coin once (she will still wake up both days regardless of outcome). Now clearly when she wakes up she will assign the probability 0.25 to each of the 4 combinations in (M,T)x(H,T).

Now suppose further that after we wake her up, for the specific case Tuesday and Heads, we tell her, "go back to sleep" but for any other combination we ask her "what is the probability for heads?", and that she knows this.

Hopefully, we can all agree that for this version of the problem, she will arrive at 1/3, by Bayes' theorem, when she gets the question.

And the only difference between this problem, and the stated problem, is that in the stated problem, she never gets to wake up on Tuesday. But I don't see how that should change her information when she is woken up. It may throw us off a bit since it distorts the meaning of an 'observation' (did she get to 'observe' what happened the second day?).

Intuitively speaking, when woken up, she learns that she was not told "go back to sleep" in the modified problem, or that she "is not asleep" in the original problem.

[/ QUOTE ]

This is a great way of looking at it and what it makes clear is that the answer should be 1/2. You don't change your answer for the probability because of the number of times you are asked the question. If you do you are changing the meaning of probability so that it gives the fequency per question asked rather than the frequency per Coin flip.

Suppose I flip a coin. You are asked what the probability is that it was heads. You are asked once if it is heads. You are asked 99 times if it is tails. Do you say the probability is 1% because the coin is heads 1% of the times the question is asked? You are changing the meaning of probablity to one based on per question asked instead of per coin flip.

PairTheBoard

PairTheBoard

06-11-2007, 01:00 PM

[ QUOTE ]
[ QUOTE ]
Suppose we do the following experiment instead. First we flip the coin. Then if it is T, we put two black tokens in an urn. If its [H], we put one black and one white token in the urn.

Now, let S take one token and guess if it is H or T. Obviously, [if] it is black, she should give the probability one third to H.

[/ QUOTE ]
I do not believe that this is a fair translation of the problem. In your urn model, seeing a black token would be genuinely new information, precisely because of the fact that she knew it was possible to see a white token.

I think a fairer translation would be this. Flip the coin and fill your urn exactly as you describe. But instead of letting S take a token, the experimenter looks inside the bag, takes out a black token, and shows it to her. In that case, the answer is still 1/2.

[/ QUOTE ]

Yes.

PairTheBoard

jason1990

06-11-2007, 01:03 PM

[ QUOTE ]
So he has this perspective of a State of Awakening. His Model is conditioned throughout on this State of Awakening. In my Bag Model, I Modeled that by Her being presented with an Unopened Bag. It looks like you Model that by the Condition, (M1 that condition approaches Certainty. So it comes to have null effect when conditioned on. I'm not sure what T1 and T2 mean in your model though. Even under no conditions, or when p=1. When p=1 I don't see how your model can distingish between an "Awakening" on Monday and an "Awakening" on Tuesday. But looking at the Sleeping Beauty from the outside we certainly can distinguish between the two. So why can't we model these distinct "Awakenings"? We can see them happening after all.

[/ QUOTE ]
I do not think it is difficult to incorporate these "states of awakening" into the model. Consider for the moment a simpler situation. A man wakes up every day for a week with amnesia. At some point he finds himself awake. What is the probability it is Wednesday? Suppose his credence is 1/7. How should he model that? Well, he is basically assuming that today is some randomly chosen day. So we model it by actually randomly choosing a day. We could say that at the beginning of the week, the experimenters randomly chose a day and "flagged" it. When the man finds himself awake, he knows it is the flagged day, but he does not know which day is flagged.

Now incorporate this into the model. The experimenters will flag an awakening. If the coin lands heads, then they must flag Monday, since it is the only awakening. This corresponds to the fact that, upon awakening and learning the coin is heads, SB's credence that it is Monday is 1.

If the coin lands tails, then they can flag either Monday or Tuesday. We must assign a probability p to the event that they flag Monday, given that the coin landed tails. This number p would represent SB's credence, upon awakening and learning that the coin is tails, that it is Monday.

The probability space is completely specified by the prior probabilties

P(heads) = 1/2
P(Monday is flagged | tails) = p.

Can you find a credence p and an event A such that P(heads | A) = 1/3?

KipBond

06-11-2007, 01:47 PM

[ QUOTE ]
Suppose I flip a coin. You are asked what the probability is that it was heads. You are asked once if it is heads. You are asked 99 times if it is tails. Do you say the probability is 1% because the coin is heads 1% of the times the question is asked? You are changing the meaning of probablity to one based on per question asked instead of per coin flip.

[/ QUOTE ]

Suppose I flip a coin. You are asked what the probability is that it was heads. You are asked 0 times if it is heads. You are asked 1 time if it is tails. Do you say the probability is 50% because the coin is heads 50% of the time?

KipBond

06-11-2007, 01:49 PM

[ QUOTE ]
The experimenters will flag an awakening. If the coin lands heads, then they must flag Monday, since it is the only awakening. This corresponds to the fact that, upon awakening and learning the coin is heads, SB's credence that it is Monday is 1.

If the coin lands tails, then they can flag either Monday or Tuesday.

[/ QUOTE ]

I'm not sure what you are trying to get us to figure out ourselves, but I just wanted to say that this bolded part is not consistent with the original question. If the coin lands tails, then they will flag both Monday & Tuesday for wakings. Just wanted to point that out -- not sure if it matters -- because I'm not sure what you are trying to get us to understand?!? /images/graemlins/smile.gif

soon2bepro

06-11-2007, 01:55 PM

[ QUOTE ]
soon2bepro: [ QUOTE ]
A priori, the chance of the coin landing heads is 1/2. A posteriori, it's 1/3. (assuming whenever she wakes up during the experiment she will be asked this)...I don't see any paradox. There's new information so the answer changes.

[/ QUOTE ]

What new evidence is she receiving then? She knows the exact conditions for the experiment before it starts, and she can infer then that later she will believe 1/3 whereas now she believes 1/2. The experiment never really needs to even be done, since she's so rational she already knows exactly what she will believe later given whatever evidence.

[/ QUOTE ]

It's not "whatever evidence". The fact that she can predict what her conclusion will be given this new evidence doesn't mean the evidence doesn't matter or that it isn't new. It is new. When they wake her up they're conditioning the results of the coin having landed tails or heads. Just because they will wake her up even if the result is heads doesn't take away the fact that they'll wake her twice that often if the coin came up tails.

As I said, it's easier to understand this in terms of gambling odds. I assume you're a somewhat efficient gambler if you're posting in this forum.

PairTheBoard

06-11-2007, 02:38 PM

[ QUOTE ]
[ QUOTE ]
So he has this perspective of a State of Awakening. His Model is conditioned throughout on this State of Awakening. In my Bag Model, I Modeled that by Her being presented with an Unopened Bag. It looks like you Model that by the Condition, (M1 that condition approaches Certainty. So it comes to have null effect when conditioned on. I'm not sure what T1 and T2 mean in your model though. Even under no conditions, or when p=1. When p=1 I don't see how your model can distingish between an "Awakening" on Monday and an "Awakening" on Tuesday. But looking at the Sleeping Beauty from the outside we certainly can distinguish between the two. So why can't we model these distinct "Awakenings"? We can see them happening after all.

[/ QUOTE ]
I do not think it is difficult to incorporate these "states of awakening" into the model. Consider for the moment a simpler situation. A man wakes up every day for a week with amnesia. At some point he finds himself awake. What is the probability it is Wednesday? Suppose his credence is 1/7. How should he model that? Well, he is basically assuming that today is some randomly chosen day. So we model it by actually randomly choosing a day. We could say that at the beginning of the week, the experimenters randomly chose a day and "flagged" it. When the man finds himself awake, he knows it is the flagged day, but he does not know which day is flagged.

Now incorporate this into the model. The experimenters will flag an awakening. If the coin lands heads, then they must flag Monday, since it is the only awakening. This corresponds to the fact that, upon awakening and learning the coin is heads, SB's credence that it is Monday is 1.

If the coin lands tails, then they can flag either Monday or Tuesday. We must assign a probability p to the event that they flag Monday, given that the coin landed tails. This number p would represent SB's credence, upon awakening and learning that the coin is tails, that it is Monday.

The probability space is completely specified by the prior probabilties

P(heads) = 1/2
P(Monday is flagged | tails) = p.

Can you find a credence p and an event A such that P(heads | A) = 1/3?

[/ QUOTE ]

I think we're getting to it here. With the example of being awoken every night of the week. The man is asked the question, "what is the probability this is monday". What if instead of randomly flagging a day to ask him we simply ask him every single day? He is asked 7 times and must answer 7 times. Each time he is asked his infomation is identical. His answer should be the same each time he is asked. If he answers 1/7 he will be giving the frequency wrt the number of times he is asked the question. The day will be monday 1/7 of the times the question is asked. No assumptions need be made about how the day was flagged to ask him because he is asked all 7 days. I believe that is part of the problem statement. If not we can make it so.

Now the coin is flipped. Heads he is just asked Monday. Tails he is asked both Monday and Tuesday. But now the question asked is, "what is the probability the coin landed heads? Or equivalently, "what is the probability you are in a week where you are not being asked Tuesday", or more simply, "what is the probability this is an "Asking" dictated by Heads?", or more simply yet, "What is the probability this is a Heads-Week?".

My contention is that he can answer in two ways. If he wants to continue being correct wrt the number of times he is being asked, like he did with his 1/7, he should say 1/3 frequency wrt number of times Asked. But if he wants to be correct wrt the coin flip he will say 1/2. That is, he is recognizing that half the times this experiment is repeated, ie half the number of coin flips, he will be getting asked this question on a Monday dictated by Heads.

I think the 1/2 wrt coin flip frequency puts the man in conformity with what we see from the outside. His 1/3 wrt number of times Asked is a personal probability measure he adopts for his situation. It has the utility of giving him the right odds if he has to bet on it each time he is asked the question.

PairTheBoard

KipBond

06-11-2007, 02:59 PM

[ QUOTE ]
I think the 1/2 wrt coin flip frequency puts the man in conformity with what we see from the outside. His 1/3 wrt number of times Asked is a personal probability measure he adopts for his situation. It has the utility of giving him the right odds if he has to bet on it each time he is asked the question.

[/ QUOTE ]

His answer is definitely 1/2 wrt coin flips -- both internal & external. It should be 1/3 wrt flip+wakings - both internal & external.

PairTheBoard

06-11-2007, 03:52 PM

[ QUOTE ]
[ QUOTE ]
I think the 1/2 wrt coin flip frequency puts the man in conformity with what we see from the outside. His 1/3 wrt number of times Asked is a personal probability measure he adopts for his situation. It has the utility of giving him the right odds if he has to bet on it each time he is asked the question.

[/ QUOTE ]

His answer is definitely 1/2 wrt coin flips -- both internal & external. It should be 1/3 wrt flip+wakings - both internal & external.

[/ QUOTE ]

Right. We can say the same thing from the outside. I suppose we could define other probability measures for these events and come up with even more answers. There's a queston then about what "THE" probability is. Is there a "Real" probability? If so, which is it? If not, doesn't that violate the Baysian philosophy that says there should be one? Or do they say that? I'm not sure.

The Baysian definition says THE probability for you should be the one whereby the odds it defines make you indifferent to the bets at those odds. But does that definition make sense when the amount of the bet is being dictated by the outcome of the bet. We recoginize that happening here. If he has to bet on heads each "Asking" then the Tails losing outcome of the Bet forces him to bet twice.

With the "Cards" example where Black cards are marked 2 and Red Cards 1, betting on Red you are Forced to bet the Number Marked; does the Baysian change his probability that Red happens just because the bet is structured illegitimately?

PairTheBoard

KipBond

06-11-2007, 04:04 PM

[ QUOTE ]
The Baysian definition says THE probability for you should be the one whereby the odds it defines make you indifferent to the bets at those odds. But does that definition make sense when the amount of the bet is being dictated by the outcome of the bet. We recoginize that happening here. If he has to bet on heads each "Asking" then the Tails losing outcome of the Bet forces him to bet twice.

[/ QUOTE ]

Well, you can either look at it as being two different events: where you are betting on one, being asked the probability of the other; OR you can look at it as getting more information regarding the outcome of an event for which you had a prior probability.

I guess the confusion comes when you try to come up with the same answer without more information, thinking you are still talking about the same event. You aren't.

You either have more information about the same event, or you have a different event. Depends on how you want to look at it.

It's really simple to see here:

I flip a coin. I ask you the P(H). It is 1/2. Now, if you know I will only ask you the P(H) if the outcome was tails, then what is the P(H)?

You can either look at it as more information regarding the outcome of the same event. Or as a separate event (a flip + asking event). Either way, you should come up with 0.

Jetboy2

06-11-2007, 05:02 PM

Why is this important enough to have 163+ responses and counting?

PairTheBoard

06-11-2007, 05:12 PM

[ QUOTE ]
I flip a coin. I ask you the P(H). It is 1/2. Now, if you know I will only ask you the P(H) if the outcome was tails, then what is the P(H)?

You can either look at it as more information regarding the outcome of the same event. Or as a separate event (a flip + asking event). Either way, you should come up with 0.

[/ QUOTE ]

I agree, If you know I only ask when Tails happens, and I ask you, then asking you gives you the information that tails happened.

But that's not our situation. The outcome hasn't been determined by the Asking. All you know is that you are going to be Asked more times if it's Tails. But you don't know if this is one of those times. Do you want a frequency per times Asked or per Coin Flip? The Betting odds Baysian definition requires a legitimate bet, not one where the outcome dictates the amount bet. So what would be the legitimate bet here? I think a the legitimate bet would be one where you are only required to make it once per coin flip experiment. In that case the answer is 1/2.

Notice in your No-Ask on Heads example it is not such a legitmately defined bet. Once again it would be a bet where you are only given a chance to make it when the losing outcome has already been determined.

PairTheBoard

KipBond

06-11-2007, 05:48 PM

[ QUOTE ]
[ QUOTE ]
I flip a coin. I ask you the P(H). It is 1/2. Now, if you know I will only ask you the P(H) if the outcome was tails, then what is the P(H)?

[/ QUOTE ]

I agree, If you know I only ask when Tails happens, and I ask you, then asking you gives you the information that tails happened.

But that's not our situation. The outcome hasn't been determined by the Asking. All you know is that you are going to be Asked more times if it's Tails. But you don't know if this is one of those times. Do you want a frequency per times Asked or per Coin Flip?

[/ QUOTE ]

I think my scenario is the same as the SB scenario. The Asking doesn't determine the outcome of the flip in either one. In both, the frequency of heads per Coin Flip is 50%. In the SB case, the frequency of heads per Asked is 1/3. In mine the frequency of heads per Asked is 0.

If you are betting on Heads in the SB scenario & bet once per flip you need 1:1 odds for 0EV.
If you are betting on Heads in my scenario & bet once per flip you need 1:1 odds for 0EV.

If you are betting on Heads in the SB scenario & bet once per Asking, you need 2:1 odds for 0EV.
If you are betting on Heads in my scenario & bet once per Asking you will always lose.

PairTheBoard

06-11-2007, 06:44 PM

Right. The question is, which defines the Baysian Bet which they say determines what subjective probablity is?

I agree that "Asking" would provide the kind of information we normally use to apply Bayes' Theorem, If the situation were as follows. Suppose SB always wakes up on Both monday and tuesday. She is asked both days unless it's a Heads-Tuesday. In that case she awakens and is not "Asked". Now you have something you can apply Bayes to. If she awakens and is Not Asked she knows it is Tuesday and the coin flipped Heads. By Asking her you have given her information.

But notice how that differs when she considers it before the experiment begins. She can see that she has to wait to enter the exeriment before she can find out that information. There will be real information waiting for her within the experiment.

Compare that to the Actual Experiment. SB looks at the experiment before it starts and sees that she will be awakened and asked. She knows that before it happens. She doesn't have to wait for it to happen to find out. From her view before the experiment starts she sees it will give her no information for how the coin landed. If it did, she could PREDICT heads as 1/3 now. But it doesn't.

Once she is in the experiment, being awakened and asked doesn't tell her any more about how the coin landed than she knew before the experiment began. That's because to her, all awakenings are the same. When she is awakened she knows there is a 1/2 chance the coin landed heads and this must be monday. She experiences all awakenings the same, but she knows not all are equally likely.

I think the fairly structured Bet by which SB should measure her subjective probability for the coin landing heads is as follows. A Single Wager is defined before the experiment begins. It is one Single Wager at even money on Heads. SB can either agree to this One Wager before the experiment or during the experiment. She can agree to it as many times as she likes during the experiment, but she is only agreeing to the One Wager. She isn't "making" two bets but instead she is consistently agreeing to the One Wager.

This allows her to hold the correct Credence of 50% for differing lengths of time - 1 or 2 days - without penalizing her for how long she might hold the correct Credence. The Credence should not depend on the length of time she is set-up to have it.

She agrees that her "Acceptance" of this One Wager while she is in the experiment will be conveyed by the Credence she gives for heads having been flipped. She can call off the bet when she gets inside the experiment if she thinks that having been awakened has given her new information that she didn't have before the experiment as to the likelihood of Heads. If she thinks this new information tells her the probablility heads flipped is now 1/3 she will say so and the Wager will not be Accepted.

However, if she thinks the probality of heads is still 1/2 she will say so and the Wager will be Accepted. Saying 1/2 more than once just confirms the Acceptance of the Wager. It doesn't make a second one.

She will say 1/2.

PairTheBoard

bigmonkey

06-11-2007, 10:10 PM

OK I have read all the posts since last I commented. Thanks for the good discussion, particularly to PairTheBoard and KipBond. What follows is basically my response to certain comments in the order they were posted. I'll put the names in bold so people can pick them out, although this post as a whole should reflect my general position to the SB paradox in light of all the comments. Apologies for not making the time earlier to keep up with this: a lot of the things I quote in this post were said some way back into the thread. I don't really expect everyone to understand the exact context of what I am saying here. Hopefully this post is an indication that I have kept up with the thread and should understand references to arguments you have made throughout the whole thread.

uberfish2, Beauty knows before the experiment that she won't wake up on Tuesday when head falls. Why wouldn't she know this? She knows all of the rules.

Snowball, I quite like your knicks vs lakers analogy, and I can imagine the SB paradox could be expressed in sich a way but I don't think what you have is it. Beauty knows she will wake up, whereas the person in the example doesn't know whether his neighbour will come over or not. So I don't see how waking up could be new information when you know it will happen in advance.

KipBond, about your re-configured experiment where you removed the ambiguity (sorry this was half the thread ago). Like my above objection to Snowball, Beauty does get information in your example from waking up, about the probability that heads landed. In the SB paradox she gets no information. I think that even if you want to say she can infer something from waking up, she could have inferred the same thing before she slept, and she might end up wih two contradictory answers before the experiment even begins, which still entails a paradox.

Siegmund,

[ QUOTE ]
The point is that a heads leads to one heads-awakening but a tails leads to two tails-awakenings. If you want to say that a heads-awakening and a heads entail each other, in the sense that one happening implies the other must have happened / must be about to happen, fine - but that's no reason for you to believe that if the coin comes up heads half the time, half of the awakenings must be heads-awakenings.

[/ QUOTE ]

Forget tails for the moment. We know that the probability of a heads is equivalent to the probability of a heads-awakening, because if and only if there is a heads, there is a heads-awakening. You are saying that the former is 1/2 and the latter is 1/3. In terms of a Venn diagram displaying probabilities, they occupy the same space. So all you are doing is confirming that there is a paradox for there are two equally valid methods for answering the same question but they contradict each other.

PairTheBoard, you discuss Elga's use of idenitfying the probability of T1 with T2 by using Q(Monday|Tails) and Q(Tuesday|Tails), and then remind us that he identifies the probability of H1 with T1 by finding Q(Heads|Monday) and Q(Tails|Monday). And you question whether Q(T1)= Q(Tails|Monday)= Q(Monday|Tails). Elga argues that Q(T1|Monday)=Q(H1|Monday), and from that infers that Q(T1)=Q(H1)=1/2. When I first saw this it seemed intuitively obvious but I wasn't entirely sure that this could be proved from the axioms of the probability calculus. I haven't seen any proof for it, but I'm willing to accept it, as it just doesn't seem like the paradox is located there. I think if that were proved to be a bad inference, it would have serious repurcussions elsehwere in how we think of probability.

[ QUOTE ]
He finally claims that Q(H1) is indeed the thing we were looking to find out about Q. That is Q(H1)=Q(coin having landed on heads). I think that's reasonable according to his identification of Q(H1) with Q(heads|monday). But what that ends up implying for our rational Beauty is that,

Q(heads)=Q(H1)=Q(heads|monday)=Q(tails|monday)=Q(T 1)=Q(monday|tails)=Q(tuesday|tails)=Q(T2)=50%

[/ QUOTE ]

I like your argument but I know that Elga wouldn't say that Q(H1)=Q(heads|monday), since its being Monday makes heads more likely for Elga (from 1/3 to 1/2). Working backwards from this I think that Elga doesn't identify H1 with Q(heads|Monday). Following on from what I just said before this quote, he just infers that because Q(H1|Monday)= Q(T1|Monday), Q(H1)=Q(T1). Note that that isn't a logical (or probabalistic) law, but because "Monday" is a necessary condition for both H1 and T1, by conditionalizing on "Monday" we are not cutting out any difference in probabilities between H1 and T1. For example, there's no probability that H1 outside of its being Monday, that T1 doesn't have.

[ QUOTE ]
That is the fallacy in Elga's logic.

[/ QUOTE ]

You then go on to argue against Elga's identification of Q(Tails and Monday) with Q(Tails|Monday), but I do not think he does this. His position is that Q(Tails and Monday)=1/3 and Q(Tails|Monday)=1/2.

You then come back to my argument where I stated that she expected 1.5 awakenings per experiment (incidentally my professor told me he thought this was a perverse way of looking at it...but still true) And you argue as follows:

[ QUOTE ]
She knows that if she takes two Week long trips she will get on average 3 awakenings, one due to heads and two due to tails. But should she think the three of them are equally likely to be the one she is experiencing right now? No. She is more likely to be exeriencing the one due to heads! She knows there is a 50% chance she is experiencing That one. There is only a 25% chance she is experiencing the one due to tails on monday. And only a 25% chance she is experiencing the one due to tails on tuesday. Just because there are three types of awakenings doesn't mean they are equally likely to be the one she is experiencing right now.

[/ QUOTE ]

I do agree with this. But I think this just re-inforces that there is some kind of paradox. You can reason to this conclusion from her stubborn belief that P(H)=P(T)=1/2. This I think is a valid argument and expresses the 1/2 answer well, but I think the argument for 1/3 also looks valid. In the SB paradox it seems it isn't enough to prove that the answer is x so it cannot be y, because the answer for y is intuitively plausible, and the work needs to be done on showing why y is intuitive but false.

[ QUOTE ]
What do you do when your Credence Function is not giving you the right betting odds?...The betting objection has been answered.

[/ QUOTE ]

I agree with everything you say between these lines, where you attempt to reconcile her credence function of (50,25,25) with the correct betting odds (33,33,33). My formal position (i.e. when I'm not gripped by the "illusion"/sense that Elga is describing a credence not a betting odds), is that the paradox is located in our Bayesian assumption that credences are equal to betting odds. When you think about it this is fairly trivial. ("I'm going to give you a bet on heads at 1:1 odds, and if it is tails you have to bet on heads again for the coin that just landed tails. Do you accept?" "No.")

Kipbond then says this in response to PairTheBoard's (50,25,25) model:

[ QUOTE ]
P(Coin is Heads) = 50%
P(Coin is Tails) = 50%

P(Coin is Heads & I then wake her on Monday) = 50%
P(Coin is Tails & I then wake her on Monday) = 50%
P(Coin is Tails & I then wake her on Tuesday) = 50%

If you want to combine these events to make a total "event space" that totals 100%, then each "flip-waking" event has a 1/3 chance of happening.

[/ QUOTE ]

I think this accurately reflects my "illusion"/sense that Elga really is analysing her credence and not her betting odds. I am currently stuck at this satge.

f97tosc then describes an experiment with black and white tokens in an urn which I don't think is representative of the SB paradox, and is quite similar to the knicks vs lakers example. I think Jason later agrees that the example is unrepresentative.

Deorum,

[ QUOTE ]
Quick Proof:

She knows:

1 tails flip = two awakenings

1 heads flip = one awakening

Through substitution, 1 tail flip = 2 heads flips

She will have 2/3 awakenings due to tails, and 1/3 due to heads (on average - remember, she knows that).

Since each "tails awakening" is only worth half of a "heads awakening" in terms of flips, and there are twice as many "tails awakenings" than "heads awakenings" she will come to the conclusion that the coin comes up 50% heads and 50% tails.

[/ QUOTE ]

I'm not sure it's right to say that a heads-flip equals one awakening and a tails-flip equals two awakenings, therefore two heads-flips equal one tails-flips. Heads-flips don't equal one awakening, they entail one awakening. So there is no substitution, and Beauty never gets to experience what "2 awakenings" is like, she is only ever aware of one at a time. But again though, I think arguing for one answer isn't enough in this debate: it's disproving the answer that you think is wrong. Otherwise what you have said follows fairly naturally from what the dogmatic 1/2 advocate would say.

You later answer to KipBond that: [ QUOTE ]
The answer to the question, "What is the probability that the coin was heads?" is 1/2. The answer to the question, "What is the probability that you were awakened this time by a 'tails flip'?" is 2/3.

These questions are not the same. The first refers to the probability of the state of the coin, while the second refers to the probability that she was awakened by a specific flip. The probability of the state of the coin and the probability that she was awakened by a specific flip are not the same.

[/ QUOTE ]

This is the basis of my argument with Siegmund. A heads always entails a heads-awakening and a heads-awakening always entails a heads therefore the probability of these seperate events must be the same. Equally the probability of a tails is equal to the probability of two tails-awakenings. There seems to be some problem using that method to calculate the relationship between a tails and a tails-awakening, but I think with the former version we can see that P(H)= P(H-awakening)

PairTheBoard then gives another analogy involving black and white balls in a bag. I don't quite understand what it means to be a black or white ball, but in the end you seem to settle on this as a conclusion:

[ QUOTE ]
She can simply say, the probability I am in a Heads Awakening is 1/2 per Experiment and 1/3 per Awakening.

[/ QUOTE ]

That does sound intuitive on first inspection. But we can substitute "I am in a heads awakening" for "a heads landed", as one entails the other. But there is only one coin flipped. And the coin could already be flipped by the time she wakes up on Monday. I'm not sure what it even means for the probability of heads to be 1/2 or 1/3 per something.

jason1990,

[ QUOTE ]
A person's betting odds are supposed to be tuned to the precise amount that protects the person from being exploited. There is no way to protect herself from exploitation here, no matter what odds she chooses. I think betting odds are not the right concept for this problem.

[/ QUOTE ]

I don't follow what you mean here. Her betting odds of 1/2 before the experiment and 1/3 during the experiment are what stop her from being exploited.

[ QUOTE ]
Elga does not do that. Elga (implicitly) builds a probability space that encodes only the 1/3 probability. His sample space (apparently) is {H1,T1,T2}. He then gives an argument for why it is reasonable to assign 1/3 to each outcome in this space. But he cannot formulate the 1/2 before probability in this space. So his argument is incomplete. It does not show that these two credences are consistent.

[/ QUOTE ]

Elga's conclusion is that this is a strange sort of counter-example that refutes the principle of reflection, so he is comfortable enough to accept some kind of paradox (at least to those who don't want to give up reflection).So I think his explanation does not require there to be any kind of coherence between the original credences and the credences Beauty has during the experiment, or indeed, both of those credence sets she has at the same time.

[ QUOTE ]
Consider for the moment a simpler situation. A man wakes up every day for a week with amnesia. At some point he finds himself awake. What is the probability it is Wednesday? Suppose his credence is 1/7. How should he model that? Well, he is basically assuming that today is some randomly chosen day. So we model it by actually randomly choosing a day. We could say that at the beginning of the week, the experimenters randomly chose a day and "flagged" it. When the man finds himself awake, he knows it is the flagged day, but he does not know which day is flagged.

[/ QUOTE ]

Sorry I do not really understand this example. What do you mean by flagging?

Soon2bepro,

[ QUOTE ]
It's not "whatever evidence". The fact that she can predict what her conclusion will be given this new evidence doesn't mean the evidence doesn't matter or that it isn't new. It is new. When they wake her up they're conditioning the results of the coin having landed tails or heads. Just because they will wake her up even if the result is heads doesn't take away the fact that they'll wake her twice that often if the coin came up tails.

[/ QUOTE ]

What do you mean by the people waking her up conditioalizing upon heads or tails? It doesn't matter what the experimenters think in this example. In what sense is the evidence new? I cannot see how it could be new at all. Before the experiment she isn't conditionalizing on whether she will wake up, i.e. thinking "If I wake up then...", she already knows she will wake up and this is part of her belief set. When she wakes up and tries to put "I am awake" in her belief set, she will find it is already there and it won't change any of her credences, or count as new evidence.

PairTheBoard's last post before this one: I agree with much of this. On the "One Wager" scenario: just to clarify, is it the case that the wager is only placed if she agrees to it at every chance she gets? Or does she just have to agree to it at the very last opportunity? ie either Monday or Tuesday) Actually I suppose she answers the same on Monday and Tuesday anyway so that makes no difference. She'll either follow Yes/No, Yes/Yes, Yes/No/No or Yes/Yes/Yes. I personally think we can rule out any case where she changes her mind since she isn't given any new evidence to compel her to. That leaves the cases where she accepts the wager on evens for heads. I think this example simply highlights that the 1/2 answer is more intuitive than the 1/3 answer. I think if we had to conclude that only one is right, giving up 1/2 would do a lot of damage to our idea of probability. Giving up 1/3 does some damage but not quite as much I don't think. Basically I think the project we are left with is explaining away the 1/3 answer to vindicate the 1/2 answer. In the history of the literature, there was Elga's argument for 1/3 and denying Reflection. Then there was Lewis basically just arguing the case for 1/2 and stating that therefore it cannot be 1/3. Then Christopher Hitchcock introduced the "Dutch-book" argument whereby Beauty is permitted to bet with 1/2 then with 1/3, and he used this as evidence for the 1/3 account. Leitgeb & Bradley then dismissed Hitchcock's argument saying that credences are not the same as betting odds. For a long period I was convinced that that final paper had solved the supposed paradox, but more recently I have been thinking that they have to explain how Elga's argument for 1/3 refers to a betting odds, because it looks a lot like a credence.

KipBond

06-11-2007, 11:21 PM

[ QUOTE ]
For a long period I was convinced that that final paper had solved the supposed paradox, but more recently I have been thinking that they have to explain how Elga's argument for 1/3 refers to a betting odds, because it looks a lot like a credence.

[/ QUOTE ]

I'll throw this out -- without a lot of thought -- chew it up & spit it out:

Elga's argument for 1/3 is both a betting odd & a credence. It's a betting odd in regards to the outcome of the coin flip event. It's a credence in regards to a flip-waking event.

jason1990

06-12-2007, 01:25 AM

"Credence" is just another word for Bayesian probability. When someone tells us their credences, we may ask two questions. Are they consistent? Are they justified?

The definition of "consistent" in Bayesian probability is formal and mathematical. A set of probabilities is consistent as long as it does not violate the axioms of probability. Equivalently, they are consistent if and only if they can be formulated in a common probability space. The question of consistency is a purely mathematical question.

Justification, on the other hand, is not a purely mathematical question. We cannot decide between two sets of consistent probabilities without appealing to something outside of mathematics. We may appeal to science, or philosophy, or intuition, or pragmatism. But we must appeal to something which is not mathematics.

In this problem, SB has a "prior credence" (before the experiment) of 1/2 for the probability of heads. Her "posterior credence" (during the experiment) is what we are debating. The candidates are 1/2 and 1/3. Let us focus for now on this question: which of these two posterior credences are consistent with the prior credence of 1/2?

A simple probability space has already been presented which shows that posterior 1/2 and prior 1/2 are consistent. The space is {H,T} with measure P(H) = P(T) = 1/2. We define W to be the event that she is awake at some point during the experiment. Then W = {H,T}. Her prior credence is simply P(H) = 1/2, and her posterior credence is P(H|W) = 1/2. Since both credences have been formulated in a common probability space, they are consistent.

Now, here is another probability space which shows that posterior 1/3 and prior 1/2 are also consistent. In this probability space, we model not only the uncertainty in the coin, but also SB's uncertainty regarding the day of the week. The sample space is S = {w1,w2,w3,w4,w5}, where

w1 = "The coin is heads and today is Sunday"
w2 = "The coin is heads and today is Monday"
w3 = "The coin is tails and today is Sunday"
w4 = "The coin is tails and today is Monday"
w5 = "The coin is tails and today is Tuesday"

Choose any p such that 0 < p < 1/2. Define q = (1 - 2p)/3. Let P(w1) = P(w3) = p and P(w2) = P(w4) = P(w5) = q. It is easy to check that this is a valid probability space. Her prior credence can be formulated as

P(heads | Sunday) = P(w1)/P(w1,w3) = p/2p = 1/2,

and her posterior credence can be formulated as

P(heads | not Sunday) = P(w2)/P(w2,w4,w5) = q/3q = 1/3.

This probability space has the unusual property that it appears to assign SB credences that are completely unnecessary. For instance, if SB found herself in some bizarre state of mind where she could not tell whether or not it was Sunday, then her credence for heads would be P(heads) = P(w1,w2) = p + q = (p + 1)/3. In the end, though, this is irrelevant to the question of consistency, as is the actual value of p. We have constructed a single probability space in which we can formulate both the prior 1/2 and the posterior 1/3 credences, so they are consistent.

What this means is that the "right" answer to this question cannot be determined by pure mathematics. We must decide which credence to use according to which we feel is more justified and/or useful. Maybe that is why this debate seems to have raged on longer than it should have. My personal opinion is that neither is particularly useful. The posterior 1/2 seems sensible, and its corresponding probability space seems more "natural." But it also seems to be completely useless. The posterior 1/3 looks useful for gambling, but in practice I will never gamble with scientists who already know the result I am gambling on, nor would I gamble with anyone while I am under the influence of amnesia drugs.

So my final answer (for now) is that they are both mathematically correct, and I (personally) have no reason to prefer one over the other.

PairTheBoard

06-12-2007, 01:46 AM

[ QUOTE ]
PairTheBoard's last post before this one: I agree with much of this. On the "One Wager" scenario: just to clarify, is it the case that the wager is only placed if she agrees to it at every chance she gets? Or does she just have to agree to it at the very last opportunity? ie either Monday or Tuesday) Actually I suppose she answers the same on Monday and Tuesday anyway so that makes no difference. She'll either follow Yes/No, Yes/Yes, Yes/No/No or Yes/Yes/Yes. I personally think we can rule out any case where she changes her mind since she isn't given any new evidence to compel her to. That leaves the cases where she accepts the wager on evens for heads. I think this example simply highlights that the 1/2 answer is more intuitive than the 1/3 answer. I think if we had to conclude that only one is right, giving up 1/2 would do a lot of damage to our idea of probability. Giving up 1/3 does some damage but not quite as much I don't think. Basically I think the project we are left with is explaining away the 1/3 answer to vindicate the 1/2 answer.

[/ QUOTE ]

As you said, all awakening experiences are identical to SB. So whatever logic she uses to arrive at a Credence once, she will duplicate to produce consistent credences in all her awakenings. So she will either always accept the Single Wager by believing there is a 50% chance she is in a heads produced experience or always reject the wager by believing there is a 1/3 chance she is in a heads produced experience. I claim she will believe 50%.

When in a tails produced week she will confirm the Single Wager on Tuesday. And she will break even on the wager. The only difference is that she Holds the Belief a longer period of time when the outcome is Tails than Heads. The Belief should not be affected by how long you hold it.

bigmonkey, do you think this paradox is in some ways still an open question in logic and probability? I find that hard to believe.

I thought of another way of looking at this. Suppose we first define a deterministic trip for SB. She sleeps for two weeks. In the first week she is awakened on monday. In the second week she is awakened every day of the week. Making it more extreme may help us see the principle.

Now she enters the experiment. She is in the midst of an awakening experience. Nobody asks her anything. But she wonders to herself, "I wonder which week this is?". Isn't she clearly going to believe that it's more likely she is having this awakening experience in week 2 than week 1. All 8 awakenings are exactly the same to her. So she should believe the one she is in is equally likely to by any of the 8. That means chances are 7/8 she is now experiencing one of the 7 awakenings in week 2. In other words, chances are 7/8 that she is in week 2.

Now compare that to this scenario. Again SB will sleep for two weeks. If heads if flipped she will be awakened on Monday in Week 1. She will sleep through week 2. If tails is flipped she will sleep through week 1 and be awakened 7 times in week 2. Call this her probabilistic trip.

Now, why does she think differently in the probabilistic trip? She finds herself in the midst of an awakening experience. Why shouldn't she use the same logic as above and decide there is a 7/8 chance she is in week 2? The difference is that in the deterministic trip she experiences all 8 awakenings in the same trip. 1 in week 1 and 7 in week 2. But in the probabalistic trip she doesn't have an awakening experience in both weeks.

If heads is flipped she only experiences week 1. And she knows there is a 50% chance that heads was flipped. Therefore, she must logically conclude there is a 50% chance that this awakening is her 1 day experience of week 1. She knows that if Tails flipped she is having an experience of week 2. In other words, if Heads flipped this must be week 1. If Tails flipped this must be week 2. 50% chance of heads so 50% chance this is week 1 and 50% chance this is week 2.

The fact that if it is week 2 she doesn't know what day it is doesn't change the probability of 50% that this is week 2. One experience of week 2 is exactly the same as another in week 2 for her. Essentially, when she experiences week 2 she has a longer experience of it where she can't measure time passing. She experiences week 2 for 7 days but cannot tell she is experiencing it for 7 days. Essentially with Tails she holds her 50% Credence for 7 days. With heads she holds her 50% credence for 1 day.

The Credence is correct. It's correctness should not be measured by how long it is held. When the Wager I described is set up properly so that it does not arbitrarily punish her for Holding her Credence longer when she has the losing end of the bet, it shows that her 50% Credence conforms to a correct probability calculation based on the information available.

Finally, when Beauty expresses her 50% Credence 7 times in Week 2, that should be considered One Belief consistently expressed over the entire week. That's One Belief good for making One Bet on heads. It should not be viewed as 7 seperate Beliefs betting heads 7 different times. That's what the 1/3 people are doing. Expressing a Belief for a longer period of time does not make it incorrect because it could make more bets in the longer time period.

PairTheBoard

KipBond

06-12-2007, 10:10 AM

[ QUOTE ]
Finally, when Beauty expresses her 50% Credence 7 times in Week 2, that should be considered One Belief consistently expressed over the entire week. That's One Belief good for making One Bet on heads. It should not be viewed as 7 seperate Beliefs betting heads 7 different times. That's what the 1/3 people are doing. Expressing a Belief for a longer period of time does not make it incorrect because it could make more bets in the longer time period.

[/ QUOTE ]

It depends what the bet is. Is it that the coin flip resulted in heads? Or is it that the coin I flipped prior to waking you was heads? The former is a single event, and betting on it twice when you are wrong is a mistake. The latter is multiple events in the case of tails, and you can bet on each one of those -- you'll win twice as often betting on tails.

jason1990

06-12-2007, 01:00 PM

Incidentally, if we take p = 1/5 in my "posterior 1/3" probability space, then we have a classical application of the indifference principle. If p = 1/5, then q = 1/5 also and all outcomes are equally likely. What this says is that if SB found herself awake, and was somehow in a state of complete ignorance, unable to tell the difference between the beginning of the experiment (Sunday) and the middle of the experiment (Monday/Tuesday), then she would know that there are only five possibilities (w1,...,w5) and would regard them as equally likely.

By the way, I think this whole wagering debate is entirely semantic. Suppose I bet $100 on the Yankees to win the next World Series. I then immediately place another identical $100 bet. You can equivalently regard that as two $100 wagers, or as a single $200 wager. It is silly to debate which perspective is "correct." They are both correct, both mathematically and logically.

Now suppose SB bets $1 on heads at even money each time she is awake. When the coin is tails, she will make two consecutive $1 wagers. If you regard these as separate wagers, then over the course of 100 experiments, she will have made 150 wagers and won 50 of them. Her probability of winning is 1/3 and she loses money because she chose the wrong odds. On the other hand, if you regard these consecutive wagers as a single $2 wager, then over the course of 100 experiments, she will have made 100 wagers and won 50 of them. Her probability of winning is 1/2 and she loses money because the bet size is correlated to the outcome.

It seems silly to me to argue over which is the "correct" way to think about this. They are both correct, both mathematically and logically.

KipBond

06-12-2007, 02:48 PM

[ QUOTE ]
It seems silly to me to argue over which is the "correct" way to think about this. They are both correct, both mathematically and logically.

[/ QUOTE ]

It's such a paradox. /images/graemlins/laugh.gif

PairTheBoard

06-12-2007, 06:55 PM

There is a simple, irrefutibable, clarifying intermediate statement of probability that I think we've been overlooking. SB can make this satement before entering the experiment. And it's what she needs to carry into the experiment with her as the Inititial Knowledge from which she makes her deductions. I'm calling this statement L_0. This statement applies equally well to our SB scenario or to any similiar one where different number of awakenings are produced by heads and tails. It only requires the Precondition I show below. This is L_0.

Precondition: Heads produces at least 1 awakening. Tails produces at least 1 awakening.

L_0. There is a 50% probability that ALL OF MY AWAKENINGS are due to Heads. There is a 50% probability that ALL OF MY AWAKENINGS are due to Tails.

Any 1/3 argument must deal with L_0. If you agree with L_0 then you must logically agree with the deductions that follow from it.

SB takes L_0 into the experiment with her. She now experiences an awakening. She applies L_0. This is her logic:

SB: I know there is a 50% chance that ALL of my awakenings are due to Heads. I am having an awakening. Therefore,

L1: There is a 50% probability that my awakening is ONE OF THE AWAKENINGS DUE TO HEADS.

similiarly she deduces,

L2: There is a 50% probability that my awakening is ONE OF THE AWAKENINGS DUE TO TAILS.

If the 1/3 people accept L_0 they must accept L1 and L2.

SB deduces further,

L1_1: By L1 there is a 50% probability that my awakening is One of the awakenings produced by Heads. Therefore, since Heads produces only one awakening on Monday, there is a 50% probability my awakening is the one awakening produced by heads on Monday.

simliliarly SB deduces,

L2_1: By L2, there is a 50% probability my awakening is One of the Awakenings due to Tails. Since Tails produces 2 awakenings, one on Monday and one on Tuesday, And, if tails was flipped I couldn't in my awakening now distinguish between Monday and Tuesday, I conclude that if Tails was flipped my awakening is equally likely to be a Tails-Monday one as a Tails-Tuesday One. Therefore, I divide the 50% probability that my awakening is One of the Two Tails awakenings by saying there is a 25% chance my awakening is the one due to Tails on Monday and a 25% chance my awakening is the one due to Tails on Tuesday.

Logically,

L_0 ==> L1 ==> L1_1
and
L_0 ==> L2 ==> L2_1

If the 1/3 people disagree they are not following the logic. SB will follow where the logic takes her.

It comes down to misdirection from Sklansky's Principle again. Just because SB cannot distingish which of the 3 conditions her awakening is in, Heads-Monday, Tails-Monday, or Tails_Tuesday, does not mean they are equally likely. She has knowledge about the experimental settup that lets her logically deduce that they are Not equally likely for her present awakening.

Kips example of 1000 Trials of the experiment producing 500 Heads-Monday awakenings, 500 Tails-Monday awakenings, and 500 Tails-Tuesday awakenings is not problem. How do you divide probabilities for SB's current awakening being among those? Easy. 50% that it's Heads-Monday. 25% that it's Tails-Monday. 25% that it's Tails-Tuesday.

Here's what that means. In half the 1000 Trials, when SB has an awakening it will be one of the 500 Heads-Mondays. In the other half of the Trials, SB will have 2 awakenings. So in these 500 Trials SB will have 1000 awakenings. The awakening she is in now is equally likely to be one of the 500 on Monday or 500 on Tuesday. Thus her awakening now has 50% chance of being Tails and then 50% of that to be Monday or Tuesday. Thus her awakening now has 25% chance of being one of the 500 Tails-Mondays and 25% chance of being one of the 500 Tails-Tuesdays. There is no problem.

Kim has the intuition that SB's current awakening is one of 1500. 50%(1500) is not equal to the 500 Heads-Monday awakenings. So something must be wrong. Here's what's wrong with that thinking. SB has a probability function for where she is when she has an awakening. She can test that by looking at 1 awakening in each of the 1000 trials and seeing which kind of awakening it is. She doesn't bias the test by looking at 2 awakenings in Trials that have been determined by Tails. She tests it by looking at 1 awakening per trial. So in 1000 trials she looks at 1000 awakenings, not 1500. 50%(1000)=500 Heads-Monday Awakenings. 25%(1000)=250 Tails-Monday awakenings. 25%(1000)=250 Tails-Tuesday awakenings. That's 500+250+250= 1000 Trials. She tests the pobability for "This Awakening" by taking 1 sample from each of the 1000 Trials.

PairTheBoard

bigmonkey

06-12-2007, 09:05 PM

KipBond:

[ QUOTE ]
I'll throw this out -- without a lot of thought -- chew it up & spit it out:

Elga's argument for 1/3 is both a betting odd & a credence. It's a betting odd in regards to the outcome of the coin flip event. It's a credence in regards to a flip-waking event.

[/ QUOTE ]

I agree that 1/3 is the correct betting odds for Beauty. We want to say that 1/2 is her correct credence (at least that seems to be the intuitive answer), and for her to have two different credences for the same event isn't acceptable. I know you just said (and others have argued solely on this point) that a coin-flip event is not the same event as a flip-waking event, but I'm pretty sure those two events have the same probability. I want to endorse a statement like this as some kind of axiom:

A1: If you believe that if A then B, and you believe that if B then A, then your credence for A must be equal to your credence for B.

And I think that we do (and Beauty does) believe that if a heads occurs then a heads-awakening occurs and vice versa. So she should have equal credence in a heads as she does in a heads-awakening. Does anyone deny A1?

jason1990,

[ QUOTE ]
Now, here is another probability space which shows that posterior 1/3 and prior 1/2 are also consistent.

[/ QUOTE ]

I don't really follow how they could be considered consistent when they are not equal. Or are you implying that when they are consistent it is because she has learned some information between the prior credence and the posterior credence?

Your later post:

[ QUOTE ]
Now suppose SB bets $1 on heads at even money each time she is awake. When the coin is tails, she will make two consecutive $1 wagers. If you regard these as separate wagers, then over the course of 100 experiments, she will have made 150 wagers and won 50 of them. Her probability of winning is 1/3 and she loses money because she chose the wrong odds. On the other hand, if you regard these consecutive wagers as a single $2 wager, then over the course of 100 experiments, she will have made 100 wagers and won 50 of them. Her probability of winning is 1/2 and she loses money because the bet size is correlated to the outcome.

[/ QUOTE ]

This is a good way of looking at it. Do you think the Beauty that says she won 50 out of 100 bets at evens is being optimistically self-deluded? What’s really important is whether she is in profit and winning a fraction of bets proportional to the odds you bet on them does not necessarily entail that you break even. This seems like an argument for the credence of 1/3 since that credence is generating more epistemic profit than the ½ credence. This is an objection I develop to PairTheBoard’s One Wager example at the very end of my post.

PairTheBoard:

[ QUOTE ]
do you think this paradox is in some ways still an open question in logic and probability? I find that hard to believe.

[/ QUOTE ]

No I don't think the existence of the SB paradox suggests that our current probability theory or logic are inconsistent. Those are extremely small possibilities I suppose, but not really worth meditating upon. There are axioms, premises and theories higher up the ladder which don't enjoy as much intuitive power as those, which we ought to analyse first to see if they are self-consistent and consistent with the more established theories.

[ QUOTE ]
Now compare that to this scenario. Again SB will sleep for two weeks. If heads if flipped she will be awakened on Monday in Week 1. She will sleep through week 2. If tails is flipped she will sleep through week 1 and be awakened 7 times in week 2. Call this her probabilistic trip.

[/ QUOTE ]

I quite like your two examples of the probabilistic and determined week. I think we can refine them even further. Suppose the deterministic two weeks is how you describe it, and the probabilistic two weeks consist in the following: if a heads lands then she is woken once in the first week and if a tails lands she is woken 15 times in the second week. Her expected number of wakings should then be the same as in the deterministic fortnight, of 8 awakenings.
Adapting the argument from the deterministic week one could say that she has credence of 15/16 that it is the second week when she wakes up in the probabilistic fortnight. But by your argument for your own probabilistic week example, she could infer that there's 1/2 a chance this is the one awakening of a heads-week, and 1/2 a chance it is one of 15 awakenings of a tails-week. Having these two arguments is really just expressing what Elga and Lewis would say in this new example. The 7/8 and 15/16 thesis sounds plausible, as does the 1/2 thesis. I'm not really advancing anything here, just stating what we all already have established.
I think my point is that the ½ method is always a valid way to form credences, and sometimes a valid way to make betting odds. The 7/8 method is always a valid way of forming betting odds and sometimes a valid way of making credences. It is external factors that determine whether the ½ method is good for betting odds and the 7/8 method good for credences so it’s not like we have to postulate some kind of “magical” function for when they can one but not the other.
My problem is that the 7/8 method looks like a good method to form credences and I think that if we conclude that it really isn’t, this ought to affect how we make a lot of credences in real life. Let’s just look at what these methods are.

The methods are for giving us a P(it’s the first week)

Common knowledge:

If and only if H then it’s Week 1
If and only if T then it’s Week 2
P(H)=P(T)=1/2
If and only if Week 1 then there is 1 awakening
If and only if Week 2 then there are 7 awakenings

By transitivity if and only if H then there is 1 awakening
If and only if T then there are 7 awakenings
½ method:

By A1, P(Week1)=P(H)=1/2

7/8 method:
By A1 P(1 awakening)=P(H)=P(7 awakenings)=P(T)=1/2
Expected awakenings=(1/2x1)+(1/2x7)=4
P(this being Week 1 awakening over the expected awakenings)=P([1/2]/4)= 1/8

If we say that the 7/8 method is not to be used in calculating credences then what are we saying? That you cannot quantify over what is expected in probability theory? That sounds perverse, and a really big blow if we accept this conclusion. Can the 7/8 method be described in a more specific way?

[ QUOTE ]
Expressing a Belief for a longer period of time does not make it incorrect because it could make more bets in the longer time period.

[/ QUOTE ]

I’m not sure your version is preferable, all in all. You just have a system where she doesn’t lose money. She’s still wrong for a longer time than she is right. She will have that 50% credence that it is Week 1 for a whole week if tails. In the long run she can expect to be wrong most of the time. And being wrong for some time can be translated into a utility, so your Beauty, despite not losing money, is still losing some kind of utility from having credence 50% in heads after she wakes, which she doesn’t lose if she has credence 1/8 in heads.

[ QUOTE ]
L_0. There is a 50% probability that ALL OF MY AWAKENINGS are due to Heads. There is a 50% probability that ALL OF MY AWAKENINGS are due to Tails.

[/ QUOTE ]

This is good thinking. When I first read this I thought it was a convincing argument for 1/2. But 1) we already have convincing arguments for 1/2 and 2) it could be argued that Beauty can use L_O and end up being wrong most of the time. As far as I’m concerned the L_O argument has strengthened the argument for 1/2 if it needed to be strengthened, but then as a result it’s just strengthened the paradox too. Can the L_O argument argument be used to explain why advocates of 1/2 can be shown to support the hypothesis that a coin to be flipped in the future has a 2/3 chance of landing heads, because that seems to be an independent flaw in the ½ argument.

KipBond

06-12-2007, 09:10 PM

[ QUOTE ]
Kips example of 1000 Trials of the experiment producing 500 Heads-Monday awakenings, 500 Tails-Monday awakenings, and 500 Tails-Tuesday awakenings is not problem. How do you divide probabilities for SB's current awakening being among those? Easy. 50% that it's Heads-Monday. 25% that it's Tails-Monday. 25% that it's Tails-Tuesday.

[/ QUOTE ]

So, after many trials of this experiment, 75% of her wakings will be on Monday? That's not right. So, most of the rest of what you said regarding the wakings can't be right either.

2/3 of the wakings are on Monday. This is why I asked you to tell me if you disagreed with this a long time ago -- because if you do, then we should start there.

From there we have:

Of the 2/3 of the wakings that are on Monday, 1/2 of them are heads, half are tails.

1/3 of the wakings are Head-Monday wakings.

1/3 of the times she is woken will be preceded by a head coin-flip.

2/3 of the times she is woken will be preceded by a tail coin-flip.

KipBond

06-12-2007, 09:24 PM

[ QUOTE ]
And I think that we do (and Beauty does) believe that if a heads occurs then a heads-awakening occurs and vice versa. So she should have equal credence in a heads as she does in a heads-awakening. Does anyone deny A1?

[/ QUOTE ]

It just so happens in this scenario that there is a 1 to 1 correlation between heads coin flips & heads-awakenings. That is not the case for tails-awakenings, though.

And let's not confuse what a "heads-awakening" is. It is a waking that is preceded by a head coin flip. It is twice as likely that a randomly chosen awakening will be a tails-waking as opposed to a heads-waking. EVEN THOUGH: it is equally likely that a single coin flip will result in either 1 heads awakening, or 2 tails awakenings.

PairTheBoard

06-13-2007, 12:31 AM

[ QUOTE ]
[ QUOTE ]
Kips example of 1000 Trials of the experiment producing 500 Heads-Monday awakenings, 500 Tails-Monday awakenings, and 500 Tails-Tuesday awakenings is not problem. How do you divide probabilities for SB's current awakening being among those? Easy. 50% that it's Heads-Monday. 25% that it's Tails-Monday. 25% that it's Tails-Tuesday

[/ QUOTE ]

So, after many trials of this experiment, 75% of her wakings will be on Monday? That's not right. So, most of the rest of what you said regarding the wakings can't be right either.

[/ QUOTE ]

No. You are ignoring the last part of my post, which explains what it means.

The (50%,(25%,25%)) are not percents of the 1500. They are percents of the 1000 Trials. Beauty is in the midst of One Awakening. Whe wants to know probabilities for the categories (H-M,(T-M,T-T)) for where she is AT THIS MOMENT. She tests the (50%,(25%,25%)) by Entering each of 1000 trials of the experiment, randomly looking at ONE awakening, and observing where it is at. When she does so she sees (500,(250,250)). The reason she does't see (500,(500,500)) is because she only takes One Sample Awakening from each trial. Why should she take two samples for trials with Tails outcomes? She is testing the probability of This ONE awakening she is in.

For bigmonkey. This is what SB says about her betting situation. She say, there is a 50% chance I am in a Heads-Monday awakening. There is a 50% chance I am in one of the Two Tails Awakenings. If I bet on Heads now and I am consistent I will bet again when I have my other Tails awakening. So I need to structure my bet to account for being put in a betting position twice in 50% of the Trials where Heads is a losing bet. I recognize the outcome of the bet forces me to bet twice when I am losing. So I demand 2-1 odds to bet on Heads. That does not change the fact that there is a 50% probability I am now in one of the two Tails awakenings.

The only parardox is 1/3 people refusing to follow the logic.

PairTheBoard

jason1990

06-13-2007, 04:55 AM

[ QUOTE ]
jason1990,

[ QUOTE ]
Now, here is another probability space which shows that posterior 1/3 and prior 1/2 are also consistent.

[/ QUOTE ]

I don't really follow how they could be considered consistent when they are not equal. Or are you implying that when they are consistent it is because she has learned some information between the prior credence and the posterior credence?

[/ QUOTE ]
They are consistent because they can be built on the same probability space. (See http://en.wikipedia.org/wiki/Probability_space ) I gave the definition of consistency here:

[ QUOTE ]
The definition of "consistent" in Bayesian probability is formal and mathematical. A set of probabilities is consistent as long as it does not violate the axioms of probability. Equivalently, they are consistent if and only if they can be formulated in a common probability space. The question of consistency is a purely mathematical question.

[/ QUOTE ]
I do not know what your mathematical background is. I toyed with the idea of trying to water down my response. But there is no escaping the fact that consistency in Bayesian probability is defined in this way. It is technical and it is mathematical. If you want to fully understand a problem in Bayesian probability, then you must deal with this definition. I think my proof is clear and it shows that both answers are consistent.

[ QUOTE ]
Your later post:

[ QUOTE ]
Now suppose SB bets $1 on heads at even money each time she is awake. When the coin is tails, she will make two consecutive $1 wagers. If you regard these as separate wagers, then over the course of 100 experiments, she will have made 150 wagers and won 50 of them. Her probability of winning is 1/3 and she loses money because she chose the wrong odds. On the other hand, if you regard these consecutive wagers as a single $2 wager, then over the course of 100 experiments, she will have made 100 wagers and won 50 of them. Her probability of winning is 1/2 and she loses money because the bet size is correlated to the outcome.

[/ QUOTE ]

This is a good way of looking at it. Do you think the Beauty that says she won 50 out of 100 bets at evens is being optimistically self-deluded?

[/ QUOTE ]
I do not understand this. Why would I think SB is deluded? Let me try to re-explain what you quoted. Suppose SB bets $1 on heads with an n to 1 payout each time she is awake. (I have generalized it beyond even money.)

When the coin is tails, she will make two consecutive $1 wagers. We can either view these as two separate $1 wagers, or as a single $2 wager.

If we choose to view it as a single $2 wager, then she makes one wager per experiment. She wins $n if it is heads, and loses $2 if it is tails. Her payout is n to 2. In this perspective, the probability of heads is 1/2. Her EV at the end of the experiment is

(1/2)(n) + (1/2)(-2) = (n - 2)/2.

If we choose to view them as two separate $1 wagers, then she makes a wager with each awakening. In any particular awakening, she wins $n if it is heads, and loses $1 if it is tails. Her payout is n to 1. In this perspective, the probability of heads is 1/3. Her EV per awakening is

(1/3)(n) + (2/3)(-1) = (n - 2)/3.

There are 3/2 expected awakenings per experiment, so her EV at the end of the experiment is

((n - 2)/3)(3/2) = (n - 2)/2.

The two answers that are being so laboriously debated are saying exactly the same thing, but from two different perspectives. The theory of Bayesian probability shows us that they are both consistent. That is, the theory of Bayesian probability does not tell us to reject either one of them. As well it should not, since they are simply two different ways of expressing the same truth. If this "principle of reflection" does indeed say that we should reject one of these answers, then this principle is artificially restrictive and should be ignored. The general theory of Bayesian probability is evidently much more robust, since it can accommodate both of these perspectives.

Here is a quote I read today which seemed particularly relevant.

[ QUOTE ]
Truth is a river that is always splitting up into arms that reunite. Islanded between the arms, the inhabitants argue for a lifetime as to which is the main river.

[/ QUOTE ]

PairTheBoard

06-13-2007, 08:23 AM

[ QUOTE ]
It is twice as likely that a randomly chosen awakening will be a tails-waking as opposed to a heads-waking.

[/ QUOTE ]

The Awakening Beauty is in right now was not "randomly choosen" between the three types of awakenings. First a coin was flipped which divided the three types into two classes. One class had two types, the other class had one type. In the case where the Two-Type class was chosen, then an awakening was randomly chosen betweeen the two types. That's not "Randomly chosen" between the threee types. That's a process involving randomness where one type is more likely to be chosen than the other two.

PairTheBoard

PairTheBoard

06-13-2007, 08:38 AM

Consider this experiment. There are two pieces of paper. One paper is colored Blue. The other paper has two colors on it, Red and Green. The two papers are placed in a hat. Beauty picks one of them out of that hat at random. She looks at the color or colors on the paper. With the Red-Green paper she looks at both colors but at any one time she is only looking at one of the two colors.

Now, we do the experiment. Beauty picks a paper out of the hat at random. She is looking at a color on the paper. What is the probability she is looking at Blue?

The correct answer is 1/2.

Do you say 1/3? Do you say she is looking at one of the 3 colors "at random" so it must be 1/3? Do you point out that if the experiment is done 1000 times she will look at Blue 500 times, Red 500 times, and Green 500 times. So that's 1500 views of colors. 1/2(1500)=750 not 500 Blue views, so the answer can't be 1/2?

PairTheBoard

KipBond

06-13-2007, 09:24 AM

[ QUOTE ]
[ QUOTE ]
Truth is a river that is always splitting up into arms that reunite. Islanded between the arms, the inhabitants argue for a lifetime as to which is the main river.

[/ QUOTE ]

[/ QUOTE ]

Just for the record... I am arguing that both are valid answers, depending on how the question is asked.

1/2 of the coin flips will be heads.
1/3 of the awakenings will be preceded by a head coin flip.

Just the facts. I think some people may have too much information clouding their understanding.

So, I agree with you, Jason. This has been very laborious indeed.

KipBond

06-13-2007, 09:30 AM

[ QUOTE ]
She tests the (50%,(25%,25%)) by Entering each of 1000 trials of the experiment, randomly looking at ONE awakening, and observing where it is at.

[/ QUOTE ]

How do you expect her to look at ONE of the wakings in ONE trial randomly?

KipBond

06-13-2007, 09:33 AM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Kips example of 1000 Trials of the experiment producing 500 Heads-Monday awakenings, 500 Tails-Monday awakenings, and 500 Tails-Tuesday awakenings is not problem. How do you divide probabilities for SB's current awakening being among those? Easy. 50% that it's Heads-Monday. 25% that it's Tails-Monday. 25% that it's Tails-Tuesday

[/ QUOTE ]

So, after many trials of this experiment, 75% of her wakings will be on Monday? That's not right. So, most of the rest of what you said regarding the wakings can't be right either.

[/ QUOTE ]

No. You are ignoring the last part of my post, which explains what it means.

[/ QUOTE ]

So, what % of her wakings WILL be on Monday?

(I'm guessing there is a 1/3 chance that you'll answer this question this time.)

KipBond

06-13-2007, 09:42 AM

[ QUOTE ]
Consider this experiment. There are two pieces of paper. One paper is colored Blue. The other paper has two colors on it, Red and Green. The two papers are placed in a hat. Beauty picks one of them out of that hat at random. She looks at the color or colors on the paper. With the Red-Green paper she looks at both colors but at any one time she is only looking at one of the two colors.

Now, we do the experiment. Beauty picks a paper out of the hat at random. She is looking at a color on the paper. What is the probability she is looking at Blue?

The correct answer is 1/2.

Do you say 1/3? Do you say she is looking at one of the 3 colors "at random" so it must be 1/3? Do you point out that if the experiment is done 1000 times she will look at Blue 500 times, Red 500 times, and Green 500 times. So that's 1500 views of colors. 1/2(1500)=750 not 500 Blue views, so the answer can't be 1/2?

[/ QUOTE ]

In your experiment, after 1000 random paper-picks, here are the outcomes:

500 Blue-Looks
250 Red-Looks
250 Green-Looks

See, that's not the same as the SB experiment:

After 1000 random coin-flips, here are the outcomes:

500 Head-Wakings
1000 Tail-Wakings

Do you see how these two experiments are completely different?

PairTheBoard

06-13-2007, 09:42 AM

[ QUOTE ]
[ QUOTE ]
She tests the (50%,(25%,25%)) by Entering each of 1000 trials of the experiment, randomly looking at ONE awakening, and observing where it is at.

[/ QUOTE ]

How do you expect her to look at ONE of the wakings in ONE trial randomly?

[/ QUOTE ]

Easy. She tells us she is entering a Trail and wants an observataion for ONE of her awakenings. If she is within a trial where she is having more than one awakening we can pick one of them for her at random and record where she is at when she is having it. If she is within a trial where she only has one awakening then that's the ONE awakening for THAT trial that we observe.

PairTheBoard

PairTheBoard

06-13-2007, 09:54 AM

[ QUOTE ]
[ QUOTE ]
Consider this experiment. There are two pieces of paper. One paper is colored Blue. The other paper has two colors on it, Red and Green. The two papers are placed in a hat. Beauty picks one of them out of that hat at random. She looks at the color or colors on the paper. With the Red-Green paper she looks at both colors but at any one time she is only looking at one of the two colors.

Now, we do the experiment. Beauty picks a paper out of the hat at random. She is looking at a color on the paper. What is the probability she is looking at Blue?

The correct answer is 1/2.

Do you say 1/3? Do you say she is looking at one of the 3 colors "at random" so it must be 1/3? Do you point out that if the experiment is done 1000 times she will look at Blue 500 times, Red 500 times, and Green 500 times. So that's 1500 views of colors. 1/2(1500)=750 not 500 Blue views, so the answer can't be 1/2?

[/ QUOTE ]

In your experiment, after 1000 random paper-picks, here are the outcomes:

500 Blue-Looks
250 Red-Looks
250 Green-Looks

See, that's not the same as the SB experiment:

After 1000 random coin-flips, here are the outcomes:

500 Head-Wakings
1000 Tail-Wakings

Do you see how these two experiments are completely different?

[/ QUOTE ]

Did you miss the part in bold. When she picks the Red-Green paper she looks at BOTH colors. She views the BOTH. So in 1500 trials, Red gets viewed 500 times and Green gets viewed 500 times. She just doesn't view them simulateously. She views them one at a time. Just like with her Tails Awakenings. She views both awakenings. Just not simulateously.

The models are equivalent for the probability statement being made.

Beauty enters the Sleeping experiment. She picks the heads Sleep and tails Sleep at random. While in the Tails Sleep she views Two Awakenings. She just doesn't view them simultaneously.

Now we see Beauty in the Sleep experiment. She is viewing an Awakening for the Sleep she picked. What is the probability she is viewing a Heads awakening? 1/2

The models are identical.

PairTheBoard

PairTheBoard

06-13-2007, 09:59 AM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Kips example of 1000 Trials of the experiment producing 500 Heads-Monday awakenings, 500 Tails-Monday awakenings, and 500 Tails-Tuesday awakenings is not problem. How do you divide probabilities for SB's current awakening being among those? Easy. 50% that it's Heads-Monday. 25% that it's Tails-Monday. 25% that it's Tails-Tuesday

[/ QUOTE ]

So, after many trials of this experiment, 75% of her wakings will be on Monday? That's not right. So, most of the rest of what you said regarding the wakings can't be right either.

[/ QUOTE ]

No. You are ignoring the last part of my post, which explains what it means.

[/ QUOTE ]

So, what % of her wakings WILL be on Monday?

(I'm guessing there is a 1/3 chance that you'll answer this question this time.)

[/ QUOTE ]

In my 2 papers example, where she views BOTH colors of the Red-Green paper, just not simultaneously. In 1000 paper picks she views Blue 500 times, Red 500 times, and Green 500 times. What percent of the Color Viewings are Blue?

The probability of you understanding this is ....?

PairTheBoard

KipBond

06-13-2007, 10:35 AM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Kips example of 1000 Trials of the experiment producing 500 Heads-Monday awakenings, 500 Tails-Monday awakenings, and 500 Tails-Tuesday awakenings is not problem. How do you divide probabilities for SB's current awakening being among those? Easy. 50% that it's Heads-Monday. 25% that it's Tails-Monday. 25% that it's Tails-Tuesday

[/ QUOTE ]

So, after many trials of this experiment, 75% of her wakings will be on Monday? That's not right. So, most of the rest of what you said regarding the wakings can't be right either.

[/ QUOTE ]

No. You are ignoring the last part of my post, which explains what it means.

[/ QUOTE ]

So, what % of her wakings WILL be on Monday?

(I'm guessing there is a 1/3 chance that you'll answer this question this time.)

[/ QUOTE ]

In my 2 papers example, where she views BOTH colors of the Red-Green paper, just not simultaneously. In 1000 paper picks she views Blue 500 times, Red 500 times, and Green 500 times. What percent of the Color Viewings are Blue?

The probability of you understanding this is ....?

PairTheBoard

[/ QUOTE ]

Are you saying that 50% of her wakings will be on Monday?

Did you take lessons from Sklansky?

KipBond

06-13-2007, 10:38 AM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Consider this experiment. There are two pieces of paper. One paper is colored Blue. The other paper has two colors on it, Red and Green. The two papers are placed in a hat. Beauty picks one of them out of that hat at random. She looks at the color or colors on the paper. With the Red-Green paper she looks at both colors but at any one time she is only looking at one of the two colors.

Now, we do the experiment. Beauty picks a paper out of the hat at random. She is looking at a color on the paper. What is the probability she is looking at Blue?

The correct answer is 1/2.

Do you say 1/3? Do you say she is looking at one of the 3 colors "at random" so it must be 1/3? Do you point out that if the experiment is done 1000 times she will look at Blue 500 times, Red 500 times, and Green 500 times. So that's 1500 views of colors. 1/2(1500)=750 not 500 Blue views, so the answer can't be 1/2?

[/ QUOTE ]

In your experiment, after 1000 random paper-picks, here are the outcomes:

500 Blue-Looks
250 Red-Looks
250 Green-Looks

See, that's not the same as the SB experiment:

After 1000 random coin-flips, here are the outcomes:

500 Head-Wakings
1000 Tail-Wakings

Do you see how these two experiments are completely different?

[/ QUOTE ]

Did you miss the part in bold. When she picks the Red-Green paper she looks at BOTH colors. She views the BOTH. So in 1500 trials, Red gets viewed 500 times and Green gets viewed 500 times. She just doesn't view them simulateously. She views them one at a time. Just like with her Tails Awakenings. She views both awakenings. Just not simulateously.

The models are equivalent for the probability statement being made.

Beauty enters the Sleeping experiment. She picks the heads Sleep and tails Sleep at random. While in the Tails Sleep she views Two Awakenings. She just doesn't view them simultaneously.

Now we see Beauty in the Sleep experiment. She is viewing an Awakening for the Sleep she picked. What is the probability she is viewing a Heads awakening? 1/2

The models are identical.

[/ QUOTE ]

So, if she picks the Red-Green paper, she looks at one of the colors, and then looks at the other color? And you want to know the probability that she is looking at Blue?

1/3 of her "lookings" she will be looking at Blue.
1/2 of the "pickings" she will be looking at Blue.

KipBond

06-13-2007, 10:39 AM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
She tests the (50%,(25%,25%)) by Entering each of 1000 trials of the experiment, randomly looking at ONE awakening, and observing where it is at.

[/ QUOTE ]

How do you expect her to look at ONE of the wakings in ONE trial randomly?

[/ QUOTE ]

Easy. She tells us she is entering a Trail and wants an observataion for ONE of her awakenings. If she is within a trial where she is having more than one awakening we can pick one of them for her at random and record where she is at when she is having it. If she is within a trial where she only has one awakening then that's the ONE awakening for THAT trial that we observe.

[/ QUOTE ]

So, the result will be a probability for a different scenario:

We flip a coin, if it is heads we wake SB on Monday. If it is tails, we wake her on Monday 50% of the time, and Tuesday 50% of the time.

KipBond

06-13-2007, 11:01 AM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Kips example of 1000 Trials of the experiment producing 500 Heads-Monday awakenings, 500 Tails-Monday awakenings, and 500 Tails-Tuesday awakenings is not problem. How do you divide probabilities for SB's current awakening being among those? Easy. 50% that it's Heads-Monday. 25% that it's Tails-Monday. 25% that it's Tails-Tuesday

[/ QUOTE ]

So, after many trials of this experiment, 75% of her wakings will be on Monday? That's not right. So, most of the rest of what you said regarding the wakings can't be right either.

[/ QUOTE ]

No. You are ignoring the last part of my post, which explains what it means.

[/ QUOTE ]

So, what % of her wakings WILL be on Monday?

(I'm guessing there is a 1/3 chance that you'll answer this question this time.)

[/ QUOTE ]

In my 2 papers example, where she views BOTH colors of the Red-Green paper, just not simultaneously. In 1000 paper picks she views Blue 500 times, Red 500 times, and Green 500 times. What percent of the Color Viewings are Blue?

The probability of you understanding this is ....?

PairTheBoard

[/ QUOTE ]

Are you saying that 50% of her wakings will be on Monday?

Did you take lessons from Sklansky?

[/ QUOTE ]

And it wasn't a wild guess that you were saying 75% of her wakings are on Monday:

PariTheBoard:
[ QUOTE ]
P(Monday) = (.5)H + (.5)(.5)T = 75%
P(Tuesday) = (.5)(.5)T = 25%

So having awoken she knows there is a 75% chance it is Monday, and a 25% chance it is Tuesday.

[/ QUOTE ]

You've now recanted that (I think)? So...

What % of her wakings are on Monday?

f97tosc

06-13-2007, 11:28 AM

[ QUOTE ]

This is a great way of looking at it and what it makes clear is that the answer should be 1/2.

[/ QUOTE ]

If the probability is 1/2 for Heads-Monday, then it must be 1/4 for Tails-Monday and 1/4 for Tails-Tuesday (right?). Now suppose that after she wakes up, she is told that it is in fact not Tuesday. Then by Bayes' theorem she aught to update her probability for Heads-Monday to 0.5/(0.25+0.5)~=0.67. How is this possible, now that the problem is symmetric between heads and tails?

[ QUOTE ]

You don't change your answer for the probability because of the number of times you are asked the question. If you do you are changing the meaning of probability so that it gives the fequency per question asked rather than the frequency per Coin flip.

Suppose I flip a coin. You are asked what the probability is that it was heads. You are asked once if it is heads. You are asked 99 times if it is tails. Do you say the probability is 1% because the coin is heads 1% of the times the question is asked?

[/ QUOTE ]

Suppose that Sleeping Beauty will keep her memory throughout the experiment, and she knows this beforehand. Now when she is asked the first time, she learns nothing, and should say 1/2 for heads or tails. But if she is asked a second time, in the original problem or in your 1/99 problem, then she will learn with 100% certainty that the outcome was tails.

Thus, if the frequency of questions is caused by the outcome of the coin flip, then observing the number of questions can provide information on heads vs tails.

In the forgetful case, she will not know for sure whether there are multiple questions or not. It might be the first question/wake-up, in which case she would have learned nothing, or it could be the second time, in which case she would have learned that it was tails with certainty (if her memory would have worked). Taken together this constitutes probabilistic, but not conclusive, evidence in favor of tails. Another way to look at it is that waking up is probabilistic evidence in favor of multiple awakenings, and as such, for a tail outcome of the toss.

[ QUOTE ]

You are changing the meaning of probablity to one based on per question asked instead of per coin flip.

PairTheBoard

[/ QUOTE ]

I would say that the probability represents her information about the coin flip, and that multiple awakenings (or even one, if she can't tell one and multiple apart) constitutes relevant information.

KipBond

06-13-2007, 01:06 PM

[ QUOTE ]
Consider this experiment. There are two pieces of paper. One paper is colored Blue. The other paper has two colors on it, Red and Green. The two papers are placed in a hat. Beauty picks one of them out of that hat at random. She looks at the color or colors on the paper. With the Red-Green paper she looks at both colors(*) but at any one time she is only looking at one of the two colors.

Now, we do the experiment. Beauty picks a paper out of the hat at random. She is looking at a color on the paper. What is the probability she is looking at Blue?

The correct answer is 1/2.

Do you say 1/3? Do you say she is looking at one of the 3 colors "at random" so it must be 1/3? Do you point out that if the experiment is done 1000 times she will look at Blue 500 times, Red 500 times, and Green 500 times. So that's 1500 views of colors. 1/2(1500)=750 not 500 Blue views, so the answer can't be 1/2?

[/ QUOTE ]

(*) PTB later clarified (I think) that she will look at one color, then look at the other color; which makes this scenario identical to the SB problem.

Now, let's consider.

A photographer will take a picture of Beauty every time she looks at a color. We do this experiment 1000 times. So, 1500 photographs are taken of Beauty looking at a color. We put all 1500 photographs in a bucket and draw one at random. What is the probability that we will draw a picture of Beauty looking at Blue?

jason1990

06-13-2007, 01:25 PM

jason1990

06-13-2007, 02:05 PM

[ QUOTE ]
L_0. There is a 50% probability that ALL OF MY AWAKENINGS are due to Heads. There is a 50% probability that ALL OF MY AWAKENINGS are due to Tails.

[/ QUOTE ]
The event "All of my awakenings in this experiment are due to heads" is simply the event "the coin in this experiment landed heads." Both perspectives assert L_0 before the experiment. You assert that SB should take L_0 into the experiment with her. The 1/3 perspective asserts that she should not. So it seems you have become the monster you despise by simply "proclaiming your premise."

PairTheBoard

06-13-2007, 03:06 PM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Kips example of 1000 Trials of the experiment producing 500 Heads-Monday awakenings, 500 Tails-Monday awakenings, and 500 Tails-Tuesday awakenings is not problem. How do you divide probabilities for SB's current awakening being among those? Easy. 50% that it's Heads-Monday. 25% that it's Tails-Monday. 25% that it's Tails-Tuesday

[/ QUOTE ]

So, after many trials of this experiment, 75% of her wakings will be on Monday? That's not right. So, most of the rest of what you said regarding the wakings can't be right either.

[/ QUOTE ]

No. You are ignoring the last part of my post, which explains what it means.

[/ QUOTE ]

So, what % of her wakings WILL be on Monday?

(I'm guessing there is a 1/3 chance that you'll answer this question this time.)

[/ QUOTE ]

In my 2 papers example, where she views BOTH colors of the Red-Green paper, just not simultaneously. In 1000 paper picks she views Blue 500 times, Red 500 times, and Green 500 times. What percent of the Color Viewings are Blue?

The probability of you understanding this is ....?

PairTheBoard

[/ QUOTE ]

Are you saying that 50% of her wakings will be on Monday?

Did you take lessons from Sklansky?

[/ QUOTE ]

Are you saying that 50% of her 3 Colors are Blue? Why don't you answer my question?

I'm saying that 50% of her Sleeps will have nothing but Monday awakenings.

PairTheBoard

KipBond

06-13-2007, 03:09 PM

[ QUOTE ]
P(heads) = 1/2,
P(she is told it is not Tuesday | heads) = 1, and
P(she is told it is not Tuesday | tails) = 1/2.

Under these assumptions, with A = "she is told it is not Tuesday," Bayes theorem gives

P(heads|A) = P(heads)P(A|heads)/[P(heads)P(A|heads) + P(tails)P(A|tails)]
= (1/2)(1)/[(1/2)(1) + (1/2)(1/2) = 2/3.

[/ QUOTE ]

You are answering the questions I asked PTB (I think).

Here's what I asked:

[ QUOTE ]
P(H|monday) = 50%
P(H|tuesday) = 0%
P(T|monday) = 50%
P(T|tuesday) = 100%

P(monday|H) = 100%
P(tuesday|H) = 0%
P(monday|T) = 50%
P(tuesday|T) = 50%

If any of those are not right, I'd like do know why. It seems intuitive to me.

Now:

P(monday|waking) = 2/3
P(tuesday|waking) = 1/3

[/ QUOTE ]

I think you just said this:

P(H|monday) = 2/3

Which would mean:

P(T|monday) = 1/3

This seems wrong to me. Did I misunderstand you? I don't understand the difference between "told that it is Monday" and "it is Monday", if 100% of the time it is Monday you are told it is Monday, and 100% of the time you are told it is Monday, it is indeed Monday.

KipBond

06-13-2007, 03:18 PM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Kips example of 1000 Trials of the experiment producing 500 Heads-Monday awakenings, 500 Tails-Monday awakenings, and 500 Tails-Tuesday awakenings is not problem. How do you divide probabilities for SB's current awakening being among those? Easy. 50% that it's Heads-Monday. 25% that it's Tails-Monday. 25% that it's Tails-Tuesday

[/ QUOTE ]

So, after many trials of this experiment, 75% of her wakings will be on Monday? That's not right. So, most of the rest of what you said regarding the wakings can't be right either.

[/ QUOTE ]

No. You are ignoring the last part of my post, which explains what it means.

[/ QUOTE ]

So, what % of her wakings WILL be on Monday?

(I'm guessing there is a 1/3 chance that you'll answer this question this time.)

[/ QUOTE ]

In my 2 papers example, where she views BOTH colors of the Red-Green paper, just not simultaneously. In 1000 paper picks she views Blue 500 times, Red 500 times, and Green 500 times. What percent of the Color Viewings are Blue?

The probability of you understanding this is ....?

PairTheBoard

[/ QUOTE ]

Are you saying that 50% of her wakings will be on Monday?

Did you take lessons from Sklansky?

[/ QUOTE ]

Are you saying that 50% of her 3 Colors are Blue? Why don't you answer my question?

[/ QUOTE ]

I have not dodged any of your questions. You have dodged mine.

No, 1/3 of her Colors are Blue.

And, as previously answered, 1/3 of the Color Viewings are Blue.

[ QUOTE ]
I'm saying that 50% of her Sleeps will have nothing but Monday awakenings.

[/ QUOTE ]

Yeah, that's wrong.

jason1990

06-13-2007, 04:01 PM

[ QUOTE ]
This seems wrong to me. Did I misunderstand you? I don't understand the difference between "told that it is Monday" and "it is Monday", if 100% of the time it is Monday you are told it is Monday, and 100% of the time you are told it is Monday, it is indeed Monday.

[/ QUOTE ]
Recall Monty Hall. You pick a door. Probability of picking correctly: 1/3. You now learn Door C is incorrect. What is your new probability? It depends on how you learned this piece of information. Unless you make assumptions on Monty's behavior, you cannot answer the question. The information alone is insufficient. The manner in which you learned it plays a role.

For SB, she learns it is Monday. Is this just some abstract piece of knowledge she happens to have? Or did the coin toss have something to do with the manner in which she learned it? What assumptions will you choose to impose on this situation? It depends entirely on your perspective. Both perspectives are logically consistent. Both perspectives lead to the same EV, measured under different time scales.

Incidentally, I would rather not continue this line of discussion unless it is placed in the context of one or more probability spaces. I think I am the only one in this thread that has posted a probability space. In fact, I posted several, in favor of both positions. I am not the only mathematician here, yet everyone seems content to discuss this in the hand-waving fashion of the philosophers. One of the most important things we can learn from all of these probability "paradoxes" (Monty Hall, Two Envelope, etc.) is that imprecise language and informal reasoning can lead to absurdities.

PairTheBoard

06-13-2007, 04:06 PM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Consider this experiment. There are two pieces of paper. One paper is colored Blue. The other paper has two colors on it, Red and Green. The two papers are placed in a hat. Beauty picks one of them out of that hat at random. She looks at the color or colors on the paper. With the Red-Green paper she looks at both colors but at any one time she is only looking at one of the two colors.

Now, we do the experiment. Beauty picks a paper out of the hat at random. She is looking at a color on the paper. What is the probability she is looking at Blue?

The correct answer is 1/2.

Do you say 1/3? Do you say she is looking at one of the 3 colors "at random" so it must be 1/3? Do you point out that if the experiment is done 1000 times she will look at Blue 500 times, Red 500 times, and Green 500 times. So that's 1500 views of colors. 1/2(1500)=750 not 500 Blue views, so the answer can't be 1/2?

[/ QUOTE ]

In your experiment, after 1000 random paper-picks, here are the outcomes:

500 Blue-Looks
250 Red-Looks
250 Green-Looks

See, that's not the same as the SB experiment:

After 1000 random coin-flips, here are the outcomes:

500 Head-Wakings
1000 Tail-Wakings

Do you see how these two experiments are completely different?

[/ QUOTE ]

Did you miss the part in bold. When she picks the Red-Green paper she looks at BOTH colors. She views the BOTH. So in 1500 trials, Red gets viewed 500 times and Green gets viewed 500 times. She just doesn't view them simulateously. She views them one at a time. Just like with her Tails Awakenings. She views both awakenings. Just not simulateously.

The models are equivalent for the probability statement being made.

Beauty enters the Sleeping experiment. She picks the heads Sleep and tails Sleep at random. While in the Tails Sleep she views Two Awakenings. She just doesn't view them simultaneously.

Now we see Beauty in the Sleep experiment. She is viewing an Awakening for the Sleep she picked. What is the probability she is viewing a Heads awakening? 1/2

The models are identical.

[/ QUOTE ]

So, if she picks the Red-Green paper, she looks at one of the colors, and then looks at the other color? And you want to know the probability that she is looking at Blue?

1/3 of her "lookings" she will be looking at Blue.
1/2 of the "pickings" she will be looking at Blue.

[/ QUOTE ]

You are evading the question. She is behind a curtain. You are informed that she has picked a paper out of the hat and is looking at a color on the paper. What is the probability she is looking at Blue?

Equvilantley. Beauty enters a Heads-Tails Sleep at random. She is experiencing an awakening in that Sleep. What is the probability it is a Heads-Monday awakening?

PairTheBoard

PairTheBoard

06-13-2007, 04:17 PM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
She tests the (50%,(25%,25%)) by Entering each of 1000 trials of the experiment, randomly looking at ONE awakening, and observing where it is at.

[/ QUOTE ]

How do you expect her to look at ONE of the wakings in ONE trial randomly?

[/ QUOTE ]

Easy. She tells us she is entering a Trail and wants an observataion for ONE of her awakenings. If she is within a trial where she is having more than one awakening we can pick one of them for her at random and record where she is at when she is having it. If she is within a trial where she only has one awakening then that's the ONE awakening for THAT trial that we observe.

[/ QUOTE ]

So, the result will be a probability for a different scenario:

We flip a coin, if it is heads we wake SB on Monday. If it is tails, we wake her on Monday 50% of the time, and Tuesday 50% of the time.

[/ QUOTE ]

What I am describing is the way she would test to see if her probability of 50% that This Awakening in This Sleep is a Heads-Monday. She enters 1000 Sleeps identically to how she entered this one and has an observation made of One Awakening in Each Sleep Experience. She then looks at the statistics for this sampling. She takes One Sample Awakening from Each of 1000 Sleeps.

PairTheBoard

jason1990

06-13-2007, 04:20 PM

[ QUOTE ]
[ QUOTE ]
This seems wrong to me. Did I misunderstand you? I don't understand the difference between "told that it is Monday" and "it is Monday", if 100% of the time it is Monday you are told it is Monday, and 100% of the time you are told it is Monday, it is indeed Monday.

[/ QUOTE ]
Recall Monty Hall. You pick a door. Probability of picking correctly: 1/3. You now learn Door C is incorrect. What is your new probability? It depends on how you learned this piece of information. Unless you make assumptions on Monty's behavior, you cannot answer the question. The information alone is insufficient. The manner in which you learned it plays a role.

For SB, she learns it is Monday. Is this just some abstract piece of knowledge she happens to have? Or did the coin toss have something to do with the manner in which she learned it? What assumptions will you choose to impose on this situation? It depends entirely on your perspective. Both perspectives are logically consistent. Both perspectives lead to the same EV, measured under different time scales.

Incidentally, I would rather not continue this line of discussion unless it is placed in the context of one or more probability spaces. I think I am the only one in this thread that has posted a probability space. In fact, I posted several, in favor of both positions. I am not the only mathematician here, yet everyone seems content to discuss this in the hand-waving fashion of the philosophers. One of the most important things we can learn from all of these probability "paradoxes" (Monty Hall, Two Envelope, etc.) is that imprecise language and informal reasoning can lead to absurdities.

[/ QUOTE ]
I hate the fact that I am getting drawn into all this hand-waving, but I feel I should give you one more informal description of this alternative perspective. But I will not defend it. Take it as it is and chew on it.

SB wakes up. The experimenter says, "I am about to tell you what day it is..." Before he can finish, she pauses to do a bit of reasoning. She thinks there is a 50% chance the coin is heads, and in this case, he is about to say Monday. She thinks there is a 50% chance the coin is tails, and in this case, he will say either Monday or Tuesday with equal likelihood. Overall, she reasons that there is a 75% chance he is about to say Monday. He then says, "Today is Monday." She now applies Bayes theorem and concludes the probability of heads is 2/3.

PairTheBoard

06-13-2007, 04:28 PM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Kips example of 1000 Trials of the experiment producing 500 Heads-Monday awakenings, 500 Tails-Monday awakenings, and 500 Tails-Tuesday awakenings is not problem. How do you divide probabilities for SB's current awakening being among those? Easy. 50% that it's Heads-Monday. 25% that it's Tails-Monday. 25% that it's Tails-Tuesday

[/ QUOTE ]

So, after many trials of this experiment, 75% of her wakings will be on Monday? That's not right. So, most of the rest of what you said regarding the wakings can't be right either.

[/ QUOTE ]

No. You are ignoring the last part of my post, which explains what it means.

[/ QUOTE ]

So, what % of her wakings WILL be on Monday?

(I'm guessing there is a 1/3 chance that you'll answer this question this time.)

[/ QUOTE ]

In my 2 papers example, where she views BOTH colors of the Red-Green paper, just not simultaneously. In 1000 paper picks she views Blue 500 times, Red 500 times, and Green 500 times. What percent of the Color Viewings are Blue?

The probability of you understanding this is ....?

PairTheBoard

[/ QUOTE ]

Are you saying that 50% of her wakings will be on Monday?

Did you take lessons from Sklansky?

[/ QUOTE ]

And it wasn't a wild guess that you were saying 75% of her wakings are on Monday:

PariTheBoard:
[ QUOTE ]
P(Monday) = (.5)H + (.5)(.5)T = 75%
P(Tuesday) = (.5)(.5)T = 25%

So having awoken she knows there is a 75% chance it is Monday, and a 25% chance it is Tuesday.

[/ QUOTE ]

You've now recanted that (I think)? So...

What % of her wakings are on Monday?

[/ QUOTE ]

Are you now deliberately misunderstanding me so as to avoid looking at the logic. Beauty enters a Random Sleep. She experiences an awakening in that Random Sleep. There is a 50% probabability it is a Heads-Monday awakening (Blue). A 25% that it's a Tails-Monday awakening (Green). And a 25% chance it is a Tails-Tuesday awakening (Red). There is a 75% chance it is a Monday Awakening (Blue or Green).

If you are going to try to misunderstand just so you can continue arguing I am going to quit the discussion with you.

PairTheBoard

KipBond

06-13-2007, 04:28 PM

[ QUOTE ]
[ QUOTE ]
So, if she picks the Red-Green paper, she looks at one of the colors, and then looks at the other color? And you want to know the probability that she is looking at Blue?

1/3 of her "lookings" she will be looking at Blue.
1/2 of the "pickings" she will be looking at Blue.

[/ QUOTE ]

You are evading the question. She is behind a curtain. You are informed that she has picked a paper out of the hat and is looking at a color on the paper. What is the probability she is looking at Blue?

[/ QUOTE ]

I'm not evading the question. I guarantee you, 100%, that I will do my best to answer any question that you pose to me as forthright as possible. It doesn't seem like you are doing the same. Anyway, here's my answer:

1/2 or 1/3 depending on how I interpret the situation.

I THINK this is what Jason means when he talks about using imprecise "hand waving" arguments, rather than mathematical probability spaces. I'm, obviously, not educated in how to do that formally, but I think I'm trying to say the same thing.

If the situation is that for every PICK of the paper, I am asked ONE TIME what the probability is that she's looking at Blue, then the answer is 1/2. That means, on average, that half of the times I am asked the question, she will be looking at Blue.

On the other hand, if I am asked the question for every LOOKING, then the answer is 1/3. That means, on average, that 1/3 of the times I am asked the question, she will be looking at Blue.

If I don't know the situation in which I'm being asked the question, then (like the Monty Hall problem) I have to make certain assumptions -- and specify those assumptions.

[ QUOTE ]
Equvilantley. Beauty enters a Heads-Tails Sleep at random. She is experiencing an awakening in that Sleep. What is the probability it is a Heads-Monday awakening?

[/ QUOTE ]

Please explain what you mean by "enters a Heads-Tails Sleep at random" if you still want me to answer this question.

jason1990

06-13-2007, 04:32 PM

[ QUOTE ]
SB wakes up. The experimenter says, "I am about to tell you what day it is..." Before he can finish, she pauses to do a bit of reasoning. She thinks there is a 50% chance the coin is heads, and in this case, he is about to say Monday. She thinks there is a 50% chance the coin is tails, and in this case, he will say either Monday or Tuesday with equal likelihood. Overall, she reasons that there is a 75% chance he is about to say Monday. He then says, "Today is Monday." She now applies Bayes theorem and concludes the probability of heads is 2/3.

[/ QUOTE ]
Incidentally, check out the bold part above. It is an application of the indifference principle. So you can see that both sides of this debate are using the indifference principle. They just use it in two different places.

KipBond

06-13-2007, 04:50 PM

[ QUOTE ]
I hate the fact that I am getting drawn into all this hand-waving, but I feel I should give you one more informal description of this alternative perspective. But I will not defend it. Take it as it is and chew on it.

[/ QUOTE ]

I'm going to go back & re-read all of your posts, Jason, to see if I can understand the formal probability spaces you've explained. I'll probably have to ask some stupid questions about them, so please bear with me.

[ QUOTE ]
SB wakes up. The experimenter says, "I am about to tell you what day it is..." Before he can finish, she pauses to do a bit of reasoning. She thinks there is a 50% chance the coin is heads, and in this case, he is about to say Monday. She thinks there is a 50% chance the coin is tails, and in this case, he will say either Monday or Tuesday with equal likelihood. Overall, she reasons that there is a 75% chance he is about to say Monday. He then says, "Today is Monday." She now applies Bayes theorem and concludes the probability of heads is 2/3.

[/ QUOTE ]

My first thought on this is: "so, shouldn't Beauty now realize that she's wrong?"

It seems intuitively obvious to me that given Beauty's understanding of the scenario, when she wakes up, she should realize there is a 2/3 chance that it is Monday. Not 3/4. Which, of course, means that there can't be a "50% chance the coin is heads" (prior to knowing it is Monday). Her calculation that given that today is Monday, the P(Heads)=2/3 should be further evidence that something is wrong.

This is more apparent if you revise the problem to only flip a coin after her initial Monday waking (since she is going to be woken up every Monday anyway). Note, however, that when you do this, the question to Beauty has to then be: "What is the probability that the coin I will flip, or have already flipped, will be (or is already) Heads?"

In this case, when you tell Beauty that today is Monday, she knows the coin hasn't even been flipped yet. So, now the P(H) HAS to be 1/2. Right?

jason1990

06-13-2007, 05:16 PM

[ QUOTE ]
Her calculation that given that today is Monday, the P(Heads)=2/3 should be further evidence that something is wrong.

[/ QUOTE ]
Why? Because knowing it is Monday gives her no information regarding the coin flip? And no information implies the probability should remain unchanged and still be 1/2? Well, this is the exact same argument originally used to justify the other side of the debate. She wakes up with amnesia and has no information other than what she originally had on Sunday, so the probability should remain 1/2. You cannot have your cake and eat it too. You cannot reject this argument initially and then use it later to say that P(heads|Monday) = 2/3 is illogical.

PairTheBoard

06-13-2007, 05:30 PM

[ QUOTE ]
If the probability is 1/2 for Heads-Monday, then it must be 1/4 for Tails-Monday and 1/4 for Tails-Tuesday (right?). Now suppose that after she wakes up, she is told that it is in fact not Tuesday. Then by Bayes' theorem she aught to update her probability for Heads-Monday to 0.5/(0.25+0.5)~=0.67. How is this possible, now that the problem is symmetric between heads and tails?

[/ QUOTE ]

It depends on how she is told doesn't it? If in every sleep, when she is having a Monday Awakening she it told it is monday then she can concude 50% heads. But does she know she is being told this every Monday Awakening? It's like the Monty Hall problem isn't it. Does the contestent Know that Monty Always eliminates one of the Goat Doors? If so, she can apply Bayes. But if she doesn't know this she can't apply Bayes. Monty may just be messin with her this time. Maybe Monty hardly ever eliminates the Goat Door but he's doing it this time just to be mischievous. With Beauty it's the same principle. Unless she knows How you are telling her it is Monday she can't apply Bayes. And why are you telling her? She doesn't get told in the actual experiment.

Suppose in my Two Papers example, Beauty is behind the curtain and calls out, I'm not looking at Green right now. What can we conclude about the probablities she is looking at Blue or Red? We can't conclude anything unless we know why she has decided this particular time to break the rules and volunteer this information. She is suppose to wait until we ask her what color she is viewing at the moment.

[ QUOTE ]
[ QUOTE ]
You don't change your answer for the probability because of the number of times you are asked the question. If you do you are changing the meaning of probability so that it gives the fequency per question asked rather than the frequency per Coin flip.

Suppose I flip a coin. You are asked what the probability is that it was heads. You are asked once if it is heads. You are asked 99 times if it is tails. Do you say the probability is 1% because the coin is heads 1% of the times the question is asked?

[/ QUOTE ]
Suppose that Sleeping Beauty will keep her memory throughout the experiment, and she knows this beforehand. Now when she is asked the first time, she learns nothing, and should say 1/2 for heads or tails. But if she is asked a second time, in the original problem or in your 1/99 problem, then she will learn with 100% certainty that the outcome was tails.

Thus, if the frequency of questions is caused by the outcome of the coin flip, then observing the number of questions can provide information on heads vs tails.

In the forgetful case, she will not know for sure whether there are multiple questions or not. It might be the first question/wake-up, in which case she would have learned nothing, or it could be the second time, in which case she would have learned that it was tails with certainty (if her memory would have worked). Taken together this constitutes probabilistic, but not conclusive, evidence in favor of tails. Another way to look at it is that waking up is probabilistic evidence in favor of multiple awakenings, and as such, for a tail outcome of the toss.

[/ QUOTE ]

You are not considering my discussion of a Deterministic Two Week Sleep in which she is awakened only on Monday the first Week but all 7 days the Second Week.

In that case, the number of awakenings does give her information about which week This Awakening is likely to be in. But when you change that experiment to a Probabilistic one, then either All of her Awakenings are in the First week or All of her Awakenings are in the second week. She knows she will have at least one regardless. So when she has one she can only conclude it is either one of the 7 in the second week (which will happen in 50% of her sleeps), or it is the one and only one in the first week (which will happen in 50% of her sleeps).

If Tails, she will say 7 times, "There is a 50% probability Tails flipped and this is one of my 7 awakenings due to tails". And 7 times she will be correct.

PairTheBoard

PairTheBoard

06-13-2007, 05:46 PM

[ QUOTE ]
[ QUOTE ]
Consider this experiment. There are two pieces of paper. One paper is colored Blue. The other paper has two colors on it, Red and Green. The two papers are placed in a hat. Beauty picks one of them out of that hat at random. She looks at the color or colors on the paper. With the Red-Green paper she looks at both colors(*) but at any one time she is only looking at one of the two colors.

Now, we do the experiment. Beauty picks a paper out of the hat at random. She is looking at a color on the paper. What is the probability she is looking at Blue?

The correct answer is 1/2.

Do you say 1/3? Do you say she is looking at one of the 3 colors "at random" so it must be 1/3? Do you point out that if the experiment is done 1000 times she will look at Blue 500 times, Red 500 times, and Green 500 times. So that's 1500 views of colors. 1/2(1500)=750 not 500 Blue views, so the answer can't be 1/2?

[/ QUOTE ]

(*) PTB later clarified (I think) that she will look at one color, then look at the other color; which makes this scenario identical to the SB problem.

Now, let's consider.

A photographer will take a picture of Beauty every time she looks at a color. We do this experiment 1000 times. So, 1500 photographs are taken of Beauty looking at a color. We put all 1500 photographs in a bucket and draw one at random. What is the probability that we will draw a picture of Beauty looking at Blue?

[/ QUOTE ]

That would be a different experiment. And it would not match what Beauty is doing in her sleep. Beauty does not pick an awakening out of a hat containing the 3 Heads-Monday, Tails-Mondy, and Tails-Tuesday awakenings. She First enters a Type of Sleep decided by Heads or Tails. In that Sleep either all of her awakenings will have been determined by heads or they will all have been determined by Tails.

PairTheBoard

PairTheBoard

06-13-2007, 05:54 PM

[ QUOTE ]
[ QUOTE ]
L_0. There is a 50% probability that ALL OF MY AWAKENINGS are due to Heads. There is a 50% probability that ALL OF MY AWAKENINGS are due to Tails.

[/ QUOTE ]
The event "All of my awakenings in this experiment are due to heads" is simply the event "the coin in this experiment landed heads." Both perspectives assert L_0 before the experiment. You assert that SB should take L_0 into the experiment with her. The 1/3 perspective asserts that she should not. So it seems you have become the monster you despise by simply "proclaiming your premise."

[/ QUOTE ]

What assumption do the 1/3 people have her carry into the Sleep. Does she assume that some of her Awakenings in THIS SLEEP will be due to Heads and some will be due to Tails? Does she think that in THIS SLEEP she might have both a Heads-Monday awakening and a Tails-Tuesday awakening? I don't think I'm being a monster when I follow the problem's premise that she will be logical. That's the problem's premise, not mine.

PairTheBoard

KipBond

06-13-2007, 05:57 PM

Kip:
[ QUOTE ]
So, after many trials of this experiment, 75% of her wakings will be on Monday?

[/ QUOTE ]

PTB:
[ QUOTE ]
No. You are ignoring the last part of my post, which explains what it means.

[/ QUOTE ]

Kip:
[ QUOTE ]
So, what % of her wakings WILL be on Monday?

[/ QUOTE ]

PTB:
[ QUOTE ]
There is a 75% chance it is a Monday Awakening.

[/ QUOTE ]

PTB:
[ QUOTE ]
If you are going to try to misunderstand just so you can continue arguing I am going to quit the discussion with you.

[/ QUOTE ]

See how I could misunderstand you?

Your "No" answer to my question threw me off. I assume you meant "Yes"? I'm really trying to understand your position. Direct answers to questions are usually the best way to avoid misunderstandings.

So, I will now assume that your answer to the question "What % of her wakings will be on Monday" is 75%. I think that is wrong, though.

PairTheBoard

06-13-2007, 06:11 PM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
So, if she picks the Red-Green paper, she looks at one of the colors, and then looks at the other color? And you want to know the probability that she is looking at Blue?

1/3 of her "lookings" she will be looking at Blue.
1/2 of the "pickings" she will be looking at Blue.

[/ QUOTE ]

You are evading the question. She is behind a curtain. You are informed that she has picked a paper out of the hat and is looking at a color on the paper. What is the probability she is looking at Blue?

[/ QUOTE ]

I'm not evading the question. I guarantee you, 100%, that I will do my best to answer any question that you pose to me as forthright as possible. It doesn't seem like you are doing the same. Anyway, here's my answer:

1/2 or 1/3 depending on how I interpret the situation.

I THINK this is what Jason means when he talks about using imprecise "hand waving" arguments, rather than mathematical probability spaces. I'm, obviously, not educated in how to do that formally, but I think I'm trying to say the same thing.

If the situation is that for every PICK of the paper, I am asked ONE TIME what the probability is that she's looking at Blue, then the answer is 1/2. That means, on average, that half of the times I am asked the question, she will be looking at Blue.

On the other hand, if I am asked the question for every LOOKING, then the answer is 1/3. That means, on average, that 1/3 of the times I am asked the question, she will be looking at Blue.

If I don't know the situation in which I'm being asked the question, then (like the Monty Hall problem) I have to make certain assumptions -- and specify those assumptions.

[ QUOTE ]
Equvilantley. Beauty enters a Heads-Tails Sleep at random. She is experiencing an awakening in that Sleep. What is the probability it is a Heads-Monday awakening?

[/ QUOTE ]

Please explain what you mean by "enters a Heads-Tails Sleep at random" if you still want me to answer this question.

[/ QUOTE ]

There are Two Types of Sleeps she can enter. One is called a Heads-Type where she will be awakened on Monday only. The other is called a Tails-Type where she will be awakened on Monday and Tuesday. She wants to enter one of these two Types of sleep randomly. So she has someone toss a coin and not tell her how it landed. She then enters That Sleep Type defined by the outcome of the coin toss.

Not only does that define it for you but it also accurately describes the Problem's Sleep that Beauty enters. And she knows it. She can repeat that process 1000 times and have One Sample of an awakening taken from each of those 1000 Trials. The statistics for those 1000 samples match the probability she gives when she considers "This Awakening" in the Problem.

Now can you answer the question?

PairTheBoard

jason1990

06-13-2007, 06:26 PM

It is hard to parse all this informal language being thrown around by both sides. My interpretation of your L_0 was that it amounted to simply declaring that 1/2 is the right answer. If you meant something else, it was not clear to me. This is why I would prefer to talk about this in the formal language of probability theory.

[ QUOTE ]
I don't think I'm being a monster when I follow the problem's premise that she will be logical. That's the problem's premise, not mine.

[/ QUOTE ]
Yes, the problem says she will be logical. It also says she thinks the probability of heads on Sunday is 1/2. So what is your definition of "logical" when it comes to probability statements? If I build a probability space and it is valid and it says the probability of heads on Sunday is 1/2, then is that sufficient to say that I am adhering to the premises of the problem? Or do you have some additional criteria that you will apply in order to judge the logic of my space? I earlier posted a probability space that encodes the 1/3 argument. It is clearly a valid probability space and it says that SB's probability of heads is 1/2 on Sunday. Is there something about this probability space that you find illogical? If so, then what measure would you put on this particular sample space?

oe39

06-13-2007, 06:29 PM

flip a coin.

if heads, mark H and flip again. if tails, mark T, check again. if still tails, mark T and flip the coin again.

repeat.

------

the result of the coin flips will be 50-50, but 2/3 of the "measurements" will be T. nothing confusing about that.

how is this a different situation?

PairTheBoard

06-13-2007, 06:33 PM

[ QUOTE ]
Kip:
[ QUOTE ]
So, after many trials of this experiment, 75% of her wakings will be on Monday?

[/ QUOTE ]

PTB:
[ QUOTE ]
No. You are ignoring the last part of my post, which explains what it means.

[/ QUOTE ]

Kip:
[ QUOTE ]
So, what % of her wakings WILL be on Monday?

[/ QUOTE ]

PTB:
[ QUOTE ]
There is a 75% chance it is a Monday Awakening.

[/ QUOTE ]

PTB:
[ QUOTE ]
If you are going to try to misunderstand just so you can continue arguing I am going to quit the discussion with you.

[/ QUOTE ]

See how I could misunderstand you?

Your "No" answer to my question threw me off. I assume you meant "Yes"? I'm really trying to understand your position. Direct answers to questions are usually the best way to avoid misunderstandings.

So, I will now assume that your answer to the question "What % of her wakings will be on Monday" is 75%. I think that is wrong, though.

[/ QUOTE ]

No. Let me be more clear. 75% of her Sample-Awakenings from the 1000 Trials, where she takes One Sample-Awakening from each of the 1000 Trials, will be Monday Awakenings. That corresponds to the probability she gives for "This Awakening" in the problem.

Clear now? 75% of her Sample-Awakenings.

PairTheBoard

PairTheBoard

06-13-2007, 06:56 PM

[ QUOTE ]
It is hard to parse all this informal language being thrown around by both sides. My interpretation of your L_0 was that it amounted to simply declaring that 1/2 is the right answer. If you meant something else, it was not clear to me. This is why I would prefer to talk about this in the formal language of probability theory.

[ QUOTE ]
I don't think I'm being a monster when I follow the problem's premise that she will be logical. That's the problem's premise, not mine.

[/ QUOTE ]
Yes, the problem says she will be logical. It also says she thinks the probability of heads on Sunday is 1/2. So what is your definition of "logical" when it comes to probability statements? If I build a probability space and it is valid and it says the probability of heads on Sunday is 1/2, then is that sufficient to say that I am adhering to the premises of the problem? Or do you have some additional criteria that you will apply in order to judge the logic of my space? I earlier posted a probability space that encodes the 1/3 argument. It is clearly a valid probability space and it says that SB's probability of heads is 1/2 on Sunday. Is there something about this probability space that you find illogical? If so, then what measure would you put on this particular sample space?

[/ QUOTE ]

This is my description of the model. I'm sure you can convert it to a mathematical one. Beauty enters one of Two Types of Sleep. One is a Heads-Type sleep where she will be awakened only on N days of the week. One is a Tails-Type sleep where she will be awakened on M days of the week. N and M are both nonzero.

L_0 simply asserts the obvious. If heads lands, all N of Beauty's awakings will be within a Heads-Type sleep. If tails lands, all M of Beauty's awakenings will be within a Tails-Type Sleep.

In the Problem N=1 and M=2. But I also looked at the case where N=1 and M=7. In that case I noted the difference between it and a Deterministic Model where Beauty sleeps for 2 weeks and is awakened once in the first week and 7 times in the second week. The 1/3 people are using an indifference principle based on that Deterministic model rather than the Probablistic one.

Also, When Beauty enters one of the two random Sleep Types and considers her position while in "This Awakening" she doesn't have to apply the indifference principle to say that the Probability she is in Sleep-Type Heads is 50%. She only applies the indiffernce principle if she is pressed to decide between Monday and Tuesday for the 50% of Sleeps where she is in Sleep-Type Tails.

If this Model does not accurately describe the Problem's Sleep for Beauty, where she finds herself in "This Awakening" then I would like to know why. The probability Beauty gives for "This Awakening" can be tested by taking One Sample-Awakening from each of 1000 independent Trials of Sleeps of this description. The only assumption would be that when the Trial's Sleep is of a Tail's Type the indiffernce principle is applied and the Monday-Tails awakening, Tuesday-Tails awakenings are taken as the single sample with equal likelihood.

PairTheBoard

PairTheBoard

06-13-2007, 09:11 PM

[ QUOTE ]
Yes, the problem says she will be logical. It also says she thinks the probability of heads on Sunday is 1/2. So what is your definition of "logical" when it comes to probability statements? If I build a probability space and it is valid and it says the probability of heads on Sunday is 1/2, then is that sufficient to say that I am adhering to the premises of the problem? Or do you have some additional criteria that you will apply in order to judge the logic of my space? I earlier posted a probability space that encodes the 1/3 argument. It is clearly a valid probability space and it says that SB's probability of heads is 1/2 on Sunday. Is there something about this probability space that you find illogical? If so, then what measure would you put on this particular sample space?

[/ QUOTE ]

I guess another Model for Beauty's Sleep is, she is always Awakened on Monday. Then a coin is tossed to determine if she will be awakened on Tuesday. Beauty finds herself in "This Awakening" of the problem after entering this Sleep Sunday at Midnight.

Now there are Two Awakening Types. A Monday Awakening and a Tails-Tuesday Awakening. Beauty finds herself in "This Awakening" of the problem. How did she come to be in "This Awakening"? If she decides on probabilities that this is a Monday Awakening or a Tuesday-Tails awakening, how can we test her probability assesment by Taking one sample Awakening from each of 1000 such Sleeps? How would the sample Awakening be picked from a Trial Sleep so defined? How do you apply Bayes' Theorem to Events where one of the Events happens with probability 1, the Monday Awakening?

If the Logic of the problem allows two consistent but different Probability Models which are both consistent with the Logic of the problem, what does that say about the problem? That it is not well defined?

PairTheBoard

KipBond

06-13-2007, 09:51 PM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Equvilantley. Beauty enters a Heads-Tails Sleep at random. She is experiencing an awakening in that Sleep. What is the probability it is a Heads-Monday awakening?

[/ QUOTE ]

Please explain what you mean by "enters a Heads-Tails Sleep at random" if you still want me to answer this question.

[/ QUOTE ]

There are Two Types of Sleeps she can enter. One is called a Heads-Type where she will be awakened on Monday only. The other is called a Tails-Type where she will be awakened on Monday and Tuesday. She wants to enter one of these two Types of sleep randomly. So she has someone toss a coin and not tell her how it landed. She then enters That Sleep Type defined by the outcome of the coin toss.

Not only does that define it for you but it also accurately describes the Problem's Sleep that Beauty enters. And she knows it. She can repeat that process 1000 times and have One Sample of an awakening taken from each of those 1000 Trials. The statistics for those 1000 samples match the probability she gives when she considers "This Awakening" in the Problem.

Now can you answer the question?

[/ QUOTE ]

As you defined, each "Sleep" has a 50% probability of being followed by a Heads-Monday awakening.

But, why equivocate on the word "Sleep"? Just say "Flip". Each "Flip" has a 50% probability of being followed by a Heads-Monday awakening.

In your definition, a single "Tails Sleep" has TWO awakenings (and thus two, "sleeps"). No need to equivocate and make this even more confusing to discuss.

But, even with the equivocation on "Sleep", your question can still be answered 1/3, as there are still 3 "awakenings" that Beauty can find herself in.

oe39

06-13-2007, 10:01 PM

is it a common problem-solving technique on the board to try to make the problem a ton more difficult than it is?

i guess i don't even understand why this is a "paradox".

the coins lands tails half the time, but is measured twice as often when it does. so, the flip result is tails 1/2 the time, while the measurement result is tails 2/3 of the time.

KipBond

06-13-2007, 10:03 PM

[ QUOTE ]
75% of her Sample-Awakenings

[/ QUOTE ]

So, what % of her TOTAL wakings will be on Monday?

PairTheBoard

06-13-2007, 10:26 PM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Equvilantley. Beauty enters a Heads-Tails Sleep at random. She is experiencing an awakening in that Sleep. What is the probability it is a Heads-Monday awakening?

[/ QUOTE ]

Please explain what you mean by "enters a Heads-Tails Sleep at random" if you still want me to answer this question.

[/ QUOTE ]

There are Two Types of Sleeps she can enter. One is called a Heads-Type where she will be awakened on Monday only. The other is called a Tails-Type where she will be awakened on Monday and Tuesday. She wants to enter one of these two Types of sleep randomly. So she has someone toss a coin and not tell her how it landed. She then enters That Sleep Type defined by the outcome of the coin toss.

Not only does that define it for you but it also accurately describes the Problem's Sleep that Beauty enters. And she knows it. She can repeat that process 1000 times and have One Sample of an awakening taken from each of those 1000 Trials. The statistics for those 1000 samples match the probability she gives when she considers "This Awakening" in the Problem.

Now can you answer the question?

[/ QUOTE ]

As you defined, each "Sleep" has a 50% probability of being followed by a Heads-Monday awakening.

But, why equivocate on the word "Sleep"? Just say "Flip". Each "Flip" has a 50% probability of being followed by a Heads-Monday awakening.

In your definition, a single "Tails Sleep" has TWO awakenings (and thus two, "sleeps"). No need to equivocate and make this even more confusing to discuss.

But, even with the equivocation on "Sleep", your question can still be answered 1/3, as there are still 3 "awakenings" that Beauty can find herself in.

[/ QUOTE ]

The "Sleep" is a week long. There are two Sleep-Types. That's the model. A Sleep is entered. A sample awakening is taken from it. If heads, there is only one choice for the sample awakening. Thus out of 500 Sleep Trials, 500 Monday Awakenings are taken as the Sample Awakening. If Tails there are two awakenings. One is chosen at random for the Sample Awakening. Thus another 250 Monday awakenings are chosen for the Sample awakening and 250 Tuesday awakenings are chosen for the sample awakening.

However, I am beginning to think the problem simply is not well defined. This "Awakening" Beauty finds herself in. How is it defined? How can it be duplicated in many trials of a well defined model? Are there two ways this can be done consistent with the logic of the problem? If so, I don't think the problem is well defined. I've shown one way how the Sampling from Independent Trials can be done. Is there another that makes sense?

As I mentioned in my last reply to Jason, you could say Beauty begins her Sleep Sunday at Midnight. She then has a Monday awakening with certainty. Then a coin is flipped. If Tails, she has another awakening on Tuesday. In this model there are Two Types of awakenings. A Monday awakening with certainty and a Tuesday awakening with probability 1/2. But how can you apply this model to test Beauty's probabilities for "This Awakening" in the problem. If you repeat Trials of this "Sleep" you have two Awakenings to pick as your Sample Awakening. How do you randomly pick between a Certain Awakening and one with probability 1/2 and identify it as your Sample Awakening?

Once again you will have 1000 Trial Sleeps. In 500 trials you will have only the Monday awakening to choose for your Sample Awakening. In the other 500 trials you will have both a Monday and Tuesday Awakening to choose for your Sample Awakening. If you choose one indifferently then you will choose another 250 Monday awakenings for your Sample and 250 Tuesday awakenings for your sample. Same as before.

If you want to argue 1/3, can you produce a model like this where 1000 Trial Sleeps can be looked at and a sample awakening taken from each with statistics recorded to show that 1/3 of the 1000 sample awakenings are due to a Heads having been flipped? I don't think you can do it. Without such a Model how is "This Awakening" in the problem one to which probabilities can be applied? What can your 1/3 probability mean other than a recognition that two losing bets vs one winning bet require 2-1 odds to break even?

PairTheBoard

KipBond

06-13-2007, 10:28 PM

Jason -

You seem to be trying to model the problem where SB believes the P(H) is 1/2 prior to going to sleep. I do not think this is part of the problem, and perhaps part of the confusion.

I have been using the following as the official version of the problem:

[ QUOTE ]
Some researchers are going to put you to sleep. During the two days that your sleep will last, they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you are first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?

[/ QUOTE ]

Reference: http://www.princeton.edu/~adame/papers/sleeping/sleeping.html

I am further assuming that the "when you are first awakened" part, should be ignored, or otherwise treated as indifferent (since you have no way of knowing when you are "first awakened").

KipBond

06-13-2007, 11:11 PM

[ QUOTE ]
If you want to argue 1/3, can you produce a model like this where 1000 Trial Sleeps can be looked at and a sample awakening taken from each with statistics recorded to show that 1/3 of the 1000 sample awakenings are due to a Heads having been flipped? I don't think you can do it.

[/ QUOTE ]

Sunday evening Beauty is put to sleep. We flip a coin. If it is Tails, we wake her on Monday. If it is Heads, we keep her asleep. Either way, we wake her on Tuesday. Upon each awakening, a record is taken as to the status of the coin (Heads or Tails). We keep repeating this until 1000 awakenings have been documented. We find that 1/3 of them were Heads awakenings. 2/3 of them were Tails awakenings.

I know you will object that I'm recording two samples of the same "Sleep" -- but my rebuttal is that we are sampling awakenings, not "Sleeps". ("Sleep" using your definition.)

Bottom line:

There is a 1/2 probability that a random coin flip will result in a Heads-awakening.
- meaning 1/2 of all coin flips will be Heads, and result in a Heads-awakening.

There is a 1/3 probability that a random awakening was caused by a Heads-flip.
- meaning 1/3 of all awakenings will be the result of a Heads-flip.

Two different sample spaces. Two different questions. Two different answers. Shouldn't be that confusing.

oe39

06-13-2007, 11:39 PM

[ QUOTE ]
[ QUOTE ]
If you want to argue 1/3, can you produce a model like this where 1000 Trial Sleeps can be looked at and a sample awakening taken from each with statistics recorded to show that 1/3 of the 1000 sample awakenings are due to a Heads having been flipped? I don't think you can do it.

[/ QUOTE ]

Sunday evening Beauty is put to sleep. We flip a coin. If it is Tails, we wake her on Monday. If it is Heads, we keep her asleep. Either way, we wake her on Tuesday. Upon each awakening, a record is taken as to the status of the coin (Heads or Tails). We keep repeating this until 1000 awakenings have been documented. We find that 1/3 of them were Heads awakenings. 2/3 of them were Tails awakenings.

I know you will object that I'm recording two samples of the same "Sleep" -- but my rebuttal is that we are sampling awakenings, not "Sleeps". ("Sleep" using your definition.)

Bottom line:

There is a 1/2 probability that a random coin flip will result in a Heads-awakening.
- meaning 1/2 of all coin flips will be Heads, and result in a Heads-awakening.

There is a 1/3 probability that a random awakening was caused by a Heads-flip.
- meaning 1/3 of all awakenings will be the result of a Heads-flip.

Two different sample spaces. Two different questions. Two different answers. Shouldn't be that confusing.

[/ QUOTE ]

this looks exactly right.

PTB: what are you doing? why all these different models?

KipBond

06-13-2007, 11:49 PM

[ QUOTE ]
[ QUOTE ]
Her calculation that given that today is Monday, the P(Heads)=2/3 should be further evidence that something is wrong.

[/ QUOTE ]
Why? Because knowing it is Monday gives her no information regarding the coin flip? And no information implies the probability should remain unchanged and still be 1/2? Well, this is the exact same argument originally used to justify the other side of the debate. She wakes up with amnesia and has no information other than what she originally had on Sunday, so the probability should remain 1/2. You cannot have your cake and eat it too. You cannot reject this argument initially and then use it later to say that P(heads|Monday) = 2/3 is illogical.

[/ QUOTE ]

Well, she should logically see that whether the coin is flipped prior to the Monday waking or after the Monday waking is inconsequential since the coin will always be flipped only once per week, and she will always be woken on Monday regardless of the outcome of the coin. Therefore, if she holds that P(H|monday) is 2/3, she now believes that a future outcome of a fair coin flip is 2/3 likely to be Heads. That is contrary to her credence that the P(Heads) of a fair coin is 1/2.

She then holds her prior credence that P(H|monday) is 1/2. Every Monday the coin has a 1/2 chance of being Heads. This is true whether the coin was flipped prior to the waking or not.

Now, she should realize that the probability that the coin is/will-be Heads given that she has been woken either Monday or Tuesday, but is uncertain of which it is, can NOT be 1/2, since the P(H|tuesday) is 0.

And, therefore, she knows her answer to the following question never can be 1/2:
"What is the probability that a randomly chosen future Monday or Tuesday waking will follow a head coin flip?"

*EDITED to clarify that final question; Ironic, that's the problem with this problem -- the question posed to SB prior/during the experiment is ambiguous.

PairTheBoard

06-14-2007, 12:53 AM

[ QUOTE ]
[ QUOTE ]
If you want to argue 1/3, can you produce a model like this where 1000 Trial Sleeps can be looked at and a sample awakening taken from each with statistics recorded to show that 1/3 of the 1000 sample awakenings are due to a Heads having been flipped? I don't think you can do it.

[/ QUOTE ]

Sunday evening Beauty is put to sleep. We flip a coin. If it is Tails, we wake her on Monday. If it is Heads, we keep her asleep. Either way, we wake her on Tuesday. Upon each awakening, a record is taken as to the status of the coin (Heads or Tails). We keep repeating this until 1000 awakenings have been documented. We find that 1/3 of them were Heads awakenings. 2/3 of them were Tails awakenings.

I know you will object that I'm recording two samples of the same "Sleep" -- but my rebuttal is that we are sampling awakenings, not "Sleeps". ("Sleep" using your definition.)

[/ QUOTE ]

So in other words you can't provide the model I asked for. In the problem Beauty is looking at "This Awakening". To test her probabilities for "This Awakening" you need repeated trials where you sample One Awakening from each, representative of the "random" "This Awakeing" she is experiencing and giving probability frequencies for. You didn't do that. Evidently you can't.

[ QUOTE ]
There is a 1/3 probability that a random awakening was caused by a Heads-flip.
- meaning 1/3 of all awakenings will be the result of a Heads-flip.

[/ QUOTE ]

You didn't show that. You showed that if you take two samples of awakenings from at Tails Sleep and only 1 sample of an awakening from a Heads sleep you have twice as many Tails Samples as Heads Samples. That only shows you have biased your sampling method.

Do a Trial. Take one sample awakening per Trial.

PairTheBoard

jason1990

06-14-2007, 01:18 AM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Her calculation that given that today is Monday, the P(Heads)=2/3 should be further evidence that something is wrong.

[/ QUOTE ]
Why? Because knowing it is Monday gives her no information regarding the coin flip? And no information implies the probability should remain unchanged and still be 1/2? Well, this is the exact same argument originally used to justify the other side of the debate. She wakes up with amnesia and has no information other than what she originally had on Sunday, so the probability should remain 1/2. You cannot have your cake and eat it too. You cannot reject this argument initially and then use it later to say that P(heads|Monday) = 2/3 is illogical.

[/ QUOTE ]

Well, she should logically see that whether the coin is flipped prior to the Monday waking or after the Monday waking is inconsequential since the coin will always be flipped only once per week, and she will always be woken on Monday regardless of the outcome of the coin. Therefore, if she holds that P(H|monday) is 2/3, she now believes that a future outcome of a fair coin flip is 2/3 likely to be Heads. That is contrary to her credence that the P(Heads) of a fair coin is 1/2.

[/ QUOTE ]
What sort of hand-waving should I reply with? Should I tell you that the coin was actually flipped at the beginning of the universe when the initial conditions were laid out after the big bang? Should I tell you that the fact that SB is not experiencing a Tuesday-awakening gives her metaphysical information about the likelihood of possible futures? (That actually reminds me of the "doomsday argument.") The fact is, the logic you are criticizing can be consistently done in a valid probability space. There is nothing wrong with it and it does not contradict her credence that a fair coin comes up heads with probability 1/2. It may contradict your common sense, but you should expect that from a "paradox."

Part of the confusion almost surely comes from putting into words that which is best understood with symbols. Maybe it is like a koan, in the sense that all of this ordinary language and commonsense thinking can lead you away from, rather than toward, a full understanding. (It might also be like a koan in the sense that if you do not think too deeply about it, then it just looks silly and you wonder what all the fuss is about.)

jason1990

06-14-2007, 01:44 AM

[ QUOTE ]
If this Model does not accurately describe the Problem's Sleep for Beauty, where she finds herself in "This Awakening" then I would like to know why.

[/ QUOTE ]
Your model is an accurate description, from a certain perspective. The other model is also an accurate description, but from a different perspective.

[ QUOTE ]
This is my description of the model. I'm sure you can convert it to a mathematical one.

[/ QUOTE ]
For N = 1 and M = 2, I see this probability space:

w2 = "The coin is heads and today is Monday"
w4 = "The coin is tails and today is Monday"
w5 = "The coin is tails and today is Tuesday"

with measure P(w2) = 0.5, P(w4) = P(w5) = 0.25. Does this look right? You can obtain this by taking the model I gave for the 1/3 argument, changing the measure, and restricting it to a subspace.

Why is your measure better than the other? You also use the indifference principle to justify your choice of measure. Why is your use of it more logical than the other? Both measures make counterintuitive claims. Yours says P(heads|Monday) = 2/3. The other says P(heads|Monday or Tuesday) = 1/3. Is yours less counterintuitive?

[ QUOTE ]
The probability Beauty gives for "This Awakening" can be tested by taking One Sample-Awakening from each of 1000 independent Trials of Sleeps of this description.

[/ QUOTE ]
Why is taking one sample from each experiment the only logical approach? Why is it illogical to sample all of them? Why are you not being illogical when you take a constant sized sample from a population whose size is randomly varying? In the end, all you can really claim is that there exists a sampling method that produces frequencies matching your conclusion. But the other side can make that claim as well. Why is your side of it more logical than the other?

[ QUOTE ]
If the Logic of the problem allows two consistent but different Probability Models which are both consistent with the Logic of the problem, what does that say about the problem? That it is not well defined?

[/ QUOTE ]
In my opinion, it is defined just well enough. Clarify it any further and it becomes trivial. Leave it as it is, and it might produce a real awakening.

PairTheBoard

06-14-2007, 03:44 AM

What do you think that "real awakening" is? What lesson does this Paradox teach us?

It looks like some of the alternate models I suggested we look at as well as many more are presented here:

Some Sleeping Beauty Postings (http://www.maproom.co.uk/sb.html)

Evidently the Problem remains open to discussion.

It looks to me that the "This Awakening" of the Problem cannot be pinned down objectively. In the more general model where there are N awakenings produced by Heads and M awakenings produced by Tails, we ask about the likelihood that "This Awakening" is in the group of N or in the group of M. Should we consider the likelihood of "This Awakening" being in one group or the other to depend on the size of the group (1/3 people) or should it depend on the likelihood the group gets produced (1/2 people)?

Either intuition can proclaim a consistent mathematical probability model to describe the intuition. For the 1/3 people it's not that the probability of Heads changes when Beauty enters her Sleep. Rather that when she enters her Sleep she is more likely to travel down the "Wider Path" of Awakenings produced by Tails. The Wide Path of Tails Awakenings biases which path she travels, rather than changing the probability that path was produced.

So it looks to me that likelihoods for "This Awakening" are simply not well defined by the problem. Both 1/3 and 1/2 perspectives are consistent and both depend on intuition. There appears to be insufficient information in the problem to determine the "correct" intuition. Thus the intuition chosen becomes a Bayes Prior producing two distinct but consistent subjective models.

I don't think it violates the Reflection Principle though. Beauty doesn't have to enter the Sleep and experience an awakening to adopt the Bayes Prior Intuition that she is more likely to be experiencing awakenings travelling the Wide Path of Awakenings produced by Tails. Or to adopt the Bayes Prior Intuition that she is equally likely to travel the narrow path of Awakenings produced by Heads.

PairTheBoard

jason1990

06-14-2007, 07:22 AM

[ QUOTE ]
It looks to me that the "This Awakening" of the Problem cannot be pinned down objectively. In the more general model where there are N awakenings produced by Heads and M awakenings produced by Tails, we ask about the likelihood that "This Awakening" is in the group of N or in the group of M. Should we consider the likelihood of "This Awakening" being in one group or the other to depend on the size of the group (1/3 people) or should it depend on the likelihood the group gets produced (1/2 people)?

Either intuition can proclaim a consistent mathematical probability model to describe the intuition. For the 1/3 people it's not that the probability of Heads changes when Beauty enters her Sleep. Rather that when she enters her Sleep she is more likely to travel down the "Wider Path" of Awakenings produced by Tails. The Wide Path of Tails Awakenings biases which path she travels, rather than changing the probability that path was produced.

So it looks to me that likelihoods for "This Awakening" are simply not well defined by the problem. Both 1/3 and 1/2 perspectives are consistent and both depend on intuition. There appears to be insufficient information in the problem to determine the "correct" intuition. Thus the intuition chosen becomes a Bayes Prior producing two distinct but consistent subjective models.

[/ QUOTE ]
I agree. But I would point out that this phenomenon is not unique to this problem. It is a feature of Bayesian philosophy in general. Anyone who thinks that a "perfect Bayesian" ought to come up with a unique probability does not properly understand the Bayesian philosophy. People want to believe that information, combined with perfect rationality, should produce unique probability assessments. But this is not how Bayesian probability works. In fact, this problem seems to do more damage to Objective Bayesianism than Sklansky's El Diablo/Brandi example ever could.

[ QUOTE ]
I don't think it violates the Reflection Principle though. Beauty doesn't have to enter the Sleep and experience an awakening to adopt the Bayes Prior Intuition that she is more likely to be experiencing awakenings travelling the Wide Path of Awakenings produced by Tails. Or to adopt the Bayes Prior Intuition that she is equally likely to travel the narrow path of Awakenings produced by Heads.

[/ QUOTE ]
I think this is a great observation. Elga's argument makes it seem SB is doing all her reasoning when she wakes up. He also only builds half the probability space -- the half that includes Monday and Tuesday. If we use a space like I constructed, then the space exists external to the experiment. The "information" she has on Sunday is that it is Sunday. When she wakes up during the experiment, the information she has is that it is not Sunday. By conditioning on these two different events, she gets two different probabilities. A space like this not only proves consistency, it also illustrates exactly which "information" is being used to change the credence.

[ QUOTE ]
What do you think that "real awakening" is? What lesson does this Paradox teach us?

[/ QUOTE ]
I think you got it.

f97tosc

06-14-2007, 07:56 AM

PTB and Jason1990. I agree with you that the "being told" part was a bit sloppy on my part, because, as the Montey Hall problem illustrates, she does not know if the experimenters may have some special protocol by which they reveal this information.

Nevertheless, complications such as these do not invalidate Bayes' theorem or the point I tried to make. Suppose instead, that she learns that the date is Monday by accidentally looking at a test protocol, and that the likelihood for this sort of thing is not affected by the coin flip outcome, and that she knows that the experimenters would never try to set her up (if we want to be terribly formal we could say that she knew beforehand that she would see this with probability epsilon in either case, and then let the epsilon go to zero). Now by applying Bayes' theorem we get 2/3 for Heads, which is incorrect.

Now one may argue over the realism of this scenario, or that this was not the original problem, but that is beside the point. The point is that we are free to make this as an assumption, and when we do, the right answer to the modified problem should emerge thru Bayes' theorem.

PTB, you were referring to some other posts so I will take a look at that at some later time and respond.

KipBond

06-14-2007, 10:38 AM

[ QUOTE ]
So in other words you can't provide the model I asked for. In the problem Beauty is looking at "This Awakening". To test her probabilities for "This Awakening" you need repeated trials where you sample One Awakening from each, representative of the "random" "This Awakeing" she is experiencing and giving probability frequencies for. You didn't do that. Evidently you can't.

[ QUOTE ]
There is a 1/3 probability that a random awakening was caused by a Heads-flip.
- meaning 1/3 of all awakenings will be the result of a Heads-flip.

[/ QUOTE ]

You didn't show that. You showed that if you take two samples of awakenings from at Tails Sleep and only 1 sample of an awakening from a Heads sleep you have twice as many Tails Samples as Heads Samples. That only shows you have biased your sampling method.

Do a Trial. Take one sample awakening per Trial.

[/ QUOTE ]

I'm pretty sure I already said that 1/2 of your "Sleeps" will be a Head-Sleep. There, I said it again. 1/2 of Beauty's "Sleeps" will be Heads-Sleeps. 1/2 of Beauty's "Sleeps" will be Tails-Sleeps. Did you hear that? 1/2... that is to say 50%... or ONE (1) out of every TWO (2) of her "Sleeps" will be Heads-Sleeps -- meaning it is a "Sleep" that follows a Head result of a fair coin flip. 1/2. 50%.

But, that's not the question in the OP. :P

KipBond

06-14-2007, 11:13 AM

Just for the record, I've argued from the beginning that there are 2 valid answers to this question. However, I do think that given a certain interpretation of the question, one answer is more correct than the other (based off of appeal to rationality & intuition).

If the question is asked in such a way as to be regarding the coin FLIP, the 1/2 answer is more correct.

If the question is asked in such a way as to be regarding the AWAKENING, the 1/3 answer is more correct.

Clearly, you can turn either question in to the other by appeal to a different rationality or non-intuitiveness, and you wouldn't be "wrong" per se... but you would be argumentative (at least).

Now, I will make one last appeal to intuition & rationality regarding the question as posed in regards to the AWAKENING:

Given the problem:

[ QUOTE ]
Some researchers are going to put you to sleep. During the two days that your sleep will last, they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you are awakened, to what degree ought you believe that this awakening followed a Head result of the coin flip?

[/ QUOTE ]

There are 3 scenarios that we can find ourselves in:

Heads & Monday
Tails & Monday
Tails & Tuesday

The probability space for these must total 100%, as there is no other alternative.

How do we divide this space? If we start with the P(H) = 1/2, and the P(monday|T) = P(tuesday|T), then we divide it as such:

P(H+M) = 1/2
P(T+M) = 1/4
P(T+T) = 1/4

This is the way the "1/2 people" choose to divide this probability space. While the "1/3 people" choose to divide it as such:

P(H+M) = 1/3
P(T+M) = 1/3
P(T+T) = 1/3

The latter case seems to contradict the P(H) = 1/2 credence. However, this can be intuitively rationalized by saying that the "waking" is more information regarding the coin flip, so will change the P(H) once the "waking" occurs. (Yes, there are arguments against this.)

However, the latter case allows for certain variations of the scenario without losing it's appeal:

Given that it is Monday, what is the P(H+M)? The "1/3 people" can now say "1/2" which seems perfectly consistent with their prior credence of P(H) = 1/2, especially since the coin need not even be flipped until AFTER Monday for the scenario to stay the same. The "1/2" people must now say that given that it is Monday, the P(H+M) = 2/3. Even if the coin has yet to be flipped. Which is just intuitively irrational.

And what if we change the scenario in such a way that along with the coin flip, we also roll a die. If the flip is Heads, we proceed as before. However, if the flip is Tails, we then wake SB EITHER Monday (if the die is even), OR Tuesday (if the die is odd), but not both. So, now what do the 1/3 people say the probability space is:

P(H+M) = 1/2
P(T+M) = 1/4
P(T+T) = 1/4

This is the same as the "1/2 people" said in the previous scenario. So, what would the "1/2 people" say now? They have to keep the space the same. Which is intuitively irrational, since our scenario will now result in half of the awakenings when the coin flip is Tails.

I'm done, now. I think I have a firm understanding of the problem -- and how it can be interpreted/modeled 2 different ways resulting in 2 different answers -- both of which can be correct for certain applications of the model.

PairTheBoard

06-14-2007, 02:39 PM

I think we can see the paradox in the Phrase, "While you are on the Path". Notice how the phrase confuses location and time. You enter one of two paths with equal probability. One path is longer than the other. You have no sense of time while on either. Now you consider "Self Locating" credences, "While you are on the path". The "While" Engages Time into your credence. The "on the path" Engages Location into your credence. The Phrase, "While you are on the path" egages both Time and Location simultaneously without defining how that should be done.

We try to apply our intuition to determine how to Merge Time and Location logically. But either way we do it produces a Counter Intuition. If we intuit we are more likely to find ourselves on the longer path, "While we are on the path", then we are In a Place with greater probability than we Entered The Place. We are On the Path with greater probability than we Entered the Path. Time "on the path" has taken precedence over Location "on the path", even though "While" indicates they should be treated equally.

If we intuit we are equally likely to find ourselves on either path, "While we are on the path", then when we go on to finish our "Self Locating" in Time our "Self Locating" probabilities for the Time We Entered the two paths are no longer equal. Location "on the path" has taken precedence over Time "on the path", even though "While" indicates they should be treated equally.

So the "paradox" is in the word "While". "While" indicates Location "on the path" and Time "on the path" should be treated equally. Yet in either of our intuitions, we find ourselves giving precendence to one over the other.

Is there an Unconventional Model which may not obey the normal Rules of Probability, but which can do what "While" implies. Treat both Location "on the path" and Time "on the path" Equally?

I suspect the construction of such a Model might very well advance the Theory of "Self Locating Probability". It may be a new kind of "probability" model. Maybe some kind of "Tensor Probability".

PairTheBoard

PairTheBoard

06-14-2007, 04:34 PM

In terms of Taking two Paths with equal probability, one path being longer than the other, and having no sense of time, "While on the Path", if you adopt the Intuition that when "Self Locating" you are equally likely to find yourself on either of the other two paths, then the Counter-Intuition which is expressed by SB as,

P(Heads|Monday) = 2/3

is saying that if "Self Location" is equally likely to be on either of the paths you were equally likely to have entered, Then "Self Location" is more likely to find yourself Closer to the Entrance of the Shorter Path.

It's a statement of likelihood for the "Self Location" and implies no contradiction to the equal likelihood of entering the path.

If you adpopt the Inutition that "Self Location" is more likely to be on the Longer Path then you have the Counter-Inutition that you are more likely to "Self Locate" on a path you were equally likely to have Entered.

Looking at it this way, I don't find one Counter-Intuition more difficult to accept than the other. They both are the result of assumptions about how "Self Location" works in a setting where Place and Time are simultaenously and ambiguously Merged.

Actually, I think we could Merge Place and Time in an infinite number of consistent ways. For example, we might decide that instead of likelihoods for "Self Location" depending on the ratio of path lengths, the likelihoods for "Self Location" should depend on the Square Root of the ratio of path lengths. I believe jason could construct a mathematical probability model for that assumption just as well. And who's to say It's not the right Intuition?

PairTheBoard

oe39

06-14-2007, 05:34 PM

pairtheboard, are you starting a blog in this thread?

KipBond

06-14-2007, 06:05 PM

I just came across this:

Sleeping Beauty and Self-Location: A Hybrid Model
(2006) Nick Bostrom
Faculty of Philosophy, Oxford University

http://www.anthropic-principle.com/preprints/beauty/synthesis.pdf

borisp

06-14-2007, 11:39 PM

Well, I haven't read the entire thread, but the problem with the paper above is simple. They (the experimenters) ask the question "What is your credence?" to a perfectly rational agent. Perfectly rational agent's response: what is your definition of credence?

It seems as if this whole "paradox" arises in the same way that every other paradox arises, from undefined terms. Sleeping Beauty is asked a question about a quantity that has no precise definition, so how could she possibly answer? Upon defining the notion of credence, it is possible for her to answer, but this of course depends on your definition, and different definitions can lead to different answers.

For example, Sleeping Beauty can be asked to guess the result of the flip. She can then guess heads or tails, and she can also assess the probability that she is correct (1/3 heads, 2/3 tails.)

But notice I've slipped in a sophisticated term: probability. This is a mathematically precise notion, and it is one possible interpretation for the notion of "credence." It is DIFFERENT from being forced to guess the outcome. Loosely, it refers to the limit of the ratio of "successes" to "trials" as "trials" tends to infinity.

So the probability that a coin toss results in heads is 1/2, and always will be. If Beauty is forced to guess on THIS coin flip, then if she guesses heads, her guess will be right with 1/3 probability, and will be wrong with 2/3 probability.

If she is asked what the probability that THIS coin is heads is, she will respond that she can't answer. One cannot meaningfully assess probability for events that have already happened, and she cannot know whether the flip has happened or not. For if she were to answer, 2/3 of the time she would be correct to answer "1/2," and 1/3 of the time a perfectly rational agent assesses probability for something that is either 0 or 1, and we all die in a massive end to the universe, since rational agents have suddenly gone crazy. Being a rational agent she would want to avoid this outcome, so she asks for clarification.

It always makes sense to ask her to guess, and also then ask her the probability that she will be right when she learns of the accuracy of her guess (an event that hasn't happened). There is no bias in the first; she may as well guess heads or tails. Upon doing so, she will correctly answer the latter question contingent on the response to the first.

As a mathematician, it is always funny to read problems that involve "perfectly rational agents." Evidently, they only exist in the minds of irrational agents that are willing to bother them with terms that are not well defined. They probably spend most of their time asking questions and being disappointed with our ambiguous answers.

PairTheBoard

06-14-2007, 11:54 PM

[ QUOTE ]
I just came across this:

Sleeping Beauty and Self-Location: A Hybrid Model
(2006) Nick Bostrom
Faculty of Philosophy, Oxford University

http://www.anthropic-principle.com/preprints/beauty/synthesis.pdf

[/ QUOTE ]

Nick Bostrom's got an interesting idea there. It's along the lines of what I suggested might be done when I coined the phrase "Tensor Probability". It's a model that has probability aspects to it while avoiding the problems that certain Bayes type conditional proability calculations seem to infer.

His main idea is that the possible "Agent-Parts" Beauty may experience in H1 = Heads & Monday, T1 = Tails & Monday, and T2 = Tails & Tuesday, have their own credence function P. But when Beauty is TOLD it is Monday, she now is experiencing a different possible Agent-Part, Agent-Part+, which may be either H11 = Heads and TOLD-Monday or T11 = Tails & TOLD-Monday. Bostrom assigns an entirely new credence function P+ for Agent-Part+, with P+(Heads)=1/2.

He maintains the Intuition of 1/2 people for the Credence Function P. The conditional credence P(Heads|Monday)=2/3 does not mean that Beauty believes Heads has probability 2/3 if she is told it is Monday. If she is told it is Monday she now becomes Agent-Part+ with credence function P+ and has credence 1/2 that the coin flips Heads.

I'm not sure I caught Bostrom's interpretaion of P(Heads|Monday)=2/3 but I think it might be along the lines of my Two Path Model of realizing that If I give equal credence to being on either of the two paths that I entered with equal likelihood, then if I am on the shorter path it is more likely I am close to the Entrance to the Path than if I were on the Longer Path. Which only makes sense, and is not really counter intuitive. The counter intuitive thing in the old 1/2 model is if the conditional probability P(Heads|Monday)=2/3 is interpreted as being forced on Beauty when she is TOLD it is Monday. Bostrom's model is not burdened by that misinterpretation.

Bostrom handles the Betting Odds problem similiarly to how jason and I have explained it here. It is not really a valid objection to the 1/2 model nor to Bostrom's Hybrid Model.

At this point I thought Bostrom's Model might be better described simply as a Defense of the 1/2 Model. But Bostrom's Model suprised me in how it handles repeated Trials of Beauty Sleeps. The 1/3 people who find the 1/2 model extremely counter intuitive for the 1000 Trials of Sleeps will find Bostrom's Model more satisfying to their intuition.

Bostrom noted as I have, that in a Deterministic Double Sleep for Beauty where she is awakened only on Monday in Week 1 but on both Monday and Tuesday in Week 2, the indifference principle should give her 2/3 credence she awakens in Week 2. This is because the 3 awakenings involve Actual Agent-Parts of Beauty. Bostrom also agrees with the 1/2 people when there is just One Sleep, that this Indifference should not be applied because the 3 Agent-Parts are not Actual but only possible. So the objective coin toss probability should be applied between the possible H1 Agent-Part and the possible (T1,T2) pair of Agent-Parts. But that's for One Sleep only.

If there are 1000 succesive Trials, or Probability Sleeps, Brostrom recognizes that now there are very likely to be 500 ACTUAL H1-type Agent-Parts, T1-type Agent-Parts, and T2-type Agent-Parts. So the Indifference Principle should be applied more like it would be if the 1000 Weeks were divided between Heads-Types and Tails-Types Deterministically. So his P depends on the number of trials because there are more Actual Agent-Parts in more trials which are statistically likely to be close to what they would be if they were deterministic. Essentially his Hybrid model produces a P(N) where N is the number of succesive Probabalistic Trials of Sleeps. He has,

P(H1) = P(1)(H1) = 1/2

While P(N)(H1 type Monday) --> 1/3 as N--> large.

He also showed how P(2) should be explicitly calculated according to his Hybrid Model.

So he gets some of the best of both the 1/2 and 1/3 intuitions while avoiding the worst counter intutuitons that their pure forms produce.

It looks like a very reasonable view to me.

PairTheBoard

borisp

06-15-2007, 12:00 AM

What does "credence" mean?

jason1990

06-15-2007, 12:21 AM

Here is an interesting connection between this problem and Markov processes. We have three states:

0 = heads-Monday
1 = tails-Monday
2 = tails-Tuesday

These represent SB's awakenings. If we repeat the experiment many times, then her sequence of awakenings is a Markov chain. It goes from 0 --> (0 or 1) with probability 1/2 each; it goes from 1 --> 2 with prob 1; and it goes from 2 --> (0 or 1) with prob 1/2. Its transition matrix is

[1/2 1/2 0; 0 0 1; 1/2 1/2 0].

From here it is easy to compute the stationary distribution, which is (1/3,1/3,1/3).

However, a Markov chain always has one unit of time between each transition, which is not the case in the actual problem. In the actual problem, she sleeps an extra day after a heads-Monday. So let us turn this Markov chain into the corresponding continuous time Markov process, where the jumps out of state 0 take twice as long as the other jumps. Such a process has generator

[-1/4 1/4 0; 0 -1 1; 1/2 1/2 -1].

And its stationary distribution is (1/2,1/4,1/4).

PairTheBoard

06-15-2007, 12:25 AM

[ QUOTE ]
What does "credence" mean?

[/ QUOTE ]

Level of conviction for a belief as measured like a probability and which works like a probability in reasoning with it.

Read the Baysian Probability entry in Wikipedia.

PairTheBoard

borisp

06-15-2007, 02:19 AM

Yes, I am well aware of Bayesian analysis and Wikipedia, but then again I have spent thousands of $ on my math textbooks, so I might as well use them instead, dontcha think? I'll probably do it in my office in the math department at Berkeley, you can stop by for office hours if you like.

But thanks for changing the word that is not well - defined from "credence" to "belief."

A rigorous definition of probability will tell you that you need a space and a measure and so on, before the notion of probability is even computable. The pages on Wiki vaguely refer to "probability" as if it is some axiomatically understood notion. This is far from correct, but for the most part, the omission of rigor is harmless for applications to the real world, like what actuaries do and so on. But in abstract discussion, it leads to idiotic "paradoxes" when things are not understood properly. (IMO, the only paradox is people's constant insistence on using terms they don't understand, and responding with juvenile "are you retarded??" type responses when queried for definitions.)

When Sleeping Beauty awakens, she is in a "tangled" event state: 2/3 of the time, the coin hasn't flipped, and 1/3 of the time, it has. She knows this because she knows how to condition her space on the knowledge that she has woken up.

Probability with regard to the upcoming coin toss is only meaningful in the former situation, and of course then it is 1/2. But before she is willing to assert this probability, she will insist that someone inform her of the nature of her reality. In the absence of that information, a rational agent is left to describe her current space completely and can do nothing more, because she can only respond to precise and well defined requests. She doesn't have "beliefs." And she can only be "perfectly Bayesian" with respect to events that are measurable subsets of well defined measure spaces. The space of "all possibilities" is not such a space.

(The philosophy guy's paper amounts to a boring and trivial description of a very simple event space, for which not all events have a meaningful notion of probability. In this sense, everything he has written is trivially correct, in that he outlines the principles that a rational agent will use in reasoning, were the agent asked to carry out a precise task or respond to a precise query.)

Once the coin has flipped, it can no longer be reasoned upon probabilistically, because it has become a unique event, never to be repeated. So one cannot compute the limit as "trials" goes to infinity, it either is heads or it isn't. In other words, the measure space here is a single point, and that point either has value 0 or 1. Asking the rational agent what the probability is will be met with a puzzled look as she bemoans her cursed existence among irrational agents. (Loosely, this uniqueness bit is the "no cloning theorem" in quantum computing, see Wiki haha)

Now, if she is asked ahead of time to guess, this is an entirely different question, and equally trivial. She would know that on average, if she decides ahead of time to always guess tails, then those times she is right, she will be rewarded with two chances to guess, and if she is wrong, she only guesses once. This actually depends on her understanding of probability with regard to the coinflip; note she develops this strategy beforehand. She is also defining an event in her space: the guess. The accuracy of her guesses can be reasoned upon probabilistically, but only before the whole scenario begins. DURING the process, she cannot know the likelihood that a particular guess is correct or not, for the same reasons as discussed above.

Rational agents can't answer questions that don't have computable answers. Without a precise notion of probability, or belief, you might as well be asking the agent to pick a number at random from the interval [0,1]. These choices will obviously be uniformly distributed about 1/2...but wait...what's the definition of random...

PairTheBoard

06-15-2007, 02:53 AM

I see.

PairTheBoard

f97tosc

06-15-2007, 04:46 AM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Suppose we do the following experiment instead. First we flip the coin. Then if it is T, we put two black tokens in an urn. If its [H], we put one black and one white token in the urn.

Now, let S take one token and guess if it is H or T. Obviously, [if] it is black, she should give the probability one third to H.

[/ QUOTE ]
I do not believe that this is a fair translation of the problem. In your urn model, seeing a black token would be genuinely new information, precisely because of the fact that she knew it was possible to see a white token.

I think a fairer translation would be this. Flip the coin and fill your urn exactly as you describe. But instead of letting S take a token, the experimenter looks inside the bag, takes out a black token, and shows it to her. In that case, the answer is still 1/2.

[/ QUOTE ]

Yes.

PairTheBoard

[/ QUOTE ]

So do the two of you agree then, that if the protocol is changed so that the white token can be revealed (for example, she is wakened up on Tue-Heads but told the weekday and coin flip in this case), then the right probability is 1/3, when she experiences another outcome (black ball/wake up and not told Tue-Heads)?

bigmonkey

06-15-2007, 04:24 PM

Sorry guys I cann't really contribute to this thread anymore, as I am totally snowed under with work. I will get around to reading all of the replies but it would be unreasonable to expect anybody to defend positions they argued for weeks or months ago. Thanks for all the contributions though, especially to PairTheBoard and KipBond. I don't think you can ever have too many arguments from analogy in this debate.
The SB paradox is one area I am focusing research on, and I'll regard this thread as much as a reference as some of the original papers concentrated on the subject.

PairTheBoard

06-15-2007, 06:33 PM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Suppose we do the following experiment instead. First we flip the coin. Then if it is T, we put two black tokens in an urn. If its [H], we put one black and one white token in the urn.

Now, let S take one token and guess if it is H or T. Obviously, [if] it is black, she should give the probability one third to H.

[/ QUOTE ]
I do not believe that this is a fair translation of the problem. In your urn model, seeing a black token would be genuinely new information, precisely because of the fact that she knew it was possible to see a white token.

I think a fairer translation would be this. Flip the coin and fill your urn exactly as you describe. But instead of letting S take a token, the experimenter looks inside the bag, takes out a black token, and shows it to her. In that case, the answer is still 1/2.

[/ QUOTE ]

Yes.

PairTheBoard

[/ QUOTE ]

So do the two of you agree then, that if the protocol is changed so that the white token can be revealed (for example, she is wakened up on Tue-Heads but told the weekday and coin flip in this case), then the right probability is 1/3, when she experiences another outcome (black ball/wake up and not told Tue-Heads)?

[/ QUOTE ]

So now you flip a coin, If heads you put a black ball and a white ball in the bag. If tails you put two black balls in the bag. You then pick a ball at random from the bag and show her what color it is. When you do this and you happen to show her a black ball she applies Bayes' Theorem to conclude there is 1/3 chance the ball came from the bag produced by Heads? ie. P(Heads|Not a White Ball) = 1/3

How does this correspond to Sleeping Beauty's situation. If her "Sleep" was defined by the 4 Agent-Parts, H1,H2,T1,T2 where she is awakened on both Monday and Tuesday regardless of the Flip, then she will construct a Credence Function P for those Agent-Parts. It will give equal likelhood to all 4 Agent-Parts. That Credence Funtion will compute P(H1|not H2) = 1/3. Going by the discussion in the Hybrid Model, that is the Reasoning she would use when experiencing any of the 4 Agent-Parts H1,H2,T1,T2. She thinks to herself, If I am not in H2 I am more likely to be in T1 or T2 than H1.

But, again using the Hybrid Language, that is different than her always being told that she is not in H2 when she is in H1,T1, and T2. Once she is actually Told the Extra Information, not-H2, she is now experiencing Different Agent-Part+'s. They are clearly different Agent-Parts than H2. H2 can't possibly be one of these Agent-Part+'s. And to be well defined, Beauty must be informed before the Sleep that she will be told this information every awakening but the Heads-Tuesday. So she enters the Sleep knowing she will experience the Agent-Part H2, and the Agent-Parts H1+,T1+, and T2+. This defines an entirely new experiment.

According to the Hybrid Model, the three Agent-Part+'s, H1+, T1+, and T2+ will be assigned a P+ Credence Function of their own. P+ is not bound by conditional probabilities produced by P, which mean something entirely different. And it looks reasonable to me to view P+ as being equivalent to the Credence function Beauty would assign to the orignial problem's Sleep, where Heads-Tuesday is just slept through. It has been taken out of the credence function P+ just as if it were being slept though. Thus P+ can reasonably give P+(H1+)=1/2 as in the original problem.

I think the Hybrid Model gives the best language for looking at what's going on.

I'll be interested to see jason's response if he is inclined to give one.

PairTheBoard

f97tosc

06-16-2007, 06:17 AM

[ QUOTE ]

So now you flip a coin, If heads you put a black ball and a white ball in the bag. If tails you put two black balls in the bag. You then pick a ball at random from the bag and show her what color it is. When you do this and you happen to show her a black ball she applies Bayes' Theorem to conclude there is 1/3 chance the ball came from the bag produced by Heads? ie. P(Heads|Not a White Ball) = 1/3

[/ QUOTE ]

But the difference between this problem and the given problem is not that Sleeping Beauty is given any "replacement black balls" or extra Heads wake up experiences. The difference is that she never experiences anything at all during the white ball condition. That the observer is kept asleep, or prevented from observing under certain outcomes, is perhaps the most unusual feature of this problem.

I would say that if we have a probabilistic problem where we will be "kept asleep" during certain outcomes, then before the experiment is realized, this should cause a divergence between the expected actual outcome, and the expected experienced outcome, assuming that we experience anything at all. That is why I see no contradiction in assigning 1/2 before, but 1/3 after she wakes up, the latter being conditioned on that it is not Tue-Heads.

However, if we do experience one of the outcomes that is not associated with being asleep, then I would say that we have exactly the same information in the sleep and no-sleep versions of the problem. In the present case, when Sleeping Beauty wakes up, she knows that it is not Tue-Heads, regardless of whether Tue-Heads is associated with staying asleep or waking up and being told that it is not Tue-Heads. Or even more concretely, if the Tue-Heads protocol was changed after she woke up, she wouldn't learn anything about the coin flip. Assigning a different belief in the two scenarios would make her beliefs depend on unrealized or future events, and therefore violate causality.

f97tosc

06-16-2007, 06:24 AM

[ QUOTE ]

I think a fairer translation would be this. Flip the coin and fill your urn exactly as you describe. But instead of letting S take a token, the experimenter looks inside the bag, takes out a black token, and shows it to her. In that case, the answer is still 1/2.

[/ QUOTE ]

I don't think this is a fair translation. To paraphrase your version, if sleeping beauty is about to pick the white ball, the experimenter replaces it with a black ball. But there is no replacement black ball experience; she doesn't get teleported back to Monday for an additional wake-up under Tue-Heads. This is why in your problem, when repeated many times would yield the same number of Heads and Tails wake-up experiences, but the given problem has only 1/3 as many Heads wake-up experiences, and why in your problem, sleeping beauty should wager on 1/2, but in the given problem she should wager on 1/3.

It is true though that in the given problem she is "kept asleep", or prevented from observing, during the white ball condition. But as I have argued in a different post, this mustn't affect her belief when she actually draws a black ball. That would violate causality.

f97tosc

06-16-2007, 08:39 AM

[ QUOTE ]

f97tosc:[ QUOTE ]
I can't say I am familiar with the reflection principle, but I suspect that the "resolution" will have to do with the fact during certain experimental outcomes, we supress Sleeping Beauty's ability to state (or even form) her belief.

[/ QUOTE ]

The reflection principle is that:

Px(A|Px+y[A]= c)=c

Where Px and Px+y are credence functions at time x and time x+y. So the principle is that given that at some later time your credence for A is c then your credence for A now should be C. If you believed now that tomorrow you would receive some important information about soemthing that will make you have credence c in A, then you should now already have credence c in A. I think it could be argued that Beauty isn't completely rational because she believes she will be given an amnesiac, and therefore cannot be in a rational cognitive state, but then people argue that she knows exactly what memories she will forget, and also the amnesiac is only given to her after going back to sleep to forget her awakening, so when she wakes up she is supposed to be fully rational.

[/ QUOTE ]

Thanks for this, and sorry for the slow reply (didn't see it before). I would say that the reflection principle goes out the window when, under certain conditions (Tuesday, in particular), her observation is made contingent on the experimental outcome. If she wakes up under tails, but not under heads, she cannot really be said to have made a normal "observation" when she wakes up. Of course in the present problem we also have the confusing business about Monday and forgetting, but the principle is the same, when she doesn't know if it is Monday or Tuesday, and the observation is incomplete on Tuesday, then that makes the observation as a whole incomplete as well. If we change the problem so that she does wake up on Tue-Heads, and gets told that it is Tue-Heads in that case, then the principle holds and we get 1/3; I think PTB and Jason1990 agree to this as well.

But even though the notion of "censored" observations causes trouble for common probability principles, I would still argue that sleeping beauty can form rational beliefs when she already knows that the censored event hasn't happened. In fact, I think that changing her belief (P=1/3) based on censorship in some future or unrealized scenario would violate causality.

jason1990

06-16-2007, 10:30 AM

[ QUOTE ]
So do the two of you agree then, that if the protocol is changed so that the white token can be revealed (for example, she is wakened up on Tue-Heads but told the weekday and coin flip in this case), then the right probability is 1/3, when she experiences another outcome (black ball/wake up and not told Tue-Heads)?

[/ QUOTE ]
If I understand you correctly, then I agree.

[ QUOTE ]
[ QUOTE ]

I think a fairer translation would be this. Flip the coin and fill your urn exactly as you describe. But instead of letting S take a token, the experimenter looks inside the bag, takes out a black token, and shows it to her. In that case, the answer is still 1/2.

[/ QUOTE ]

I don't think this is a fair translation. To paraphrase your version, if sleeping beauty is about to pick the white ball,

[/ QUOTE ]
The SB I described is never about to pick the white ball. She is never about to pick any ball.

[ QUOTE ]
and why in your problem, sleeping beauty should wager on 1/2,

[/ QUOTE ]
The SB I described would never wager. She knows the test protocol, so she knows a bet on heads will effectively pay out 1 to 2. Being rational, she requires P(heads) > 2/3 before she wagers. Since P(heads) = 1/2, she declines the wager.

[ QUOTE ]
That would violate causality.

[/ QUOTE ]
The "violates causality" argument is also used to say that P(heads|Mon) = 2/3 is irrational. Essentially, the argument says that flipping the coin Mon night changes nothing, and in that case you cannot say P(heads|Mon) = 2/3 since the flip has not yet occurred.

Personally, I do not think this necessarily violates causality. For instance, I could postulate that SB is a determinist. In that case, the coin flip is entirely determined by the state of the universe on Sunday. Either that state is Sh, which will inevitably lead to heads, or it is St, which will inevitably lead to tails. If she takes her experience of a Monday awakening and uses it to alter her beliefs about that prior state of the universe, she is not violating causality.

Frankly, though, I do not know how long I can continue to play devil's advocate for both sides. I am not a professional philosopher.

I realize the thread is long, so perhaps I should restate my stance. The conclusion I finally reached was that each argument represents the adoption of a consistent prior distribution. Therefore, the mathematics of probability cannot determine the "right" answer. I realize this is a fairly trivial observation. However, I think it is one that laypeople often overlook. Many casual readers probably believe there is a mathematically correct answer to this problem.

In other words, I am simply observing that the arguments presented by both sides amount to philosophical justifications of their priors. (Of course, this is also a fairly trivial observation, since most/all of the papers on this subject seem to be in the philosophy literature.) I also stated that I personally find neither side's arguments to be overwhelmingly convincing.

f97tosc

06-16-2007, 11:36 AM

I agree that the central difficulty here is making the right defnition and probability model, not solving it, and that this has some philosophical flavor too it. I would say though, that I don't think there is any fundamental ambiguity in this problem. In the "bent coin" problem it is not so clear how we rigorously model a coin that is "bent in an unknown way", because different people may have different ideas of what this may mean. In this problem, however, the problem is quite well-specified, anybody reading the problem knows exactly what Sleeping Beauty experiences and knows and does not know in a given situtation. The problem has some unorthodox assumptions that make things difficult, but I don't think it is ill-specified. Any disagreement is of our own making!

And I am just surprised that you are willing to throw out causality in favor of an intuition or a simpler model. I hope that you agree that there is no proof that "waking up at least once" captures all the information in waking up. Without causality, however, it is unclear how anybody could ever form a rational belief about anything. Other than that I agree that it is time to close the debate and I will summarize my position in a different post.

Cheers,

f97tosc

f97tosc

06-16-2007, 11:51 AM

This has been an interesting discussion but I feel it is starting to take up too much time for me so I am going to wrap up my position.

The central difficulty of this problem is that under certain conditions, sleeping beauty sleeps and cannot observe what happens; the question is then how she should rationally form beliefs in the cases when she is awake. As I understand it, discussants on "both sides" of the issue are in agreement that if the test protocol is "uncensored" (i.e., she is woken up and told the coin flip under Tue-Heads), then the probability of heads (when she wakes up and not told this) is 1/3, and the reflection principle holds as well. It is not particularly strange that the reflection principle may go out the window if we censor the observer's ability to state or form beliefs under certain conditions. When this happens we must keep separate probabilities of what may happen, and probabilities of what may be experienced (of course, the problem is about what is experienced).

I think the best way to deal with the unusual censorship phenomenon (in this problem and potentially in others, when the observer may be kept asleep, killed, etc) is to build a formal, uncensored probability model. Then from there we argue, based on basic principles of causality, symmetry and information, that under those outcomes when the observer actually does observe, it doesn't matter whether there would or would not be censorship under some other outcomes. This point is difficult to prove rigorously, but as Jason1990 pointed out, when we translate an informally stated word problem to a mathematical one there must always be some element of the translation that cannot be completely formalized. I would just say that the principle of causality is a stronger guide than is the assumption that there is no information in waking up, which is just that: an assumption. I concede that the informationless event "wake up more than once" makes for nice and easy models, and even that it may sound somewhat appealing, but there is simply no proof that this event captures all the information of the situation. Frankly, I think that many posts in the thread are just rephrasing, restating and assuming the no-information claim over and over - but we can't assume away that which is being disputed. To me, it is just as appealing that there is information in waking up, based on the biases associated with the different wake-up frequencies. I wish that everybody could at least be open to the possibilities that there may or may not be information in waking up, and then reach a conclusion based on other properties of the problem. As I see it, none of the points below have been given an adequate explanation by those who favor 1/2.

1. Comparing to the uncensored problem and invoking causality, as discussed above, suggest P(Heads|Wake up unknown day) = 1/3.

2. Long-term frequency. Repeating the experiment many times we have 1/3 of the wake-ups in the heads case. PTB wrote "Just because there are three types of awakenings doesn't mean they are equally likely to be the one she is experiencing right now." No, but by what information does she determine that the present one is more or less likely? Having no distinguishing feature she must assign the same probability to them all, and for this to be consistent with the frequency (which she knows) it must be 1/3.

3. Wagering. Similarly, Sleeping Beauty should make wagers she should bet on 1/3. Any other probability assignment would enable somebody else with no more information than her to exploit her with wagers.

4. Modified problems/Bayes' theorem (this is related to 1.). If we change the problem, by adding or removing certain pieces of information, only 1/3 gives sensible results. 1/2 would make the belief contingent on future and/or unrealized events and therefore violate causality. This can also be seen by writing the following equation - it is impossible to make sensible right hand side assignments that lead to 1/2.

P(Heads|Wake up unknown day) = P(Heads|Tuesday)xP(Tuesday) + P(Heads|Monday)xP(Monday)

5. Information content of answer. How can it be that two reasonable approaches lead to different answers? One explanation is that neither path does anything wrong, but that one path makes less use of the available information. IMO, conditioning on "waking up at least once" is not wrong, just incomplete. But how do we know that it is not the 1/3 answer that is incomplete? Because 1/3 - 2/3 answer has higher information content (lower enthropy), and is closer (lower average error) to the actual flip outcome, than is 1/2-1/2.