Sleeping Beauty Paradox

[PairTheBoard]

[ 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]

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]

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]

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]

[ 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]

[ 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, (M<p or T<p). As p-->1 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]

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]

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]

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.

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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]

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.