The envelope problem, and a possible solution
 

Recently I read a post on the 2 envelopes paradox ( Here).

I will not elaborate on what the problem is, because it is stated and discussed clearly in the provided link and I ask you to please read the OP. I want to discuss a solution put forward by a friend of mine.

A friend of mine argued the following:
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If we were te simulate the envelope problem, it would show that looking at the amount of money in your envelope is not going to affect the EV of always switchin envelopes. Which would be in line with the argument that switching envelopes is neutral EV. He asked me, "if we were to play this game, do you expect that looking at the amount would change anything to you EV?" And I answered: "I do not see how always switching can possibly become anything other than neutral EV, when the only change you make is that you are merely passivily looking at the value now". After which he followed:

Most likely the argument that switching is EV neutral is the correct one. The other argument then cannot be correct, so somewhere in the argument there must be flaw. He argued that the point of determination is key. The amounts in the envelope are determined before the start of the game, not after you learn the value.

If the value of the second envelope was determined after you had learned the value of the first (for example by flipping a coin) then idd switching would be +EV. This is comparable to saying, I give you $100, you can keep it, or I flip a coin; heads you double it, tails you lose half.

However, this is not the case in the envelope problem he said. The values are assigned to the envelopes on beforehand. Even though intuitively it may feel like a switch might double your money or cut it in half after you learn the value; in reality there is only one possible outcome. For example, if you look at the value in envelope 1, and it is $100. Then intuitively it might feel like you have a 50% shot at $50 or $200, but you don't have it. Say the ammount in the second envelope was $50, it would not be possible to get $200 on that trial. The flaw would boil down to using the $100 as new information in determining your EV when it doesn't provide any new information really. So, even though the method of determining your EV is right in argument 1, it is not applicable in that situation.

What I am wondering is, how is this line of reasoning? He is fully convinced that it is the solution to the problem, and a colleague of mine basically suggests the same solution.
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I still am not sure, I'd appreciate some imput.

Re: The envelope problem, and a possible solution
 

there was a link to a decent explanation to the problem in the original thread. the set up of the problem is that one envelope has $N and the other envelope has $2N in it. the trick is that the value of N is unknown.

but the participant must have some belief about how likely different values of N are. this belief just describes how likely the participant thinks different values for the smaller envelope are. and if the player thinks some values are more likely than others (maybe he believes extremely high values are unlikely) then it is not true that after learning the contents of one envelope the other envelope has a 50/50 chance of being the big envelope.

for example, if you open the envelope and discover a large amount of money, you might think that it is probably the 2N envelope. if you open the envelope and discover a small amount of money, you might think it is probably the N evelope.

so, after opening the first envelope, switching will sometimes have +EV and sometimes -EV. but before you learn the contents of the first envelope, you are indifferent between switching and not switching.

Re: The envelope problem, and a possible solution
 

[ QUOTE ]
If we were to simulate the envelope problem . . .

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How?

Paul

Re: The envelope problem, and a possible solution
 

I got frustrated with Aaron in this dialogue before. Sure am glad you started it up again. [img]/images/graemlins/smile.gif[/img]

Here's how my dialogue would go when I hear the problem.

Paradoxer: "I put a random amount X in one envelope and 2X in the other."
Me: "Wait, how was X chosen?"
P: "I'm not telling you. Now, you open and find $100. Should you switch?"
M: "Man, I dunno. I can tell you the correct move if you describe the opponent."
P: "I'm not going to tell you. So, what's the best move?!"

It's retarded. Of course you and I can arrive at different answers if we try to fill-in the gaping hole differently. That doesn't show that the problem's clever; all that shows is that the problem has gaping holes.
-Sam

P.S. Assuming that X is chosen equally likely from the set of all positive integers is flat wrong. You're not allowed to assume that, because it doesn't make sense. [img]/images/graemlins/smile.gif[/img]

Re: The envelope problem, and a possible solution
 

thanks for the reply guys.

@econ: Depending on the amount of money in the envelope, someone might indeed use this information for switching which I expect to influence EVj. However, as I understand it, stated in the original problem, it's not a matter of psychology. If you are indeed allowed to estimate EV according to the method of argument 1, switching would always be +EV. And if you believe arguemtn 2 to be true, it wouldn't matter. It's therefore that I simplified the problem to always switching.

@Paul: we could simply play it, or write a program to play it out. Note that as stated you could play this game without ever really having to actively participate. So you could run a lot of trials. From your response it seems a bit like you believe that how you simulate the problem is going to matter. If this is the case, I'd like to hear those thoughts.

@Sam: Sam, you refer to the sampling method. But at this point I can't really deduce why you believe the manner of sampling is important. You said: "Assuming that X is chosen equally likely from the set of all positive integers is flat wrong. You're not allowed to assume that, because it doesn't make sense." Alright, if we then do not make that assumption, how would that impact the results?

Re: The envelope problem, and a possible solution
 

I don't like the idea that explaining why always switching is equivalent to always not switching is a solution. It's not complete. As bigpooch mentioned in the other thread (and others have mentioned in past threads), you can do better than either always switching or always not switching. This should be added to the statement that you can't assume the conditional expectation is 1/2.

Choose a value t. If you see an amount less than t, switch. If you see an amount greater than t, don't switch. Conditioned on the case that both amounts are less than t, you break even compared with either always switching or never switching. Conditioned on the case that both amounts are greater than t, you break even again. However, if one amount is greater than t, and one amount is less than t, you always end up with the greater amount. If this happens with positive probability, switching with threshold t beats both the strategy of always switching and the strategy of never switching.

Instead of letting t be constant, you can let t be a positive real random variable with positive probability on each interval [x,2x]. Another way to describe this is that you switch probabalistically, with the probability of switching strictly lower at 2x than x, with limits of 0 and 1 as x goes to infinity and 0, respectively. This random threshold will have a positive probability of being between the two amounts no matter what distribution is chosen, and this will beat the strategy of always switching or never switching, even though you can't be sure by how much it will be better.

In fact, you can describe the strategy of always switching as using a threshold of +infinity, and you can describe the strategy of never switching as using a threshold of 0. You can see these thresholds have 0 probability of being between the amounts.

Mike Caro got this wrong, and the Wikipedia article doesn't mention that you can do better than either.

Re: The envelope problem, and a possible solution
 

[ QUOTE ]
@econ: Depending on the amount of money in the envelope, someone might indeed use this information for switching which I expect to influence EVj. However, as I understand it, stated in the original problem, it's not a matter of psychology. If you are indeed allowed to estimate EV according to the method of argument 1, switching would always be +EV. And if you believe arguemtn 2 to be true, it wouldn't matter. It's therefore that I simplified the problem to always switching.


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My argument isn't based on psychology, it's based on Bayesian statistics. According to Bayesian statistics, statisticians can have believes about random variables before they gather any data on them. These beliefs are called "priors."

The only prior that allows arguement 1 to be correct is believing that the value of the smaller envelope is distributed uniformly from 0 to infinity. But this distribution doesn't make sense . . . it means you think $5 Billion is as likely a value as $5.

For any other "prior," argument 1 fails.

Re: The envelope problem, and a possible solution
 

[ QUOTE ]

@Sam: Sam, you refer to the sampling method. But at this point I can't really deduce why you believe the manner of sampling is important. You said: "Assuming that X is chosen equally likely from the set of all positive integers is flat wrong. You're not allowed to assume that, because it doesn't make sense." Alright, if we then do not make that assumption, how would that impact the results?

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If you make the assumption that it is a positive number randomly chosen, and you get an odd amount of money in your envelope (1/4 of the time you will), you know to switch.

Re: The envelope problem, and a possible solution
 

thanks for the replies again guys. You'll have to bear with me, I'm no mathmatician or statistician (is that a word?) so I'm slow here.

@econ: high econ I don't know anything about priors, so I'll need some further explanations on that. You said:
"but the participant must have some belief about how likely different values of N are." and went on to describe how the value might influence the way in which he determines ev.
In the original post it is stated that:
"Argument 1: It's +EV to switch. You had a 50/50 chance of picking the high or low envelope so there's a 50% chance that the other envelope is the high and a 50% chance it's the low. Therefore, EV of switch = 0.5*(+100) + 0.5*(-50) = +25."
nowhere here do I read anything about how the size of the value influences the decision to switch. Rather, I read that upon learning the value, switching is EV.

@Phzon: you said: "I don't like the idea that explaining why always switching is equivalent to always not switching is a solution. It's not complete. As bigpooch mentioned in the other thread (and others have mentioned in past threads), you can do better than either always switching or always not switching."

I merely refer to the fact that argument 1 as stated implies that switching will always be positive EV upon learning the value. Argument 2 says it will never matter. You might be able to do better by sometimes switching, but imo that was not stated in the original problem. For that reason I simplified to always switching. Similar to my respons to Econ, choosing a switch strategy might well influence your EV, however as I understand it, that was not waht was stated in the original problem. Correct me if I'm wrong.

@Tom: Are you saying that the sampling method will unintentionally provide information on the value of envelope 2 and that in this manner the sampling method is significant to the problem?

Again, my question really comes down to it that my friend believes it's not a paradox, rather that argument 1 is based on a fallacy. How did he do? is he right?

Re: The envelope problem, and a possible solution
 

The question in the OP is: Is switching +EV? The answer is, it depends. It depends on (i) the amount you see and (ii) the way in which those amounts were chosen by the person who put the money in there.

Both arguments are flawed. Argument 2 is flawed because it doesn't even answer the question. It answers the question, is always switching +EV? Argument 1 is flawed because, after seeing the first amount, there is not a 50% chance you have the higher amount. The probability depends on (i) and (ii) above. If you make assumptions about (ii), then you can calculate the EV. Sometimes it will be positive, sometimes it will be negative.

Your friend is wrong. He did not identify the flaw in Argument 1. Specifically, your friend thinks that the fact that the money was determined before you chose has something to do with the EV. But we know from our poker experience that this is a silly argument. The deck is shuffled and the order of the cards is determined prior to any of our bets. But that does not affect our EV calculations, which are based on the fact that we don't know this order, not that the order is yet to be determined.

Re: The envelope problem, and a possible solution
 

THank you for your reply Jason. You make a couple of very interesting points. You also seem to argue that the size of the value and the sampling method are of critical importance. A number of others have argued this, but I sill can't see the point. I suspect that I lack specific knowledge to fully appreciate those arguments. if you would be so kind, I'd really appreciate it if anyone could make a bit more of a detailed post such that laymen like me can understand.

The point about the relation to poker and point of determination is exactly the point that led me to question this explanation myself. However, there are some differences between pokersituations and this situation that might be of influence. I will not yet go into those here. But I do want to discuss that, because initially I was going to relate this post to exactly that. Which is why I started a new post before, instead of replying in the original thread.

Re: The envelope problem, and a possible solution
 

In the envelope problem, to come to any conclusion other than that of argument #2 requires you to make one or more assumptions which are not given in the original problem. If you cannot derive any useful information from viewing the content of one of the envelopes, then it doesn't matter if you switch or not. Argument #2 is flawed if and only if you somehow gain information beyond that presented in the original problem by viewing the contents of one of the envelopes.

The flaw in argument #1 is that it assumes that, regardless of the amount that we see in the envelope we open, there is a 50-50 chance that the amount in the other envelope is double this amount, as opposed to half this amount. This statement cannot possibly be true for all values that we may encounter in the envelope.

Put simply, when we see that our envelope contains $100, we know that the other envelope contains either $200 or $50. We do not know, however, the chances that it contains either amount.

Re: The envelope problem, and a possible solution
 

@BBB: thank you for your reply. You said: "The flaw in argument #1 is that it assumes that, regardless of the amount that we see in the envelope we open, there is a 50-50 chance that the amount in the other envelope is double this amount, as opposed to half this amount. This statement cannot possibly be true for all values that we may encounter in the envelope."

It is possible to make a distribution such that those chances were equal, according to your rational it would then always be +EV to switch. I think this is not right. The reason I think this can be illustrated by making explicit how I expect the game to look.

The way I understand it, you have two envelopes, one containing value N and the other the value 2N. One envelope is picked randomly, and then there is the option to switch or not. Let's say we always switch (for reasons I stated earlier). Most would probably agree that whether you switch is rather trivial.

In the second condition the only change we make is that we flip open one of the envelopes and reveal it's contents. Then we switch again.

Nowhere in the original problem is it stated that the value you're shown influences the probability that the other value will be higher or lower. It simply states that you had 50% chance to pick either N or 2N, and that now that you know one value, you have a 50% chance that you hold either N or 2N. Reasoning on from there they then state that your EV is 50% * value/2 + 50% * value*2.

If argument 1 is wrong, which I feel must be the case, then I think that it is because somewhere a logical step wasn't allowed.

For those who state that sometimes switching might affect EV, do you think that this would still be the case if you were switching blind. And do you think that seeing the value matters?

If not, then you probably agree that agrument 1 is probably flawed, where is it flawed?

Re: The envelope problem, and a possible solution
 

Okay, here's a little thought experiment. Imagine that I am the one placing the money in the envelopes and you are the one choosing. There are infinitely many methods that I may use to decide what money to place. I will present two of them for discussion.

The first I call Method 1. What I do is roll two dice. I multiply the sum by 100 and place the result in an envelope. I then flip a coin. If it's heads, I place twice that amount in the other envelope. If it's tails, I place half that amount in the other envelope.

The second is Method 2. It's the same as Method 1, except this time I roll only one die.

Now you open an envelope and it contains $800. What is the probability that the other envelope contains $1600?

First, suppose you somehow know that I used Method 1. Since you can see $800, you know that I either (a) rolled a 4 and flipped a head, (b) rolled an 8 and flipped a tail, or (c) rolled an 8 and flipped a head. Only in case (c) will the other envelope have $1600. The probabilities of these events are (a) 3/72, (b) 5/72, and (c) 5/72. Hence, the conditional probability of (c), given that you see $800, is 5/13. So the EV of switching is

(5/13)(1600) + (8/13)(400) = 861.5 &gt; 800

and it is +EV to switch.

Next, suppose you somehow know I used Method 2. Then it is impossible for the other envelope to contain $1600, so the EV of switching is only 400, and it is -EV to switch.

Of the infinitely many other methods I might use, some will generate neutral EV, some +EV, and some -EV. So the EV depends on the method, which is not known to us, since it's not stated in the problem. This is why the only right answer is "it depends" (both on the method and on the amount of money you see in the first envelope).

Now, one might be tempted to argue that since we don't know the method, then it's EV neutral to switch. But to say this, one would have to assign probabilities to all of the different methods and do so in such a way so that the resulting overall EV was 0. Such an assignment of probabilities is clearly not given in the problem.

Finally, for BBB's sake, let me repeat what I said before about Argument 2. It is EV neutral to always switch. Argument 2 is correct about that. But the problem did not ask about the EV of always switching. It asks about the EV of switching.

P.S. Suppose I use Method 2 and you open an envelope and discover $200. Then you can compute that the probability of the other envelope containing $400 is 1/2. So it may sometimes happen that the chance of the other envelope being greater is 50-50. But (under Method 2), this will only happen when you see certain values. It will not always be the case that the chances are 50-50. One of the main points of this puzzle is that there is no selection method that will always generate a 50-50 chance in this situation.

Re: The envelope problem, and a possible solution
 

Actually, (at least in part) the trick is that the EV for envelopes where it 'makes sense' to switch is undefined or infinite.

For example, let's say I keep flipping a coin, and count h, the number of consecutive times it comes up heads. Then I put $(3^h) in one envelope, and $(3^(h+1)) in another. What's the expected value of an envelope?

Re: The envelope problem, and a possible solution
 

The EV of the envelope is not relevant. What matters is the conditional EV of switching, which is always well-defined and finite. It "makes sense" to switch if the conditional probability, p, that the other envelope is larger, satisfies p&gt;1/3. The value p is always a well-defined number between 0 and 1.

Re: The envelope problem, and a possible solution
 

[ QUOTE ]

Nowhere in the original problem is it stated that the value you're shown influences the probability that the other value will be higher or lower. It simply states that you had 50% chance to pick either N or 2N, and that now that you know one value, you have a 50% chance that you hold either N or 2N. Reasoning on from there they then state that your EV is 50% * value/2 + 50% * value*2.


[/ QUOTE ]

Hi Naobisdad,

All we are told in the original problem is that one of the envelopes contains twice the other. This tells us that once we open one envelope, that the other contains either half as much or twice as much. It tells us nothing about the chances that the other envelope contains twice as much, and to simply assume that these chances are 50% is not correct.

As an example, suppose that the designer of the game opted to put 50 and 100 in the envelopes (but that we don't know this). Clearly, it is -EV to switch in this case if we open the envelop with 100, as there would be a zero percent chance that it would be 200, and a 100 percent chance that it would be 50. Basically, the two envelopes contain X and 2X. The EV of switching no matter what will be -X if we chose the 2X envelope, and X if we chose the 2X envelope. Since there's a 50% chance of chossing either envelope, the net EV of switching no matter what is zero.

As a more involved example, suppose that we're told that the numbers were chosen by randomly selecting some integer k between 0 and K (where K is some positive integer that we are not told), and then placing $2^k in one envelop and $2^(k+1) in the other. Then, clearly, we should swtich if our envelope contains $1. But, what about for all other values? Suppose our envelope contains $2 or more, and we've decided to switch no matter what. Then, if our envelope contained $2^K, we would automatically lose $2^(K-1) by switching. If our envelope contained less than $2^K but more than $1, chances would be 50-50 that the other envelope contained more, so we would expect to gain on average 1/2 the amount in our envelope by switching (i.e., argument #1 from the original post applies for those cases). So, the total EV of switching no matter what would be: Sum(i=1..K-1){1/2*2^i} - 2^(K-1) + 1 (the +1 at the end is because we always gain $1 when we see $1 in the envelope and switch). The first term is a geometric series (minus the i=0 term); the summation is: 1/2*(1-2^K)/(-1)-1/2, or 2^(K-1)-1. So the total EV of a switch = 2^(K-1)-1 + 2^(K-1) - 1, with equals zero. So, actually, in this setup, we should switch if the envelope we open contains $1 (since the EV for switching in this case is +$1), and we should keep our envelope otherwise (since the net EV for all other cases is -$1). Note that in this example, argument #1 (50% gain by switching) holds for most of the cases, but this is offset by the 50% loss which happens when switch the largest possible amount, which offsets the 50% gains in the cases of the smaller amounts.

To prove that, without inferring any other information about the problem, that switching is EV neutral, consider this: Suppose the deal is that we're splitting the money with another player, with whom we were not allowed to discuss a strategy beforehand. After we choose, the envelopes will be resealed and presented to the other player. He will be handed the envelope that we chose, will open it, and will decide which envelope to keep. If it were +EV to always switch, then we would of course have switched. But from his perspective, it would also be +EV to switch. Except that clearly it cannot be +EV to switch twice, since you end up back at the same envelope. So, absent any other information, switching must be EV neutral.

Re: The envelope problem, and a possible solution
 

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The EV of the envelope is not relevant. What matters is the conditional EV of switching, which is always well-defined and finite.

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Let's say I pull the envelope, and find that it contains 8 trillion dollars. Now, there is only about 1 trillion dollars in currency in circulation, so there's a problem. If the EV goes to infinity, then, for any scarce resource, there is a non-zero probability that the envelope contains more of the resource than exists, ergo, if the amount in the envelope is close to that (or larger) it makes hardly any sense to switch envelopes.

[ QUOTE ]
It "makes sense" to switch if the conditional probability, p, that the other envelope is larger, satisfies p&gt;1/3. The value p is always a well-defined number between 0 and 1.

[/ QUOTE ]

Actually, that's incorrect. It makes sense to switch if the average expected value (EV) of the other envelope is larger than the value of the held envelope.

Re: The envelope problem, and a possible solution
 

Okay this is what i'm thinking. Let's say you pick an envelope and recieve 100 dollars. Now we know for a fact that we (before we began) had an EV of +75 (50 and 100)OR an EV of +150(100 and 200). From now on switching will not change our original EV (either +75 or +150) it will only tell us which one we had. So when/if we switch we don't actually change our EV, we only believe we do. So instead of thinking of it as gambling 50 to win 100, you're actually gambling an EV of +75 OR +150 to win an EV of +75 or +150. You win and lose nothing by switching. EV=0

This is still somewhat flawed and doesn't really answer the paradox comepletely, but hey i tried :P

Re: The envelope problem, and a possible solution
 

[ QUOTE ]
[ QUOTE ]
It "makes sense" to switch if the conditional probability, p, that the other envelope is larger, satisfies p&gt;1/3. The value p is always a well-defined number between 0 and 1.

[/ QUOTE ]

Actually, that's incorrect. It makes sense to switch if the average expected value (EV) of the other envelope is larger than the value of the held envelope.

[/ QUOTE ]
If x is the value of the held envelope and p is as I defined it, then the EV of the other envelope is

2xp + 0.5x(1 - p) = x(0.5 + 1.5p),

which will be greater than x if and only if

1 &lt; 0.5 + 1.5p,

which is true if and only if p&gt;1/3.

Re: The envelope problem, and a possible solution
 

[ QUOTE ]
Recently I read a post on the 2 envelopes paradox ( Here).

I will not elaborate on what the problem is, because it is stated and discussed clearly in the provided link and I ask you to please read the OP. I want to discuss a solution put forward by a friend of mine.

A friend of mine argued the following:
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If we were te simulate the envelope problem, it would show that looking at the amount of money in your envelope is not going to affect the EV of always switchin envelopes. Which would be in line with the argument that switching envelopes is neutral EV. He asked me, "if we were to play this game, do you expect that looking at the amount would change anything to you EV?" And I answered: "I do not see how always switching can possibly become anything other than neutral EV, when the only change you make is that you are merely passivily looking at the value now". After which he followed:

Most likely the argument that switching is EV neutral is the correct one. The other argument then cannot be correct, so somewhere in the argument there must be flaw. He argued that the point of determination is key. The amounts in the envelope are determined before the start of the game, not after you learn the value.

If the value of the second envelope was determined after you had learned the value of the first (for example by flipping a coin) then idd switching would be +EV. This is comparable to saying, I give you $100, you can keep it, or I flip a coin; heads you double it, tails you lose half.

However, this is not the case in the envelope problem he said. The values are assigned to the envelopes on beforehand. Even though intuitively it may feel like a switch might double your money or cut it in half after you learn the value; in reality there is only one possible outcome. For example, if you look at the value in envelope 1, and it is $100. Then intuitively it might feel like you have a 50% shot at $50 or $200, but you don't have it. Say the ammount in the second envelope was $50, it would not be possible to get $200 on that trial. The flaw would boil down to using the $100 as new information in determining your EV when it doesn't provide any new information really. So, even though the method of determining your EV is right in argument 1, it is not applicable in that situation.

What I am wondering is, how is this line of reasoning? He is fully convinced that it is the solution to the problem, and a colleague of mine basically suggests the same solution.
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I still am not sure, I'd appreciate some imput.

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I agree with your friend and his colleague under the assumption that the amounts in the envelopes are Fixed. That is, it is NOT the case that the game can be repeated by picking NEW amounts in the envelopes according to some unknown probability distribution. Instead, the game can only be repeated by offering the SAME envelopes to a number of NEW people, none of which know their contents. Under this assumption, each time a Person opens an envelope and sees an amount, the conditional probability of the second envelope being double the amount he sees is either 100% or 0% - he just doesn't know which.

PairTheBoard

Re: The envelope problem, and a possible solution
 

[ QUOTE ]
I got frustrated with Aaron in this dialogue before. Sure am glad you started it up again. [img]/images/graemlins/smile.gif[/img]

Here's how my dialogue would go when I hear the problem.

Paradoxer: "I put a random amount X in one envelope and 2X in the other."
Me: "Wait, how was X chosen?"
P: "I'm not telling you. Now, you open and find $100. Should you switch?"
M: "Man, I dunno. I can tell you the correct move if you describe the opponent."
P: "I'm not going to tell you. So, what's the best move?!"

It's retarded. Of course you and I can arrive at different answers if we try to fill-in the gaping hole differently. That doesn't show that the problem's clever; all that shows is that the problem has gaping holes.
-Sam

[/ QUOTE ]
I understand your frustration because I refuse to answer the question. I think the problem is people assume there must be a right answer, and then use that to reason, usually to the idea that you shouldn't switch. My objection is that there is only a right answer if you assume a lot of mathematical machinery, perhaps consistent finite prior beliefs or the existence of an expected value.

It's not just theoretical hair-splitting, as the FX example and years in Purgatory example show. You can make reasonable assumptions that make it +EV to switch. The useful thing is to understand why both the switching and indifference arguments have force, and when to follow one versus the other.

I also understand it's frustrating that life forces us to play a game without clearly specified rules or odds. But don't blame me for that one. The point of this example is not to prove it's a clever question, it's to help you think about real decisions in which you don't how amounts were selected for the envelopes.

Don't argue with me, argue with the Clash:

Clash: "Should I stay or should I go? If I go there will be trouble."
SamIam: "Wait, how was the amount of trouble chosen?"
Clash: "If I stay there will be double."
SamIAM: "Man, I dunno. I can tell you the correct move if you describe the opponent."
Clash: "So you gotta let me know. Should I stay or should I go?"

Sorry SamIAM, you have to choose.

Re: The envelope problem, and a possible solution
 

Your intuition is correct of course and it's EV neutral - at least for meta-beings in a universe where all amounts of money are equally likely.

It's funny seeing Aaron's avatar here instead of a finance forum where I usually read his wisdom. I assume he's making the point that once you switch to real money in this world and real people with utility functions then it's not so clear-cut. Suppose the amount in the envelope was 1/2 of all the money in the world for example?

Re: The envelope problem, and a possible solution
 

Aaron, there is something that puzzles me about your reply regarding argument 1. Under the assumption that the method used for figuring EV in this argument is correct, then it would seem to me, that without any further assumptions that switching would always be EV.

Or am I now implicitly making the assumption that the value (say N) is an element of a set of positive real numbers?

My point is that as stated, argument 1 seems to argue that knowing the ammount will lead to a change in EV irrespective of the size of the ammount in the envelope. That is, the player will change strategy despite the fact that he/she didn't really get new information. Also, as stated, it would seem that unless N was a negative value, switching would always be EV.

What am I missing?

Re: The envelope problem, and a possible solution
 

Hey PTB,

you said: "I agree with your friend and his colleague under the assumption that the amounts in the envelopes are Fixed."

COnsider the following. If I made 100 pairs of envelopes with variable ammounts, such that each pair of envelopes contains N and 2N values. We play the game. You choose an envelope at random. And you then switch blind, or you don't. Would you predict that now switching affects EV?

Also would you think that your EV would be affected if now after picking an envelope you are now shown the value that ethe envelope contained?

Re: The envelope problem, and a possible solution
 

Hi, Masquerade. From your statement I deduce that you then agree that argument 1 must be flawed somewhere. Where do you think that is?

Re: The envelope problem, and a possible solution
 

Hello, Jason. thank you again for an elaborate reply.

As I understand it the crux of your post is that if your know the value in one envelope, then given the sampling method this knowledge will affect the chance that the value in the other envelope will be either double or half the amount you now see.

however, in the original problem I don't believe it is argued that the value is used in that way. Rather, a very straightforward, simple modification of estimating EV is used. And the probability with which the amount in the other envelope is N or 2N seems dependent on the probability with which you pick an envelope.

Then you said: "It is EV neutral to always switch. Argument 2 is correct about that. But the problem did not ask about the EV of always switching. It asks about the EV of switching."

I still don't understand, and others have argued this point also and I didn't understand then either, if argument 1 is correct, why then would you ever NOT switch?

Re: The envelope problem, and a possible solution
 

Hi, BBB. You said: "All we are told in the original problem is that one of the envelopes contains twice the other. It tells us nothing about the chances that the other envelope contains twice as much, and to simply assume that these chances are 50% is not correct."

Do you then imply that we choose an envelope in a non random way? Or do you think that those probabilities do not depend on the probability with which you choose the initial envelope? I would think that I have 50% chance op picking up N or 2N.

"The EV of switching no matter what will be -X if we chose the 2X envelope, and X if we chose the 2X envelope. Since there's a 50% chance of chossing either envelope, the net EV of switching no matter what is zero."

I agree here, and as I read it, it is the pointe of your post. Although in the more involved example you do point to the fact that the size of the ammount can matter.

If you conclude that barring other information switching is neutral EV, then you must believe that as stated in the original problem the reasoning in argument 1 is false. However from your post I do not understand how you think this is so.

Re: The envelope problem, and a possible solution
 

Previous envelope 2+2 discussion

Lori

Re: The envelope problem, and a possible solution
 

[ QUOTE ]
Hey PTB,

you said: "I agree with your friend and his colleague under the assumption that the amounts in the envelopes are Fixed."

COnsider the following. If I made 100 pairs of envelopes with variable ammounts, such that each pair of envelopes contains N and 2N values. We play the game. You choose an envelope at random. And you then switch blind, or you don't. Would you predict that now switching affects EV?

Also would you think that your EV would be affected if now after picking an envelope you are now shown the value that ethe envelope contained?

[/ QUOTE ]

No and No.

You ask, where does argument 1 break down? It's in the conclusion once seeing the amount A in Env 1, that chances are 50-50 that Env 2 has 2A. With Fixed envelope amounts the chances are either 0% or 100% you just don't know which. That's the principle that negates argument 1. If the amounts come from a population of 100 envelope pairs - or some other unknown probabilty distribution - then the conditional probabilty that Env 2 has 2A is still something other than 50% in general, and just like above you don't know what it is. It's the same principle negating argument 1. The principle is just more apparent in the Fixed Amount case.

PairTheBoard

Re: The envelope problem, and a possible solution
 

[ QUOTE ]
Hi, BBB. You said: "All we are told in the original problem is that one of the envelopes contains twice the other. It tells us nothing about the chances that the other envelope contains twice as much, and to simply assume that these chances are 50% is not correct."

Do you then imply that we choose an envelope in a non random way? Or do you think that those probabilities do not depend on the probability with which you choose the initial envelope? I would think that I have 50% chance op picking up N or 2N.

[/ QUOTE ]

NarobisDad,

You do have a 50% chance of initially picking up on N or 2N. And clearly, once you choose an envelope, there is a 50% chance that the other envelope contains N more than the one you chose, and 50% that it contains N less. If you do not gain or infer any usable information upon viewing the contents of your envelope, then there is still a 50% chance that the other envelope contains N more than yours, and a 50% chance that it contains N less. If you see $100, you now know that N is either $50 or $100. So argument 1 concludes: Since there's a 50% chance that the other envelope contains N more than mine (which is true whether N is $100 or $50), then there's a 50% chance that N contains $200, and a 50% chance that it contains $50. The first part of the preceding statement is correct, BUT THE CONCLUSION IS NOT VALID. The conclusion is only correct if N is equally likely to be $100 or $50 (which may or may not be the case).

If we have some basis of information on which to determine the probability that N is $100 versus the probability that it is $50, for example using what we know in general about how people such as our benefactor might be willing to put in envelopes and give away, it turns out that we should clearly switch if we determine that the probability that N is $100 is more than half the probability that N is $50, and we should clearly not switch if N is more that twice as likely to be $50 as it is to be $100. But to simply guess that N is just as likely to be $100 as it is to be $50 and going from there is totally baseless and meaningless. That would be like if I told you I had a coin in my pocket that was not necessarily fair and I asked you what were the chances that it would come up heads if I flipped it. Unless you had some kind of information on which to determine what the coin might be like, it would be totally meaningless for you to gess 1/2 just because that would be the answer if the coin were fair.

Re: The envelope problem, and a possible solution
 

No, I'm saying something different.

The actual envelope problem doesn't come up a lot. The point of this paradox is not to figure out what to do if it happens, it's to show you that two common reasoning devices can conflict, so you have to be careful using them.

For example, one common argument for "resolving" the envelope paradox starts by assuming you have a consistent prior distribution on amounts that might be in the envelope, then demonstrating that if you always switch, that distribution must have infinite expectation. If you expect infinity, any finite amount will be a disappointment, so you switch.

That's fine, but in reality we don't have consistent prior beliefs about everything that might happen. Assuming we do in this case is hiding from the paradox, not resolving it.

We often have problems of the sort, do I stop now or try for more at the risk of losing what I have? We rarely have consistent prior beliefs that would give us a full Bayesian solution. But we still have to decide.

Consider Maurice Kratchik's original formulation that started all of this. He knows nothing about fashion or style. He's wearing a necktie and another fashion-impaired nerd offers him a bet. They'll show their neckties to a style expert, and the with the worse necktie will win both ties. He reasons, "I will either lose my necktie or win a better tie, my chances of winning are equal, so I should take the bet." Of course, the other person can reason the same way.

Now where is the consistent set of Bayesian beliefs? Is there some "tie quality" that would make you take the bet or not? Is it entirely obvious that this bet cannot make both people better off? Does it imply infinite expectations to take the bet?

If we didn't have the paradox to teach us, most of us would unhesitatingly agree that (a) when you pick one of two envelopes with different amounts at random, there is a 50% chance that you'll get the one with the higher amount, and (b) if you have a 50/50 chance to win $2 or lose $1 you have positive expected EV. The fact that these "obvious" statements can conflict is important. Deciding which one is more reliable in the envelope example is not important, unless you get offered a lot of bets about amounts in envelopes.

Re: The envelope problem, and a possible solution
 

I'm not sure why your "No" was in reply to my post Aaron.

[ QUOTE ]
For example, one common argument for "resolving" the envelope paradox starts by assuming you have a consistent prior distribution on amounts that might be in the envelope, then demonstrating that if you always switch, that distribution must have infinite expectation.

[/ QUOTE ]

I think you misspoke here. I don't think this makes sense. I do believe it has been shown that if the assumed prior distribution has finite expectation then always switching is EV neutral. Also, if the prior distribution does not have finite expectation then it's possible for Always Switching to produce infinite expectations. But then so does Never Switching. Even then, One infinity does not improve over the other.

[ QUOTE ]
That's fine, but in reality we don't have consistent prior beliefs about everything that might happen. Assuming we do in this case is hiding from the paradox, not resolving it.


[/ QUOTE ]

I have no idea what you mean by "consistent prior beliefs".
However, assuming a prior distribution for the envelopes to have finite expectation is probably pretty safe since there's only a finite amount of money in the world.



[ QUOTE ]
"I will either lose my necktie or win a better tie, my chances of winning are equal, so I should take the bet." Of course, the other person can reason the same way.


[/ QUOTE ]

Or they might reason, "I will either win the better tie which I don't have or lose the better tie that I do have."

You also give an example in another post of betting $1000 against 500 pounds that the dollar will - I believe - drop against the pound. You claim the EV is positive for both parties. The key here is that you are making computations based on a variable unit measure.

[ QUOTE ]
If we didn't have the paradox to teach us, most of us would unhesitatingly agree that (a) when you pick one of two envelopes with different amounts at random, there is a 50% chance that you'll get the one with the higher amount, and (b) if you have a 50/50 chance to win $2 or lose $1 you have positive expected EV. The fact that these "obvious" statements can conflict is important.

[/ QUOTE ]

I don't think these statements are in "conflict". If you pick an envelope (a) is true. (b) is true on it's own merits. But if you open an envelope and see $2 what's not true is that you have a 50-50 chance to win $2 or lose $1. That statement is not implied by the previous one. It's not a matter of conflicting modes of reason or chosing which is more reliable. It's a matter of being careful about what implies what.


The point I would drive home here is that when you talk of probablities and expectations, think of what experiment or game you could be repeating under identical conditions whereby you could see the probabilties/expectations prove out over numerous trials. This is why I gave the most simple example where the prior distribution for the envelopes is simply the delta distribution. The conditions that brought about the envelope amounts were predetermined and Fixed. If we want to talk probablities and expectations on that basis we can do so by using the same envelope amounts and letting numerous people play the game, none of whom know the contents. In that scenario it's easy to see how the paradox evaporates.

PairTheBoard

Re: The envelope problem, and a possible solution
 

For your first point, these are two different ways of saying the same thing, but with an essential difference of order. I say the common argument is "if you always switch, that distribution must have infinite expectation." You say "if the assumed prior distribution has finite expectation, then always switching is EV neutral."

The difference is I start my summary of this argument with the assumption of a consistent prior distribution, which I object to. You start with "it has been shown" and bury the word "assumed" as an adjective in the middle.

This gets to your second point. We agree (I think) that if you have consistent prior beliefs about everything, as required for Bayesian statistical calculations, that either switching is EV neutral or you have infinite prior expectation. It's not clear that infinite prior expectation is irrational, despite there being only a finite amount of money in the world. For one thing, maybe there's an infinite amount in the universe. For another, you don't an infnite realization to support an infinite expectation.

But that's theory stuff, my concern is that people don't have consistent prior beliefs. Bayesian statistics achieves consistency by ignoring things that are important to real decision-making. What is a young person's prior about his lifetime income or amount of marital happiness? Isn't the choice of going to college or dumping a girlfriend somewhat like the envelope choice? Do you think understanding the Bayesian "resolution" of the paradox will help him decide?

Sometimes our best information about possible outcomes comes from the first choice we are offered. We have no idea what things cost until someone offers us something. We instinctively turn it down, because we realize the chance of the first offer being the best is pretty low. We get a few more offers, until someone offers us something that appears to be a low price relative to what we've been seeing from the others. Isn't this a pretty good description of how you make some decisions? Would you be surprised if I proved that it shows you have an infinite prior expectation?

Yes, the USD/GBP example involves two people measuring expectation in different units. I think this is very common in real decisions. Even if we are both betting money, if our credit ratings are different, then we are not computing in the same units. And since the money is presumably a means to some end, if there is price uncertainty about what we plan to spend it on, there's also a difference.

I don't say (a) and (b) imply each other, I say that if you assume both, you get the envelope paradox. The Bayesian resolution rejects (a). It says that after seeing the amount in the envelope you pick you have to change your estimate of the probability it is the higher amount. Bayesians define probability as subjective belief, and subjective belief can change even when objective facts don't (looking at the amount in the envelope can't change whether it is or isn't the higher amount). That's fine for feeling intellectually consistent, but rejecting (a) puts you at a big disadvantage when faced with practical choices. It's a very useful principle and refusing to use it on the grounds of theory is foolish. But the paradox demonstrates that it's not universally reliable, which is why the paradox is valuable.

The other common "resolution" of the paradox rejects (b). It insists on zero EV for a zero sum exchange. Before you look at your envelope the EV of exchange is clearly zero, objectivists don't believe looking at the amount can change the EV. What they give up is the idea that $1 is always $1. $1 before you look at the envelope is something different from $1 afterwards. Again, they've jettisoned some important common sense in order to get to the result they know is correct.

The Bayesians get consistency, at the cost of making some stupid decisions, the objectivists get the right decisions, at the cost of consistency. But both of them have to torture reality in the process. Both ways of thinking are useful sometimes, but we have the envelope process to remind us how shaky their philosophic foundations are.

Re: The envelope problem, and a possible solution
 

[ QUOTE ]
For your first point, these are two different ways of saying the same thing, but with an essential difference of order. I say the common argument is "if you always switch, that distribution must have infinite expectation." You say "if the assumed prior distribution has finite expectation, then always switching is EV neutral."


[/ QUOTE ]

And I insist that my statement makes sense while yours does not. You say, "if you always switch, that distribution must have infinite expectation". That's simply not true. The prior distribution can have finite expectation AND you can always switch. You have mispoken. Your statement is incomplete. I suppose I should guess that what you mean to say is, "If a prior distribution is assumed and it's assumed that always switching is not EV neutral then the prior distribution cannot have finite expectation". That statement is implied by mine. I agree the form is different. I don't think the difference is essential.

I agree this is one approach to resolving the paradox - whether the person taking it adopts your statement of the fact or mine. It speaks to one of the hidden assumptions people mistakenly make when they think that envelope amounts can be chosen from a uniform distribution on an infinite scale of money. It's really an elaboration on that point and thus succeeds in breaking down the paradox at that false premise.

However, I don't think it gets to the heart of the paradox. One of the best explanations I've seen for this was made by BBB when he said,

[ QUOTE ]
BBB -
If we have some basis of information on which to determine the probability that N is $100 versus the probability that it is $50, for example using what we know in general about how people such as our benefactor might be willing to put in envelopes and give away, it turns out that we should clearly switch if we determine that the probability that N is $100 is more than half the probability that N is $50, and we should clearly not switch if N is more that twice as likely to be $50 as it is to be $100. But to simply guess that N is just as likely to be $100 as it is to be $50 and going from there is totally baseless and meaningless. That would be like if I told you I had a coin in my pocket that was not necessarily fair and I asked you what were the chances that it would come up heads if I flipped it. Unless you had some kind of information on which to determine what the coin might be like, it would be totally meaningless for you to gess 1/2 just because that would be the answer if the coin were fair.

[/ QUOTE ]


You said,
[ QUOTE ]
This gets to your second point. We agree (I think) that if you have consistent prior beliefs about everything, as required for Bayesian statistical calculations, that either switching is EV neutral or you have infinite prior expectation. It's not clear that infinite prior expectation is irrational, despite there being only a finite amount of money in the world. For one thing, maybe there's an infinite amount in the universe. For another, you don't an infnite realization to support an infinite expectation.


[/ QUOTE ]

I still don't know what you mean by "consistent prior beliefs about everything". You have yet to explain this. It may be that you are not dealing in the realm of mathematics. If that's the case you should be more clear about where you depart from mathematics and explain more precisely just what you are saying in whatever realm you are dealing. I believe there is some controversy about certain so called "Baysian" approaches which lack sound mathematical foundations. I suppose if this is where you're at there's not much basis for argument with you. We will have to take your pronouncements on authority.


[ QUOTE ]
But that's theory stuff, my concern is that people don't have consistent prior beliefs. Bayesian statistics achieves consistency by ignoring things that are important to real decision-making. What is a young person's prior about his lifetime income or amount of marital happiness? Isn't the choice of going to college or dumping a girlfriend somewhat like the envelope choice? Do you think understanding the Bayesian "resolution" of the paradox will help him decide?


[/ QUOTE ]

"Isn't the choice of going to college or dumping a girlfriend somewhat like the envelope choice?"

I see no relationship between the two.

[ QUOTE ]
Sometimes our best information about possible outcomes comes from the first choice we are offered. We have no idea what things cost until someone offers us something. We instinctively turn it down, because we realize the chance of the first offer being the best is pretty low. We get a few more offers, until someone offers us something that appears to be a low price relative to what we've been seeing from the others. Isn't this a pretty good description of how you make some decisions? Would you be surprised if I proved that it shows you have an infinite prior expectation?


[/ QUOTE ]

You would have to explain your model more precisely for us to see its implications. I don't see how it relates to the envelope paradox.


[ QUOTE ]
Yes, the USD/GBP example involves two people measuring expectation in different units. I think this is very common in real decisions. Even if we are both betting money, if our credit ratings are different, then we are not computing in the same units. And since the money is presumably a means to some end, if there is price uncertainty about what we plan to spend it on, there's also a difference.


[/ QUOTE ]

Maybe people do this. But I don't see how the observation that people often do bad math relates to the envelope paradox.

[ QUOTE ]
I don't say (a) and (b) imply each other, I say that if you assume both, you get the envelope paradox.

[/ QUOTE ]

Let's be clear here. Your statement of (a) and (b) are

(a) when you pick one of two envelopes with different amounts at random, there is a 50% chance that you'll get the one with the higher amount

(b) if you have a 50/50 chance to win $2 or lose $1 you have positive expected EV.

Now you say, "if you assume both, you get the envelope paradox"

There's no need to assume (b). (b) is simply a true statement. Also, if by (a) you mean that if you chose one of the two closed envelopes there's a 50-50 chance it will have the larger amount then that is simply a true statement as well. It's you who are missing the heart of the paradox. It's right here. It's when someone opens an envelope, sees $2 in it and says that (a) implies the premise of (b) that they are making a mistake.

Instead, you remain unclear as to what you mean by (a) so that you can conclude,

[ QUOTE ]
The Bayesian resolution rejects (a). It says that after seeing the amount in the envelope you pick you have to change your estimate of the probability it is the higher amount.

[/ QUOTE ]

There's no need to reject (a) unless you were unclear as to what it meant. If you are clear with (a) and mean that if you chose one of the two closed envelopes there's a 50-50 chance it will have the larger amount then there is a repeatable experiment you have in mind that can prove out this statement of probabilty. The experiment is for numerous people to pick one of the two envelopes at random. The more people who pick, the closer you will see half picking the higher amount and half the lower amount. There's a 50-50 chance that any one of them will pick the higher amount.

Consider this 2 envelope game. One envelope contains a dollar and the other envelope contains nothing. Again you have an
(a) when you pick one of two closed envelopes at random, there is a 50% chance that you'll get the one with the dollar

and a
(b) The expectation for a person picking an envelope at random is 50 cents.

Now a person picks an envelope, opens it sees it contains nothing and is given the option of switching. (a) remains true as clearly stated. (b) remains true. There's no deep philisophical conflict here between subjective Baysians and Objectivists. There's simply elementary conditional probabilty saying that once seeing the content of the first envelope the probablity for the contents of the second changes from the apriori probabilty. The exact same thing happens in the Original Two Envelope Problem. It's just not as obvious because people tend to percieve all envelope amounts as being equally likely - which we know they are not under any apriori conditions.


PairTheBoard

Re: The envelope problem, and a possible solution
 

I think we each have stated our positions as clearly as we can, and still think the other misunderstands us. Not much point in going forward, but I will pick out a few sentences that demonstrate our disagreement.

You seem to treat this as a paradox designed by people who have envelopes and don't know whether or not to switch. It is in fact a paradox designed to demonstrate that common statistical reasoning can lead to contradictory conclusions.

[ QUOTE ]
I still don't know what you mean by "consistent prior beliefs about everything".

[/ QUOTE ]
Bayesian statistics defines probability as subjective belief. It assumes you can always determine a prior belief before the experiment (in this case, before the envelope is opened). The information in the experiment (in this case, the amount in the envelope) adjusts that belief to the posterior distribution. To a Bayesian, showing that no consistent prior distribution justifies switching regardless of the amount you observe, shows that always switching is irrational. My objection is that in real decisions, people often don't have consistent prior distributions.

[ QUOTE ]
I don't see how the observation that people often do bad math relates to the envelope paradox.

[/ QUOTE ]
It's not bad math not to have a consistent prior distribution. It's impossible to have consistent beliefs about everything. It's not bad math to compute expected value in different units from someone else. Everyone has different units if you look carefully enough.

[ QUOTE ]
There's no need to assume (b). (b) is simply a true statement. Also, if by (a) you mean that if you chose one of the two closed envelopes there's a 50-50 chance it will have the larger amount then that is simply a true statement as well.

[/ QUOTE ]
You insist both statements are true and regard them as not only beyond argument, they are so obvious, they are beyond the need to state as assumptions. The paradox was designed to teach you they cannot be both true all the time.

There are senses in which each of them are true, but they are different senses.

[ QUOTE ]
if you chose one of the two closed envelopes there's a 50-50 chance it will have the larger amount then there is a repeatable experiment you have in mind that can prove out this statement of probabilty. The experiment is for numerous people to pick one of the two envelopes at random. The more people who pick, the closer you will see half picking the higher amount and half the lower amount. There's a 50-50 chance that any one of them will pick the higher amount.

[/ QUOTE ]
This is the frequentist argument. If that's your definition of probability, then (b) is not always true. This is the criticism that led to Bayesians to reject that definition of probability. Bayesians achieve consistency by rejecting (a) instead.

I understand the desire to make both (a) and (b) always true, but no one has discovered a way to do it.

When you have nothing in one of the envelopes, (b) no longer applies. It only covers cases of $1 and $2, not $1 and $0. You can't reconstruct the paradox in this case.

Re: The envelope problem, and a possible solution
 

[ QUOTE ]
You seem to treat this as a paradox designed by people who have envelopes and don't know whether or not to switch.

[/ QUOTE ]

Not at all.

[ QUOTE ]
It is in fact a paradox designed to demonstrate that common statistical reasoning can lead to contradictory conclusions.


[/ QUOTE ]

I would say it's designed to show the need to be careful when applying mathematical tools like expected value and probabilties. When you speak in those terms you need to have a clear mathematical model in mind to which you are applying them.

[ QUOTE ]
Bayesian statistics defines probability as subjective belief.

[/ QUOTE ]

This is why Mathematical Probabilists and Statisticians don't attend conferences of such so called Bayesians. I would define probability as a branch of mathematics and expect the full rigor of mathematics in its application. There's nothing subjective in such rigor.

[ QUOTE ]
To a Bayesian, showing that no consistent prior distribution justifies switching regardless of the amount you observe, shows that always switching is irrational. My objection is that in real decisions, people often don't have consistent prior distributions.


[/ QUOTE ]

I really don't know much about the Bayesians. My impression is that they sort of create a prior distribution out of thin air, do sampling, then apply actual mathematical Bayesian techniques to adjust the original distribution depending on the sample. They then claim to have evidence for the real original distribution when what they really have are conclusions based on their original assumptions.

If that's the case I don't blame you for having a bone to pick with them. However I don't think it's central to the two Envelope Paradox. In the 2E paradox people are simply not being careful in how they apply the mathematical tools of probabilty and expectation.

However, if you'd like to explain more fully and clearly what Bayesians do and give examples of real life situations where you would disagree with their techniques due to "no consistent prior distribution" - whatever that means - I'd be interested in reading it.

[ QUOTE ]
It's not bad math to compute expected value in different units from someone else. Everyone has different units if you look carefully enough.


[/ QUOTE ]

You misquoted me here. It was "variable" units I objected to. Computing expected value with a unit of measure that changes over time and treating it as if it's fixed. That's not only bad math but presenting conclusions from such a calculation to make your case is downright misleading.

[ QUOTE ]
"PTB -
There's no need to assume (b). (b) is simply a true statement. Also, if by (a) you mean that if you chose one of the two closed envelopes there's a 50-50 chance it will have the larger amount then that is simply a true statement as well."

You insist both statements are true and regard them as not only beyond argument, they are so obvious, they are beyond the need to state as assumptions. The paradox was designed to teach you they cannot be both true all the time.


[/ QUOTE ]

Not at all. The paradox was designed to force you to make the statements in a clear way and to understand what you mean by them. When you do so they are both true as I showed in my last post. The paradox forces you to see where you mistakenly use them to jump to a false conclusion as I also showed in that post. If you are going to quote me, quote this rather than the phrase above out of context.

[ QUOTE ]
PTB -
" if you chose one of the two closed envelopes there's a 50-50 chance it will have the larger amount then there is a repeatable experiment you have in mind that can prove out this statement of probabilty. The experiment is for numerous people to pick one of the two envelopes at random. The more people who pick, the closer you will see half picking the higher amount and half the lower amount. There's a 50-50 chance that any one of them will pick the higher amount."


This is the frequentist argument. If that's your definition of probability, then (b) is not always true. This is the criticism that led to Bayesians to reject that definition of probability. Bayesians achieve consistency by rejecting (a) instead.


[/ QUOTE ]

I am not defining probabilty by way of frequencies. Probabilty has rigorous mathematical definitions. However, out of those definitions comes the law of large numbers which produces such frequencies. This is mathematical probabilty which is what we are usually talking about in this forum. If the Bayesians work with something else they call subjective probability then you need to clearly explain what they do and why it matters.

You say, "This is the frequentist argument. If that's your definition of probability, then (b) is not always true."

Yet look at (b) as you stated it,

(b) if you have a 50/50 chance to win $2 or lose $1 you have positive expected EV.

I challenge you to show this as not true under a standard mathematical definition of probabilty - where the law of large numbers - aka frequency - holds.

You say, "This is the criticism that led to Bayesians to reject that definition of probability. Bayesians achieve consistency by rejecting (a) instead."

Then what is a Bayesian definition of a probabilty space? I can give you a standard mathematical definition of a probabilty space for which the law of large numbers - aka frequencies - will hold. Having rejected this, What is the Bayesian definition of a probabilty space? Do they actually have something they can do math on?


[ QUOTE ]
I understand the desire to make both (a) and (b) always true, but no one has discovered a way to do it.


[/ QUOTE ]

I think I have done it simply by stating them clearly and understanding what they mean. I agree though that there's no way that all misinterpretations of them can be made to be true.

[ QUOTE ]
When you have nothing in one of the envelopes, (b) no longer applies. It only covers cases of $1 and $2, not $1 and $0. You can't reconstruct the paradox in this case.

[/ QUOTE ]

You missed my point here. If the Bayesians reject (a) on principle then they should reject it in this case as well. Yet it's clear there's no good reason to reject a clearly stated and well understood (a) that says all participants choosing one of the two envelopes at random have a 50-50 chance of choosing the $1 envelope. Just because Mr Lucky opens an envelope, sees it's empty, and now has a better than 50% chance of getting the $1 envelope if he switches there's no good reason to reject the statement (a) that future participants have a 50-50 chance of choosing a $1 envelope. Yet the Basesians should do so according to you on the same principle they rejected it in the 2E case, because (a) can't always be true? Because (a) is no longer true for Mr Lucky? Right. Because (a) can no longer be true when misconstrued.


PairTheBoard

Re: The envelope problem, and a possible solution
 

AAron, I would love to see you explaining more about "bayesians" point of view here. For now I agree with PTB post and my impressions are that your position on that matter is inconsistent in several places.

Best wishes

Re: The envelope problem, and a possible solution
 

Dear, PTB. You state that the place where argument 1 breaks down is at the part where it assumes 50% probablility. However, unless the probabilities now take on specific values, switching will still be either +EV or -EV.

If this is true, then that leaves a bit of a strange situation. I could switch, and then after I did, switch again on the basis of the exact same reasoning, and then again, and again, and in the end I'd have to be rich.

This is basically stating what has been stated before, argument 1 would create +EV situations from both ends of the switch.

How do you reason your perspective deals with this problem?

@BBB, thank you as well for your elaborate explanation. As far as I can see my reply to PTB is also valid as a response to your post and I ask you the same thing.
If I missed something in your post that would show that the above does not relate well to your story, please let me know.

Re: The envelope problem, and a possible solution
 

Practicing Bayesian statisticians use the same rigorous mathematics (in particular, the same definition of a probability space) as we probabilists do. Here's a toy example inspired by poker.

Let m be my winrate in BB/100 and s my standard deviation. I could model the outcome of a single 100 hand session by

X = m + sN,

where N is a Normal(0,1). Suppose I know that s=15, but I don't know m. However, I have "beliefs" about m. I believe that m=2. Of course, I'm not certain of it, it could be higher or lower. But I am fairly certain that I am at least a winning player. I might decide to model these beliefs by saying that

m = 2 + M,

where M is a Normal(0,1), independent of N. In "reality", m is a fixed, deterministic number, but I am representing it as a random variable in order to model my beliefs. Similarly, in the envelope paradox, the amounts in the envelopes are fixed, deterministic values, but we may choose to represent them as random variables in order to model our beliefs about the contents of those envelopes.

Now suppose I play a session of poker and lose 100 BB. I now need to adjust my "beliefs". This, of course, is analogous to the act of opening one of those envelopes. The adjustment is done by computing the conditional law of m, given X, and plugging in the observed value X=-100. We get the conditional CDF by computing

P(m &lt;= x | X).

Upon computing this expression and differentiating, we find that the conditional law of m is again normal. According to my hasty calculations, its mean is

(225/226)*2 + (1/226)*X

and its variance is simply 225/226 (regardless of the value of X). Plugging in X=-100, our updated "belief" about our winrate is that it is 350/226=1.55 and our "belief" has a standard deviation of about 0.998.

Obviously, the process gets more complicated when we try to apply it to a sequence of sessions, and we might find it useful to apply the mathematical tools of filtering theory. One aspect of this that I find interesting, as it relates to poker, is that you can get analogs of confidence intervals which appear much more realistic over a smaller sample size. Of course, those estimates always contain some residue of our original "beliefs". For what it's worth, that residue goes away in the limit as the sample size increases.