A problem with the standard solution of the two envelopes paradox

In summary: The problem is that if we don't switch, our expected profit is $15. If we switch, our expected profit is $20 * P(other=20|one=10) * P(one=10) + $10 * P(other=10|one=20) * P(one=20) which is $20 + $10 = $30. This is the same as the $15 that we would have earned if we had not switched. Therefore, it doesn't matter whether we switch or not.
  • #1
Fredrik
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The problem

You have two indistinguishable envelopes in front of you. Both of them contain money, one of them twice as much as the other. You're allowed to choose one of the envelopes and keep whatever's in it. So you pick one at random, and then you hesitate and think "Maybe I should pick the other one". Let's call the amount in the envelope that you picked first A. Then the other envelope contains either 2A or A/2. Both amounts are equally likely, so the expectation value of the amount in the other envelope is

[tex]\frac{1}{2}\cdot 2A+\frac{1}{2}\cdot \frac{A}{2}=A+\frac{A}{4}=\frac{5}{4}A[/tex]

Since this is more than the amount in the first envelope, we should definitely switch.

This conclusion is of course absurd. It can't possibly matter if we switch or not, since the envelopes are indistinguishable. Also, if the smaller amount is X, then both expectation values are equal to

[tex]\frac{1}{2}\cdot X+\frac{1}{2}\cdot 2X=\frac{3}{2}X[/tex]

so it doesn't matter if we switch or not.

The "paradox" is that we have two calculations that look correct, but only one of them can be. The problem isn't to figure out which one is wrong, because it's obviously the first one. The problem is to figure out what's wrong with it.The solution?

There are several online articles about this problem. Some of them are just wrong, and the rest all agree that what I'm about to say solves the problem. (I'm actually presenting a simplified version of their argument here. This simplified argument has the advantage that it can be understood by people who don't know Bayes's theorem).

If the first envelope contains A, then there's some probability P that we were given envelopes that contain A and 2A. The probability that we were given envelopes that contain A/2 and A is then 1-P. The expectation value of what's in the other envelope is

[tex]P\cdot 2A+(1-P)\cdot \frac{A}{2}[/tex]

We got the result 5/4*A because we assumed that P=1/2, but is that really true? P depends on something like a function Q such that Q(x) is the probability that the smaller amount is x. We can think of the function Q as representing the method that was used to prepare the envelopes. If we know Q, we can calculate P.

[tex]P=\frac{Q(A)}{Q(A)+Q(A/2)}=\frac{1}{1+\frac{Q(A/2)}{Q(A)}}[/tex]

This means that P is only 1/2 if Q(A)=Q(A/2). Since the expression we're using for the expected value is correct no matter what A is, then this result is true no matter what A is. So we have Q(x)=Q(x/2) for all x. But then the sum of all Q(x) is infinite. This sum must be 1 since it's the sum of the probabilities of all possible results, so we have a contradiction.

This proves that there's no way to prepare the envelopes so that the probability P is 1/2 (which is necessary to make the expected value equal to 5/4*A).

This is where all the articles stop, and my problem begins. It bothers me that there's also no way to prepare the envelopes so that the expectation value is equal to A (which is necessary to reach the conclusion that it doesn't matter if we switch). If we assume that the expectation value is A, the condition we get on Q is Q(A)=1/2*Q(A/2). This is even more absurd than Q(A)=Q(A/2).

It's seems to me that something is very wrong here. Am I missing something here, or is the standard solution of the two envelopes problem completely incorrect?
 
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  • #2
There aren't 3 amounts A, 2A and A/2. There are 2 amounts, say $10 and $20. Say you picked the one with $10, then the probability that the other has $20 is 1:

P(other=20|one=10) = P(other=20 and one=10)/P(one=10) = 0.5/0.5 = 1.

If you picked the one with the $20, then the probability that the other has $10 is again 1.

If you don't switch, then your expected profit is $15. If you were to switch, your expected profit would be:

$20 * P(other=20|one=10) * P(one=10) + $10 * P(other=10|one=20) * P(one=20)

= $20 * 1 * 0.5 + $10 * 1 * 0.5 = $15,

which is identical to your "X" calculation.
 
  • #3
Perhaps I'm naive but it seems to me that the initial formula is wrong because A's value is not consistent. A is initially the cheaper envelope's value but later in the same formula it is the more expensive envelope's value. You can't take A out as a factor.

So the formula should be:

(1/2)A_1 + (1/2)A_2
= (1/2)(A_1 + A_2)
= (1/2)(3A) (with A the lesser value)
= 3a/2
 
  • #4
I don't see how you can gloss over a Bayesian piece of mathematics, since the key here is the distinction between prior and posterior distributions.
 
  • #5
matt grime said:
I don't see how you can gloss over a Bayesian piece of mathematics, since the key here is the distinction between prior and posterior distributions.
I will present the Bayesian argument here so that you can see that it adds nothing interesting to what I've already said:

We can't take for granted that the EV of the amount in the other envelope is 1/2*2A+1/2*A/2, because the correct expression is actually

P(we chose the smaller amount|the envelope we chose contains A) * 2A
+ P(we chose the larger amount|the envelope we chose contains A) * A/2

and the conditional probabilities in this expression aren't well defined because they depend on the prior distribution. So to make progress here, we have to assume that a prior distribution has been specified.

Suppose that a discrete distribution has been specified. This is effectively a function Q with the property that Q(x) is the probability that the smaller amount is x. Now we can calculate the conditional probabilities using Bayes's theorem.

I will write P(small|A) instead of P(we chose the smaller amount|the envelope we chose contains A).

P(small|A) = P(A|small) * P(small) / ( P(A|small) * P(small) + P(A|large) * P(large) )

[tex]=\frac{Q(A)\cdot\frac{1}{2}}{Q(A)\cdot\frac{1}{2}+Q(\frac{A}{2})\cdot\frac{1}{2}}=\frac{1}{1+\frac{Q(\frac{A}{2})}{Q(A)}}[/tex]

This is only 1/2 if Q(A)=Q(A/2).

Etc.

As you can see, the Bayesian approach is just a more difficult way to find the condition Q(A)=Q(A/2).
 
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  • #6
verty said:
Perhaps I'm naive but it seems to me that the initial formula is wrong because A's value is not consistent. A is initially the cheaper envelope's value but later in the same formula it is the more expensive envelope's value.
You're making the same mistake as the Wikipedia article. Once we have defined A to be the amount in the first envelope, that's what A is. To say that A has different values in two places is equivalent to saying "let's ignore this calculation and do another one instead".
 
  • #7
EnumaElish said:
...which is identical to your "X" calculation.
The challenge isn't to find a way to get the correct result, it's to figure out what's wrong with the calculation that tells us to switch.

EnumaElish said:
There aren't 3 amounts A, 2A and A/2. There are 2 amounts, say $10 and $20.
Did you mean that this is what's wrong with it? Then why is it wrong? (And why don't any of the mathematicians that have written about it think it's wrong).
 
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  • #8
wouldn't you agree that X would = mystery envelope since the amount in the selected envelope in unknown. And since you have a 50-50 chance of having selected the 10$ (A) amount or the 20$(B) amount it would be somthing like:

X <or> A/.5 + B/.5 if you want to make a huge problem out of it because you can then break this down more and more and make it look that much more complicated.

But simply: X is an unopened envelope that may have 10$ or 20$ in it but unopened-----

X = 0

Y the opened envelope would look like... Y = 10$ or Y = 20$ which would be the amount you would find in the envelope...You have to opend it sometime so no matter what envelope you select you would have a 50-50 chance of it being Y=10$ and you would have a 50-50 chance of it being Y=20$...:) In this equation you should never come up with Y=15$ that is just plain wrong.
 
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  • #9
Why are you changing the notation? I defined "X" to be the smaller amount, but in your post you're using the symbols "A" and "Y" for that quantity. You're also using "X" for the expected value of the amount in the second envelope. And I assume that when you write "/.0.5", you mean "*1/2".
 
  • #10
Fredrik said:
The problem
Let's call the amount in the envelope that you picked first A. Then the other envelope contains either 2A or A/2. Both amounts are equally likely, so the expectation value of the amount in the other envelope is

[tex]\frac{1}{2}\cdot 2A+\frac{1}{2}\cdot \frac{A}{2}=A+\frac{A}{4}=\frac{5}{4}A[/tex]
That is plain wrong.

If the value in the envelope is A then the only information one can derive is that the other envelope contains 2A if and only if A is the low value and it contains A/2 if and only if A is the high value. Since the person who picks does not know if A represents the low or the high value he cannot use both 2A and A/2 in his calculations at the same time.
 
  • #11
MeJennifer said:
Since the person who picks does not know if A represents the low or the high value he cannot use both 2A and A/2 in his calculations at the same time.
Why not? I'm not saying that you're wrong, only that you haven't proven that you're right.

Mathematicians who have considered this problem do not agree with you. They seem to agree that the expected value can be expressed as (some probability)*2A+(some probability)*A/2.

MeJennifer said:
the other envelope contains 2A if and only if A is the low value and it contains A/2 if and only if A is the high value.
Yes, so there's a non-zero probability for both of these possibilities. If the probability of the first possibility is P, then the probability of the second possibility is 1-P. You can calculate these probabilities by using Bayes's theorem, or directly from the initial distribution. You can see the details in my posts above.
 
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  • #12
Fredrik said:
Why not? I'm not saying that you're wrong, only that you haven't proven that you're right.

Mathematicians who have considered this problem do not agree with you. They seem to agree that the expected value can be expressed as (some probability)*2A+(some probability)*A/2.


Yes, so there's a non-zero probability for both of these possibilities. If the probability of the first possibility is P, then the probability of the second possibility is 1-P. You can calculate these probabilities by using Bayes's theorem, or directly from the initial distribution. You can see the details in my posts above.
You can only infer that if you picked the highest value the other envelope contains half and if you picked the lowest value the other envelope contains double. But the point is that you do not know what you picked.
So yes you can call it A but you still do not know what A is in relation to the value in the other envelope.
By using 2A and A/2 you simply state all possibilites as if the choice was not already made. But the point is that the choice was already made so either 2A or A/2 is invalid.
 
  • #13
The "paradox" is that you're mixing up numbers with random variables.

Let A denote the amount in the envelope you choose. (A is a random variable!)
Let B denote the amount in the envelope you didn't choose. (B is a random variable!)

It is true that P(B = 2A) = P(2B = A) = 1/2.

However, once you fix a particular monetary value, say a, then

P(B = 2A | A = a) is (usually) not 1/2.
 
  • #14
Hurkyl said:
The "paradox" is that you're mixing up numbers with random variables.

Let A denote the amount in the envelope you choose. (A is a random variable!)
Let B denote the amount in the envelope you didn't choose. (B is a random variable!)

It is true that P(B = 2A) = P(2B = A) = 1/2.

However, once you fix a particular monetary value, say a, then

P(B = 2A | A = a) is (usually) not 1/2.
Could not have expressed it better.
 
  • #15
Fredrik said:
The challenge isn't to find a way to get the correct result, it's to figure out what's wrong with the calculation that tells us to switch.Did you mean that this is what's wrong with it? Then why is it wrong? (And why don't any of the mathematicians that have written about it think it's wrong).
Read what MeJennifer and verty have posted.
 
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  • #16
Fredrik said:
Mathematicians who have considered this problem do not agree with you. They seem to agree that the expected value can be expressed as (some probability)*2A+(some probability)*A/2.
Can you post who these mathematicians are, and what they have written?
 
  • #17
Note that if we have the following problem:

Someone is given a sum of money, call it A. Now a proposition is made to this person. He has the option of gambling A, if he wins he gets 2A but if he loses he will only receive A/2. His odds are 50% of receiving 2A and 50% of receiving A/2.

Now here the mentioned formula is applicable:
[tex]\frac{1}{2}\cdot 2A+\frac{1}{2}\cdot \frac{A}{2}=A+\frac{A}{4}=\frac{5}{4}A[/tex]

He in fact should take the bet!

So when you realize what the difference is with this problem and the original problem then perhaps you will realize why this formula is incorrect for the originally mentioned problem.
 
  • #18
http://www.maa.org/devlin/devlin_0708_04.html

might be of interest. Notice the citing of Bayes' theorem to do precisely what you want to do; and knowing that method X is wrong doesn't tell you why it is wrong or what is more correct.

You mostly appear to be arguing on the same side - the calculation is wrong becuase you're using the wrong distributions on the posteriors, or what someone I think has called the Q function needed to take account of you not knowing how the envelopes were set up. Though that may not have been what they meant.
 
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  • #19
here check this site out it may sove your problem just fine...:)

http://www.math.hmc.edu/funfacts/ffiles/20001.6-8.shtml
 
  • #20
EnumaElish said:
Read what MeJennifer and verty have posted.
Verty just made a mistake. MeJennifer has brought up a few valid points, but she hasn't solved the problem.

EnumaElish said:
Can you post who these mathematicians are, and what they have written?
There are lots of references in the Wikipedia article. Devlin's article is the one I found the most useful. http://www.maa.org/devlin/devlin_0708_04.html .
 
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  • #21
matt grime said:
http://www.maa.org/devlin/devlin_0708_04.html

might be of interest. Notice the citing of Bayes' theorem to do precisely what you want to do; and knowing that method X is wrong doesn't tell you why it is wrong or what is more correct.

You mostly appear to be arguing on the same side...
Not only that, we're saying the same thing. The article that you found is the article that taught me the stuff I wrote about in posts #1 and #5 in this thread. I call this the "standard solution" of the two envelopes paradox because lots of other articles about this problem are using the same method.

I explained in post #1 why I find it difficult to accept this method as a valid solution of this problem (and I will explain it again in my answer to Hurkyl below).
 
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  • #22
Hurkyl said:
The "paradox" is that you're mixing up numbers with random variables.

Let A denote the amount in the envelope you choose. (A is a random variable!)
Let B denote the amount in the envelope you didn't choose. (B is a random variable!)

It is true that P(B = 2A) = P(2B = A) = 1/2.

However, once you fix a particular monetary value, say a, then

P(B = 2A | A = a) is (usually) not 1/2.
I agree with all of this, and I like your notation. It's a good idea to distinguish between the random variable A and its value a.

What you said is the starting point of what I call the "standard solution" of this problem. I presented the full argument in posts #1 and #5.

This is a very short summary of the "standard solution", using your notation:

The expected value of what you called B is

E(B) = P(B=2A|A=a) * 2a + P(2B=A|A=a) * a/2

Bayes's theorem (or a more direct argument) tells us how these conditional probabilites depend on the initial distribution (which I assume to be discrete so that I can introduce a function Q such that Q(x) is the probability that the smaller amount is x).

P(B=2A|A=a) = P(A=a|B=2A) * P(B=2A) / ( P(A=a|B=2A) * P(B=2A) + P(A=a|2B=A) * P(2B=A) )

[tex]=\frac{Q(a)\cdot\frac{1}{2}}{Q(a)\cdot\frac{1}{2}+Q(\frac{a}{2})\cdot\frac{1}{2}}=\frac{1}{1+\frac{Q(\frac{a}{2})}{Q(a)}}[/tex]

This is only 1/2 if Q(a)=Q(a/2). The authors of the articles I've read agree that this proves that the probabilities are not 1/2. If the probabilities aren't 1/2, then E(B) is not 5/4*a.

What bothers me about this argument is that if this whole line of reasoning is valid, then we should be able to show that E(B)=a. But if E(B)=a, then Q(a)=1/2*Q(a/2) and that's certainly no less absurd than Q(a)=Q(a/2).
 
  • #23
Why should we be able to show this from that line of reasoning? As the article points out, the expected winnings depend on the distribution, which we don't know. All we know is there is no distribution which can yield a prob of 1/2 for all a.
 
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  • #24
The expected value of what you called B is

E(B) = P(B=2A|A=a) * 2a + P(2B=A|A=a) * a/2
That's not right; this computes the expected value of B given that A = a. Or, notationally,

E(B | A = a) = P(B=2A|A=a) * 2a + P(2B=A|A=a) * a/2​

However, it is true that E(B) = E(A).
 
  • #25
Fredrik said:
What bothers me about this argument is that if this whole line of reasoning is valid, then we should be able to show that E(B)=a.

matt grime said:
Why should we be able to show this from that line of reasoning? As the article points out, the expected winnings depend on the distribution, which we don't know. All we know is there is no distribution which can yield a prob of 1/2 for all a.
First of all, I'm not saying that we should be able to show this. I don't know if we should or not. What I'm saying is that the reason why the article is concerned with probabilities =1/2 is that no other value makes the expected value equal to 5/4*a. That means that the people who write these articles believe that the formula for the expected value is valid. If it is, and the calculation yields something other than a, then it matters if we switch or not! But it clearly can't matter, since the envelopes are indistinguishable.

Observation 1: If the expected value is 5/4*a, Q must satisfy Q(a)=Q(a/2).
Observation 2: If the expected value is a, Q must satisfy Q(a)=1/2*Q(a/2).

If we can dismiss the result 5/4*a simply because we have made observation 1, can't we also dismiss the result a, simply because we have made observation 2?
 
  • #26
Hurkyl said:
That's not right; this computes the expected value of B given that A = a. Or, notationally,

E(B | A = a) = P(B=2A|A=a) * 2a + P(2B=A|A=a) * a/2​

However, it is true that E(B) = E(A).
OK, thanks for pointing that out. I allowed myself to use my old sloppy terminology in the reply to Matt that I just posted, but I won't do that again.
 
  • #27
Fredrik said:
First of all, I'm not saying that we should be able to show this. I don't know if we should or not. What I'm saying is that the reason why the article is concerned with probabilities =1/2 is that no other value makes the expected value equal to 5/4*a. That means that the people who write these articles believe that the formula for the expected value is valid. If it is, and the calculation yields something other than a, then it matters if we switch or not! But it clearly can't matter, since the envelopes are indistinguishable.

Observation 1: If the expected value is 5/4*a, Q must satisfy Q(a)=Q(a/2).
Observation 2: If the expected value is a, Q must satisfy Q(a)=1/2*Q(a/2).

If we can dismiss the result 5/4*a simply because we have made observation 1, can't we also dismiss the result a, simply because we have made observation 2?

They don't choose 1/2 so as to *make* the expected winnings 5/4 times as big. They make that 'naive' guess because it is 'common sense'. With this 'common sense' guess they get the answer they do. You have another case of mixing up prior and posterior things.It is also nonsense, but that is besides the point. Or possibly the main point, depending on how you view these things.

Please point out who is saying that the expected value after switching (given we know a) is a? No one is. The expected value after switching depends on the distrobution that we do not know. IT could go up, could go down. But it depends on the distribution that we do not understand.

The claim in the paradox is that always goes up for all observed a, whcih is not true.
 
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  • #28
matt grime said:
They don't choose 1/2 so as to *make* the expected winnings 5/4 times as big. They make that 'naive' guess because it is 'common sense'.
Why would they do that? "Common sense" is irrelevant here. They are trying to "solve" the two envelopes problem. To "solve" it means to show the readers exactly what's wrong with the calculation that says that we should switch. That's why they "chose" 1/2, not because it's common sense, but because it's the probability that appears in the calculation that says that we should switch.

matt grime said:
Please point out who is saying that the expected value after switching (given we know a) is a? No one is. The expected value after switching depends on the distrobution that we do not know. IT could go up, could go down. But it depends on the distribution that we do not understand.
How do you think Devlin (for example) came up with the formula for what he calls the "expected gain"? (It would be more proper to call it "expected gain given A=a". I will do that from now on). Devlin claims that the expected gain given A=a, is

P(B=2A|A=a) * 2a + P(A=2B|A=a) * a/2 - a

I'm using Hurkyl's notation here. The sum of the first two terms is E(B|A=a) and the last term is of course -E(A|A=a). So Devlin's claim is that the expected gain given A=a can be computed as

E(B|A=a) - E(A|A=a)

Devlin's "solution" to the the problem is this: The reason that we thought that the expected gain can be calculated as 1/2*2a+1/2*a/2-a = 1/4*a is that we used the prior probabilites instead of the posterior probabilites, i.e. that we tried to calculate the expected gain given A=a as

P(B=2A) * 2a + P(A=2B) * a/2 - a

instead of using the correct expression above.

Devlin also explains that the conditional probabilites in the correct expression for the expected gain given A=a depend on the prior distribution. He then shows that they can't be 1/2 for all a.

I will try one more time to explain why this is less than satisfying to me. Devlin claims that he has solved the problem. To solve it means to identify the mistake in the argument that told us to switch. Devlin says that that mistake is that we used the prior probabilites instead of the posterior probabilites. That implies (in the strictest logical sense of the word) that if we had used the posterior probabilites, we would have found that it makes no difference if we switch or not. However, it is not possible to prove that it makes no difference if we switch or not by calculating the expected gain given A=a. So what Devlin claims is the solution of this problem isn't a solution of this problem.

I guess that answers what I've been asking all along. :smile: The "standard solution" of the two envelopes problem is incorrect, because it's concerned with how the calculation of the expected gain given A=a is done. This is irrelevant, because a calculation of this quantity can't tell us what the correct decision is.

Edit: A calculation of the expected gain given A=a can tell us what the correct decision is, but only if we know both the value of a and the prior distribution (i.e. the method was used to prepare the envelopes). If we know these things, the problem is a completely different problem than the one we started with.
 
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  • #29
Fredrik said:
What bothers me about this argument is that if this whole line of reasoning is valid, then we should be able to show that E(B|A=a)=a.
(my edit in blue)

Fredrik said:
Devlin says that that mistake is that we used the prior probabilites instead of the posterior probabilites. That implies (in the strictest logical sense of the word) that if we had used the posterior probabilites, we would have found that it makes no difference if we switch or not.

I don't follow either of these assertions.
 
  • #30
Fredrik said:
Observation 1: If the expected value is 5/4*a, Q must satisfy Q(a)=Q(a/2).
Observation 2: If the expected value is a, Q must satisfy Q(a)=1/2*Q(a/2).
Here's how I'd go about this without invoking Bayes.

The present calculation proceeds in 3 steps:
(1) We pick an envelope and claim its expected value is a.
(2) Then we calculate the exp. value of the other: a/2 + (a/2)/2 = 5a/4.
(3) Finally we compare the two expected values: a < 5a/4.

There must be some probability p that would make the expected value of the first envelope equal to a. To be consistent with Step (2), we should use p=1/2. By observation 2, p cannot be 1/2; contradiction.

Notice, I did not use Obs. 1, I am not sure that you need to use both Obs. 1 and 2. Using both may amount to making the same point twice.
 
  • #31
Fredrik said:
Why would they do that? "Common sense" is irrelevant here. They are trying to "solve" the two envelopes problem. To "solve" it means to show the readers exactly what's wrong with the calculation that says that we should switch. That's why they "chose" 1/2, not because it's common sense, but because it's the probability that appears in the calculation that says that we should switch.

You're confusing things I have said. The reason why it appears paradoxical is becuase 1/2 appears to be the correct probability to use from a naive vewi point. The explanation is that this is not using any correct posterior distribution.

Devlin's "solution"

try using 'explanation of the paradox'

Devlin also explains that the conditional probabilites in the correct expression for the expected gain given A=a depend on the prior distribution. He then shows that they can't be 1/2 for all a.

just remember you said this..

I will try one more time to explain why this is less than satisfying to me. Devlin claims that he has solved the problem. To solve it means to identify the mistake in the argument that told us to switch. Devlin says that that mistake is that we used the prior probabilites instead of the posterior probabilites. That implies (in the strictest logical sense of the word) that if we had used the posterior probabilites, we would have found that it makes no difference if we switch or not.

No it doesn't. As you yourself point out, different posteriors give different expected gains on the switch depending on the observed value.

Edit: A calculation of the expected gain given A=a can tell us what the correct decision is, but only if we know both the value of a and the prior distribution (i.e. the method was used to prepare the envelopes). If we know these things, the problem is a completely different problem than the one we started with.

According to you the expected gain is zero, always. Which is incorrect.There are key points in the paradox.

1) the expected gain given the observed amount must be by a factor of 5/4
2) this is independent of making the observation, thus every swap multiplies your winnings by 5/4.

These just are not true statements.

If we did not open the envelope, the the expected value in it is 5/4*(smaller value), which is coincidentally the same as the expectation of switching, hence there is no value in changing. But given we do observe the amount in the envelope this is no longer valid. We don't know what is the correct answer because we don't know the posterior.
 
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  • #32
matt grime said:
The reason why it appears paradoxical is becuase 1/2 appears to be the correct probability to use from a naive vewi point.
No, it isn't. The reason is that a correct calculation of E(B|A=a)-E(A|A=a) will tell us what the correct decision is in a different situation. See my answer to Hurkyl below for more details on this.

matt grime said:
According to you the expected gain is zero, always. Which is incorrect.
That's not what I'm saying. I'm just saying if it isn't, then the details about how the probabilities depend on the prior distribution is irrelevant, and so is the fact that there's no prior distribution that makes the probabilities 1/2 for all a. See my answer to Hurkyl below for more details on this.

matt grime said:
1) the expected gain given the observed amount must be by a factor of 5/4
The result of the naive calculation of the expected gain given A=a is 1/4*a, because the the result of the naive calculation of the expected value of B given A=a is 5/4*a.

matt grime said:
If we did not open the envelope, the the expected value in it is 5/4*(smaller value), which is coincidentally the same as the expectation of switching,...
It's 3/2*(smaller value).

matt grime said:
...hence there is no value in changing. But given we do observe the amount in the envelope this is no longer valid. We don't know what is the correct answer because we don't know the posterior.
Knowledge of the amount in the first envelope changes nothing. We still know for sure that there's no value in changing (assuming that we have no information about the prior distribution). It is only if we learn both the amount and the prior distribution that the problem changes into another one. In this other (very different) problem we can expect to gain E(B|A=a)-a by switching.
 
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  • #33
Hurkyl said:
Fredrik said:
What bothers me about this argument is that if this whole line of reasoning is valid, then we should be able to show that E(B|A=a)=a.
(my edit in blue)
I know there are inaccuracies like this one in the posts I made before post #26. (You're quoting #22). I understand what I'm supposed to call things now. You made that clear in #24.
Hurkyl said:
Fredrik said:
Devlin says that that mistake is that we used the prior probabilites instead of the posterior probabilites. That implies (in the strictest logical sense of the word) that if we had used the posterior probabilites, we would have found that it makes no difference if we switch or not.
I don't follow either of these assertions.
Neither does Matt. I don't see why. You even say that you don't follow either of my assertions. That surprises me since there are only two sentences in the text you quoted and the first one is obviously true. Just look at what Devlin wrote:

Devlin said:
The correct expected gain calculation is:

P(B=2A|A=a) * 2a + P(A=2B|A=a) * a/2 - a​

The paradox above arose because you assumed that

P(B=2A|A=a) = P(A=2B|A=a) = 1/2​

...

To summarize: the paradox arises because you use the prior probabilities to calculate the expected gain rather than the posterior probabilities.
Those are his words. I have only changed the notation.

What I said in the second sentence you quoted is not obvious, but still not very difficult to see. The paradox arose because a naive (and incorrect) calculation of E(B|A=a)-E(A|A=a) yielded the result 1/4*a, which is positive. Now, why would a positive result cause a paradox? It doesn't. Not by itself. Only a person who believes that a calculation of E(B|A=a)-E(A|A=a) can tell us what the correct decision is in the situation that was described in the specification of the problem will conclude that a positive result is a paradox!

If you just tell this person that he got the probabilites wrong in the calculation of E(B|A=a)-E(A|A=a), have you really explained the paradox to him? Absolutely not. You also have to tell him that even if he gets the probabilities right, the result will only tell him what the correct decision is in a completely different situation (a situation in which both the prior distribution and the value a is known).

Now, if you tell him that, then that's all he needs to understand what caused the paradox. That whole thing about the probabilities not really being 1/2 is completely irrelevant. It adds nothing to his understanding of the resolution of the paradox. (It adds to his general knowledge, and is something he will probably find interesting, but that doesn't make it relevant).
 
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  • #34
I completely fail to see what it is that you state is missing from the explanation in, say, Devlin, that is actually missing. All of the things you claim are required for a full explanation of the paradox are in that linked article.

If you get the probabilities correct, then it tells you the correct answer in *every* situation, well, the *only* situation that there is. It really isn't hard: if you use the correct probabilities then you get the correct answer. If you don't you get junk. The explanation of the paradox is that the probabilities that are shoved in are nonsense. Devlin, et al, tell you what the correct things to put in are. What on Earth is the issue here? Since the probabilities to be put in are completely unknown, how can any solution tell you the expected earnings?
 
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  • #35
Fredrik said:
Neither does Matt. I don't see why. You even say that you don't follow either of my assertions. That surprises me since there are only two sentences in the text you quoted and the first one is obviously true.


No, it isn't. The 'first one' was that you asserted that the expected gain from swapping is zero. It isn't. We don't know what the expected gain is except in terms of a posterior distribution that we do not know.
 

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