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A problem with the standard solution of the two envelopes paradox

  1. Dec 7, 2006 #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?
     
    Last edited: Dec 7, 2006
  2. jcsd
  3. Dec 14, 2006 #2

    EnumaElish

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    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.
     
  4. Dec 14, 2006 #3

    verty

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    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
     
  5. Dec 14, 2006 #4

    matt grime

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    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.
     
  6. Dec 15, 2006 #5

    Fredrik

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    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).
     
    Last edited: Dec 15, 2006
  7. Dec 15, 2006 #6

    Fredrik

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    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".
     
  8. Dec 15, 2006 #7

    Fredrik

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    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).
     
    Last edited: Dec 15, 2006
  9. Dec 15, 2006 #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.
     
    Last edited: Dec 15, 2006
  10. Dec 15, 2006 #9

    Fredrik

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    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".
     
  11. Dec 15, 2006 #10
    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.
     
  12. Dec 15, 2006 #11

    Fredrik

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    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.
     
    Last edited: Dec 15, 2006
  13. Dec 15, 2006 #12
    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.
     
  14. Dec 15, 2006 #13

    Hurkyl

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    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.
     
  15. Dec 15, 2006 #14

    EnumaElish

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    Could not have expressed it better.
     
  16. Dec 15, 2006 #15

    EnumaElish

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    Read what MeJennifer and verty have posted.
     
    Last edited: Dec 15, 2006
  17. Dec 15, 2006 #16

    EnumaElish

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    Can you post who these mathematicians are, and what they have written?
     
  18. Dec 16, 2006 #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.
     
  19. Dec 16, 2006 #18

    matt grime

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    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.
     
    Last edited: Dec 16, 2006
  20. Dec 16, 2006 #19
  21. Dec 16, 2006 #20

    Fredrik

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    Verty just made a mistake. MeJennifer has brought up a few valid points, but she hasn't solved the problem.

    There are lots of references in the Wikipedia article. Devlin's article is the one I found the most useful. Link.
     
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