#### 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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