One of the envelopes contains 2 glort's worth of dollars, and the other envelope contains 4 glorts' worth of dollars. I don't know the exchange ratio of glorts to dollars, except that it is positive.

Now it is clear that the expected value is 3 glorts' worth of dollars, regardless of the envelope picked. It's still possible that the game master is using some variant of "if the value of the glort is low for the duration of this trial, put the larger amount of money in envelope B; if the value of the glort is high, put the larger amount of money in envelope a" that favors one envelope over the other.

It is easy to prove that this implication is false, by considering a simple counter-example courtesy of VincentYu.

Even simpler counter example: X and Y take values 1 and 100, independently and with equal probability. Then E(X/Y) = 1/4 ( 1/1 + 1/100 + 100/100 + 100/1), which is basically 25.5. Ditto for E(Y/X).

Expectation is a mean, and in means, large terms dominate - for E(X/Y), the (X=100,Y=1) situation dominates, while for E(Y/X), the (X=1,Y=100) situation dominates.

In fact, if X and Y are independent, identically distributed, strictly positive and non-trivial (ie not just constants), then I think that we always have E(X/Y) > 1.

Nice.

I think you reached the wrong conclusion in your final paragraphs. Can you show how the expected value calculation could be different for "the amount of money contained in the other envelope expressed in terms of the amount of money in this envelope [which is in dollars, BTW]" versus "the amount of money in the other envelope expressed in dollars"?

I see that Wikipedia says that there is no generally accepted solution to the paradox. That almost certainly means people are interpreting it differently. I'll give my opinion. But let me recast the problem in more pointed terms.

Mr Moneybags writes a check for some real positive amount of money and puts it in an envelope. Then he writes a check for double the amount of the previous check and puts it in another envelope. He shuffles the envelopes. You pick one and open it, and you find $N. He offers to let you switch to the other just this once. Should you?

On first glance it appears that the other envelope could have $N/2 or $N*2, and that you are entirely ignorant about which is the case, so your ignorance priors are 1/2 for each possibility. Then the expected value calculation comes up all wrong! Call the hypothesis that you opened the lower value envelope L and the hypothesis that you opened the higher value envelope H.

E(switching)=P(H)*(N/2)+P(L)*(N*2)=(1/2)*(N/2)+(1/2)*(N*2)=(5/4)*N.  

But that's nonsense. My opinion as to what went wrong was that it's false that you are entirely ignorant after opening the envelope, because you can now do an update on your ignorance prior.

We'll start, of course, with Bayes' formula P(H|N)/P(L|N) = P(N|H)/P(N|L) * P(H)/P(L) * P(N)/P(N). The last ratio cancels. The second-to-last ratio also cancels because the shuffling of the envelopes made it equally likely you'd open either one. We know basically nothing about the distribution that the lower value was first drawn from, so a convenient ignorance prior on it is that it was some uniform distribution over the interval $A to $B. Consequently, the higher value would be described by a uniform distribution over the interval 2*$A to 2*$B. The distributions may or may not overlap. I'll take each case separately.

Common start: A=0.

P(N=k|H)={1/3 for 0<k<=B; 1 for B<k<=2B}
P(N=k|L)={2/3 for 0<k<=B; 0 for B<k<=2B}
Integrate each with respect to k from 0 to 2B, then plug in
P(H|N)/P(H|L) = (4/3*B)/(2/3*B) = 2
and convert back to probabilities
P(H|N)=2/3, P(L|N)=1/3.

Separate start: A>0 and 2A<B.

P(N=k|H)={0 for A<k<2A; 1/3 for 2A<=k<=B; 1 for B<k<=2B}
P(N=k|L)={1 for A<k<2A; 2/3 for 2A<=k<=B; 0 for B<k<=2B}
Integrate each with respect to k from A to 2B, then plug in
P(H|N)/P(H|L) = P(N|H)/P(N|L) = (4/3*B-2/3*A)/(2/3*B-1/3*A) = 2
and convert back to probabilities
P(H|N)=2/3, P(L|N)=1/3.

No overlap: B<2A.

P(N=k|H)={0 for A<=k<=B; 1 for 2A<=k<=2B}
P(N=k|L)={1 for A<=k<=B; 0 for 2A<=k<=2B}
Integrate each with respect to k from A to 2B, then plug in
P(H|N)/P(H|L) = P(N|H)/P(N|L) = (2B-2A)/(B-A) = 2
and convert back to probabilities
P(H|N)=2/3, P(L|N)=1/3.

So viewing your prize, you suddenly feel it's twice as likely that you got the higher amount. Bizarre, huh? And your expected value calculation becomes

E(switching)=P(H|N)*(N/2)+P(L|N)*(N*2)=(2/3)*(N/2)+(1/3)*(N*2)=N,

so there's no point switching after all. Even weirder is that, before you open the envelope, this is a situation where you know for sure what you will believe in the future, but you nevertheless can't make that update till then.

This is a really good exposition of the two envelopes problem. I recall reading a lot about that when I first heard it, and didn't feel that anything I read satisfactorily resolved it, which this does. I particularly liked the more precise recasting of the problem at the beginning.

(It sounds like some credit is also due to VincentYu.)

The flaw in the argument is simply that it assumes E(X/Y) > 1 implies that E(X) > E(Y).

I didn’t understand this sentence very well at first, because the inequality on the right is two steps removed from the one on the left. I find this version clearer:

The flaw in the argument is simply that it assumes E(X/Y) > 1 implies that E(X) / E(Y) > 1. (If E(X) / E(Y) > 1, that would imply that E(X) > E(Y).)

Here's a solution to a more general version of the problem:

Let's say that the red envelope contains N times as much money as the blue envelope with probability p, and the blue envelope contains N times as much money as the red envelope with probability (1 - p).

Without loss of generality, N is at least 1.

If N = 1, both envelopes contain the same amount, and there is no point in switching.

If N > 1, let the variable s represent the smaller amount of money between the amounts of the two envelopes. So one envelope contains s, and the other envelope contains Ns.

Scenario 1: The blue envelope contains s, and the red envelope contains Ns. This scenario occurs with probability p.

Scenario 2: The red envelope contains s, and the blue envelope contains Ns. This scenario occurs with probability (1 - p).

Assume s is not less than 0.

The expected amount of money in the blue envelope, E(B) = sp + Ns(1 - p) = s(p + N - Np).

The expected amount of money in the red envelope, E(R) = s(1 - p) + Nsp = s(1 - p + Np).

If s = 0, both envelopes contain the same amount, and there is no point in switching.

If s > 0, compare the expectations of the amounts of money in the envelopes in terms of s:

E(B) > E(R) when

s(p + N - Np) > s(1 - p + Np)

p + N - Np > 1 - p + Np

2p - 2Np > 1 - N

2p(1 - N) > 1 - N

2p < 1, because (1 - N) < 0

p < 1/2

Similarly, E(B) < E(R) when p > 1/2, and E(B) = E(R) when p = 1/2.

The ratios of the expected amounts of money in each envelope depend only on p when s > 0 and N > 1.

Both envelopes contain the same amount of money when s = 0, N = 1, or both.

So if your degree of belief that the red envelope contains more money than the blue envelope is the same as your degree of belief that the blue envelope contains more money than the red envelope, don't bother switching, unless you need to kill time (which you already knew intuitively, but Q.E.D.)

I got the idea of using the variable s to represent the smaller of the amounts in the two envelopes from R Falk 2008: "The Unrelenting Exchange Paradox".