[andrewshawcare]

Once you pick an envelope, you no longer stand to only gain money.

You can lose money and that has to be reflected in the potential value of each envelope.

After the first selection, you must express the envelope value as the potential of what each envelope holds (the probabilities from the initial selection) which makes selecting again a wash.

Let A = 50 Envelope 1 is 100 Envelope 2 is 25

First selection 1/2(100) + 1/2(25) = 62.5

Great! Do it! 62.5 is bigger than 0, which is the expected value of not playing at all.

After that, it makes no difference to switch (the envelope value is recursive):

1/2(62.5) + 1/2(62.5) = 62.5

or

1/2(1/2(2A) + 1/2(A/2)) + 1/2(1/2(2A) + 1/2(A/2)) = 1/2(2A) + 1/2(A/2) = 5/4A

and the trick is that’s now the potential expected value you have (you now have 5/4A in your envelope), so switching is a wash. You never really “had” A as a value to compare against (by evaluating that 5/4A is bigger than A), we just get tripped up with the doubling and halving at the outset.

Said another way, the fact that 5/4A is bigger than A is irrelevant, no envelope contains A, they both contain the expected value of 5/4A.

[lisper]

> Let A = 50 Envelope 1 is 100 Envelope 2 is 25

No. The problem specifies that E1 is twice E2, but here you have E1 = 4 x E2.

A is the amount in one of the envelopes, so if A=50 then either E1 is 50 or E2 is 50, and the other E is 25 or 100. But under no circumstances can E1=100 and E2=25 at the same time.

[andrewshawcare]

Impossible. Schrodinger’s envelope, then. You can’t reflect the probability to be 2A AND 1/2A if A represents the value of an envelope.

The value of the other envelope must include the probability that it is the original A (not some sleight of hand new A’ that, itself, is based on an expected value).

[jph]

Excellent answer. Worth adding to the wiki page.