I've been reading another of the references - Bruss, F. Thomas (1996). "The fallacy of the two-envelope problem". Mathematical Scientist. 21 (2): 112–119. There's no link provided, but I tracked it down (in the form of a scan) to -
https://www.researchgate.net/publication/266706577_The_fallacy_of_the_two_envelopes_problem
It's helpful stuff, and below is my attempt to rework elements of it into a useful overview, and apply it to our specific example (100,?).
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First, let's look at a game that does indeed give the infamous 5/4 expectation from switching envelopes -
There are two envelopes, labelled A and B.
An unknown amount, call it U, is placed in envelope A.
Based on the toss of a fair coin, one of two possible amounts (2U and U/2) is placed in envelope B.
The player is given envelope A and the option to switch. The expected value of envelope B is:
E(B) = 0.5(2U + U/2) = 1.25U.
So, in our example, A is revealed to contain 100 (U=100), and E(B) = 125.
Switching from A to B gives an average gain of +25.
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But now let's look at the actual two-envelope problem -
There are two amounts, S (for 'smaller') and 2S (twice as large).
They are placed inside identical envelopes, one of which is randomly given to the player.
Let's call the unknown amount in that envelope U.
There is a 50% chance that U=S and a 50% chance that U=2S. -
P(U=S) = P(U=2S) = 0.5
The expected value of the randomly chosen envelope the player currently holds is -
E(U) = 0.5(S+2S) = 1.5S
The expected value of switching to the other envelope (replacing U with S half the time, and with 2S the other half the time) is -
E(switch) = 0.5(S-U) + 0.5(2S-U) = 1.5S - U
We've seen that the expected value of U, E(U), is 1.5S. So E(E(switch)) = (1.5S - 1.5S) = 0. [The expected value of the expected value of switching is zero.]
We also, of course, knew from the beginning that the expected value of
both envelopes was 1.5S, there being nothing to distinguish between them. But that's beside the point.
In our example, the player has opened the envelope to reveal that U=100. Half the time this will be the smaller amount (U=S=100), half the time it will be the larger amount (U=2S, U=100, S=50).
At this point, there is an opportunity to make the well-known blunder all over again. It's tempting to take the bottom-line formula, 1.5S - U, and evaluate it when U=100, like so -
When U=100,S=50, contribution to E(switch) = (1.5S - 2S) = 75 - 100 = -25
When U=S=100, contribution to E(switch) = (1.5S - S) = 150 - 100 = +50
Summing these gives an overall EV of +25. D'oh!
We have once again blended the two distinct scenarios. We should instead say -
E(switch) = 1.5S - U
When U=2S, contribution to E(switch) = 1.5S - 2S = -0.5S
When U=S, contribution to E(switch) = 1.5S - S = +0.5S
So E(switch) = 0, regardless of the value of S.
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