[dukemagic]

Huh I kind of assumed this thread had died off so I didn't check on it. Good to see all of the responses.

I was satisfied with the explanation of my friend and of what I think Midge was saying above:

Prior to opening either envelope, assign one envelope the value X and the other 2X. When you draw one it doesn't matter whether or not you switch, since you are 50/50 of having either one and so you stand to either gain X or lose X by switching. That is, if you have X and you switch, you gain X to now have 2X; the opposite is true if you have the envelope with 2X.

Why can't that be true after you open the envelope? You still have either X or 2X, with equal probability. The problem I was making before was then assigning new values to the second envelope of .5x and 2x, which you obviously can't do correctly, because that assigns a fourth possible value to the 2nd envelope (now you have 4 values - .5X, X, 2X, and 4X, whereas you started with only X and 2X).

So it seems to me that a simple explanation is that switching after opening is neutral EV. You stand to gain either X (if you had X and switch to 2X) or lose X (if you had 2X and switch to X). Since we are keeping the values in the envelopes consistent throughout the problem and you are indeed initially 50/50 to have either envelope, does this not make switching neutral EV?

Sorry if I'm rehashing a point we already disproved or something - I don't have enough practice with probability notation and math to understand the later responses.