Suppose that someone put some money (say, an arbitrary unit) into one envelope and twice as much (two units) into another. You get to choose an envelope, and since you are ignorant of their contents you have, intuitively, a 50% chance of getting 1 (arbitrary unit), and a 50% chance of getting 2 (of them). Let the amount in your envelope be X, and the amount in the other be Y, so that either X = 1 = Y/2, which I call case A, or else B is the case, and X = 2 = 2Y.
......You are asked if you want to exchange your envelope for the other. In case A you have (whether or not you look inside your envelope) X = 1, so you would definitely gain X = 1 on switching; and if you would have lost, you would have been in case B, with X = 2. In case B you would definitely lose 1/2 of that on switching (and if you would have gained, you would have been in case A). And it was 50-50 which envelope you chose (as above), so it is similarly 50-50 which of A and B is the case, and so your expected gain on switching is:
......50% of X (= 1 = Y/2) – 50% of X/2 (= 1 = Y) = 0
......No paradox yet, but if you ignore the parenthetical identities (which should not, intuitively, affect that equality) then the remainder of that equation says that you have a 50% chance of gaining X (the definite amount you are holding) and a 50% chance of losing X/2, and that those two cancel out; so a puzzle is, how could X = X/2? But the paradox dissolves with the seeing of that question as silly. Athough X is a definite amount, only one of the ‘X’s in that equation names it (since it can't be that both B and A are the case; by contrast, if someone flipped a coin fairly, gave you X on heads, and took X/2 off you on tails, your expectation would indeed be X/4).
......A puzzle remains: Why do we find the above so puzzling? (To be continued (a clue: we are like this!).)
Figs - These were an unexpected treat. The fig tree is only a few years old, brought here in a pot after I bought it from Wilkinsons for £3.50 in January 2017, ...
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