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The wiki page doesn't explain the flaw in the logic. I think it is this: Let X denote the value of the larger envelope, and A the value of the envelope we hav
by Dove 16y ago
The wiki page doesn't explain the flaw in the logic. I think it is this:
Let X denote the value of the larger envelope, and A the value of the envelope we have in hand.
If we have the larger envelope, the other envelope is worth A/2 (where A=X).
If we have the smaller envelope, the other envelope is worth 2A'. (where A'=X/2)
So far so good, and this is all legal. The error comes when we try to add:
(0.5 A) + (0.5 A'/2) /= 5/4 A
The A's are not the same. You can't add them as if they were.
- sp332 16y agoThe wiki page used to have the solution. Apparently there is some arguing going on about sources, but this explanation is correct: http://en.wikipedia.org/wiki/Talk:Two_envelopes_problem#Solution_to_original_paradox_is_much_simpler http://en.wikipedia.org/wiki/Talk:Two_envelopes_problem#Solu... Also, yanowitz in this HN thread has a good (equivalent) explanation.
- btilly 16y agoAnd yet, that explanation isn't correct.
- crpatino 16y agoYes, I think you are correct. I would word it this way: Let: {X, 2X} := The amount of $$$ in each envelop A := 1st envelop has $2X B := 2nd envelop has $2X; The flaw in the paradox is to assume that events A and B are independent, when in fact they are mutually exclusive. Now, we can rewrite the whole "(0.5 A) + (0.5 A'/2) /= 5/4 A" nonsense as: P(A|B)2X + P(B|A)X = 02X + 0X = 0 Which is zero because it only covers half of the probabilities space, the other half is where all the density of the probability function lies. P(A|B')2X + P(B|A')X = 0.52X + 0.5X = 3X/2; which is what we would expect by intuition, halfway between the 2 prizes.
- lbrandy 16y agoi've realized my math was wrong and removed my comment :( i hate this problem