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Old 05-17-2005, 03:53 AM
NMcNasty NMcNasty is offline
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Join Date: Feb 2004
Posts: 2
Default Re: Paradox of the wallet

This is the same as the exchange paradox (the one with the envelopes). The problem arises in that you are using X as a variable for two different values. Since in one case X is larger than in the other, you would basically be saying that X > X which is logically impossible. If you use numbers the paradox is a lot more clear:
Case 1: Player has $1, opponent has 2$. Since you already have $1 you only gain $2 in this case.
Case 2: Player has $2, opponent has 1$. You lost the $2 you started with.
Don't add or subtract from what you started with because the money is yours. You aren't starting with zero, then getting the money in your own wallet, then adding or subtracting from that number. You are simply adding or subtracting from zero because the exchange will net you either a gain or a loss.

So your EV is Pr(case 1)*($2)+ Pr(case2)*(-$2).
Its a premise of the original problem that its a coinflip as to which case will happen, although that part wasn't apparently clear. So, the equation becomes
.5*2+.5*-2 = 0.
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