Paradox of the wallet

[chaosuk]

'.....this game would appear to be -EV for both players. Clearly this can not be true either.'


there is an important distinction to be made between a game being negative EV for both players and both players making negative EV decisions. It's down to the observer.

It's easy to illistrate, it happens in poker all the time. One player decides to make a very rare bluff, but the chances of it winning are very remote since his oppo will almost certainly call. His decision to bet has -EV. His opponent however, also made a -EV call since his opponent bluffs so rarely. Both players made -EV decisons based on the balance of the hands they contained. Without perfect information both players can make +EV decision or -EV decisions.

signing off.

[alThor]

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Player A reasons that he has a 50% chance of winning this game and it is thus +EV for him to play. Player B reasons the same...how can this be?

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This is related to the paradox where there are two envelopes, one containing twice as much as the other. The bottom line is that we cannot assume that the probability of winning is independent of the amounts in our wallets.

alThor

[NMcNasty]

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.

[PairTheBoard]

Put simply, when you win you win the larger amount, but when you lose you lose the larger amount.

Same thing with the envelope switching problem.

PairTheBoard

[Jazza]

[ QUOTE ]
[ QUOTE ]
if player A wins, won't he win less than X not more than X?

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No, since you only win if the other player has more than you. Interestingly, though, if you switch the game around and say you win when you have more than the other player, this game would appear to be -EV for both players. Clearly this can not be true either.

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oh my bad, i thought the case was the later

i should learn to read good..