Paradox of the wallet

Thythe · #1 05-16-2005, 02:31 PM

Probably posted on here before, but after 15 seconds of searching I didn't find it...

Two people take out their wallets and set them on the table. If they each decide to play, the money is counted up, and whoever has the most money of the two, gives it to the other. Assume that each player has a random amount of money for this purpose. Player A reasons that if he plays and loses, he will only lose the money in his wallet now known as $X. If he wins, however, he will win more than in his wallet, or $(X+Y) where Y is greater than or equal to 1 cent. 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?

LetYouDown · #2 05-16-2005, 02:47 PM

Can we assume a finite range of amounts of money for each player? Otherwise I don't think this problem is solve-able.

Paul2432 · #3 05-16-2005, 03:11 PM

I think part of the problem here is the statement that player A feels he has a 50% chance of winning. Nothing in the problem statement suggests this will be the case. Because play is voluntary, suppose player B adopts the strategy of only playing when his wallet is empty. Then player A has a 0% chance of winning.

Paul

disjunction · #4 05-16-2005, 03:37 PM

[ QUOTE ]

Can we assume a finite range of amounts of money for each player? Otherwise I don't think this problem is solve-able.

[/ QUOTE ]

Bingo. If you assume a random finite distribution from $1 to $x, your probability of winning depends on the number you picked and the it doesn't seem that the "paradox" will work.

If you assume a random number from 0 to infinity, your probability of winning is not 50% and the problem becomes weird.

gaming_mouse · #5 05-16-2005, 05:20 PM

[ QUOTE ]

If you assume a random number from 0 to infinity,

[/ QUOTE ]

You cannot have a uniformly distributed random number between 0 and infinity. IMO, this is the "flaw" in the argument.

disjunction · #6 05-16-2005, 06:01 PM

[ QUOTE ]
You cannot have a uniformly distributed random number between 0 and infinity. IMO, this is the "flaw" in the argument.

[/ QUOTE ]

Yeah, I was going to put in parenthesis "(whatever that means"), but I thought it was redundant with me saying the problem becomes weird. It does not present a flaw in my argument, because I gave the finite case, and the problem remains undefined for the infinite case.

EDIT: Just realized that you're probably referring to the original post. BTW, "problem becomes weird" is a technical term [img]/images/graemlins/smile.gif[/img]

gaming_mouse · #7 05-16-2005, 06:07 PM

[ QUOTE ]
Just realized that you're probably referring to the original post.

[/ QUOTE ]

yeah, i was. i was basically just clarifying what you were getting at with "weird."

chaosuk · #8 05-16-2005, 09:44 PM

If both players play sub-optimally w.r.t game theory.

Suppose, in a similar game, both players are randomly assigned a number between 1 and 10 assigned to them. They must independtly decide whether or not to put up an amount of cash to bet against the other persons number, highest wins. Both players always play 7,8,9,10. Both players are given a 9. They are both making +EV bets by agreeing to play, since they are both going to win more often than they lose. But of course they are both playing sub-optimal strategies, since they could both improve their game by not playing the 7, since it never wins. Then of course we only end up both players only playing the 10. Do you think they need a blind structure?

chaos

Jazza · #9 05-16-2005, 10:20 PM

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

Thythe · #10 05-16-2005, 10:29 PM

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

[/ QUOTE ]

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.