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I just had an interview question involving a two-player game of throwing poker chips closest to the wall. Players take turns throwing chips (of which you start with zero, but chips can be "borrowed"), and each turn a player can choose to pass. The game ends when both players pass. The winner is the player who throws closest to the wall, and he or she wins 20 poker chips. The goal is to maximize the number of chips you end up with. For example, if player 1 throws 14 chips before both players pass and they are closest to the wall, they will end up with 6 chips since they receive 20 chips for winning and have to repay the "borrowed" 14 chips.

The interviewer asked what the optimal strategy for maximizing the number of chips you end up with is and I couldn't come up with anything concrete. Does anyone know the strategy? Is the game even worth playing? Does knowing the distributions of the distance of the players throw change anything?

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I'm going to assume that on every turn, the current player believes they have a nonzero probability of being the closest if they borrow and throw. (This is a VERY strong assumption!)

The game will never end since $20-\text{borrowed chips}>-\text{borrowed chips}$. The optimal strategy for both players is:

On every turn, if the current player is not the closest, borrow and throw. If they are the closest, pass.

Note that due to my strong assumption, this strategy is optimal.

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  • $\begingroup$ This cannot be correct. If I play against an opponent who I know will always throw if my chip is closer to the wall then I know I cannot win and the optimal strategy is not to play. $\endgroup$ – Tim Mar 25 at 6:52

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