Re: Is this game solveable?
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Not correct.
If you always put 1 chip, and the opponent always puts all his chips, the probability of win is 0.681
(1/21 + ... + 1/40).
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That is the wrong sum. (Don't just add those probs.)
Under your strategies, player 2 (always all in) wins if and only if he wins 20 hands in a row. The likelihood of that is
(20/21)*(21/22)*...*(38/39)*(39/40) = 20/40 = 1/2.
This is a symmetric zero-sum game. Therefore, under optimal play, each player will have an equal chance of winning. Proof: Suppose not; then the losing player could simply imitate the winning player's strategy, giving him equal odds.
But this does not prove that all strategies are equally good. That is left as an exercise to the reader. [img]/images/graemlins/smile.gif[/img]
alThor
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