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If given the cost to play, and the average win. Can I calculate the edge? (probability of winning)

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I don't think this question is appropriate for this site. – BBischof Aug 16 '10 at 18:52
Indeed it might get better answers at – Justin L. Aug 16 '10 at 19:00

It should intuitively feel like it's not (theoretically) possible -- the cost to play and the average win are determined by a bookie (or a casino, etc.), whereas the probability of winning is determined by the game itself. Given two games, one with a probability of winning $p_1$ and the other with a probability of winning $p_2$, the bookie can adjust the returns on bets so as the cost of play and average win remain constant. For example:

  • Consider a tossing coin game, the player bets 1 dollar, if the coin is "heads" it will return 2 dollars and if "tales" then there is no return. So the expected win is +1-1=0.

  • Now consider a die rolling game, where the player bets 1 dollar, if the die rolls 6, then 6 dollars are returned, otherwise there is no return. Here the expected win is +5-1-1-1-1-1=0.

In both cases the cost to play is 1 dollar and in both cases the expected win is 0, but the probability of winning is different (1/2 vs. 1/6).

In practice, however, you might be able to infer an approximate probability of winning based on past experiences of the bookie, familiarity with the game being played, etc.

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