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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.2010-08-16
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    Indeed it might get better answers at http://stats.stackexchange.com/2010-08-16

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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.