[BruceZ]

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two players are playing a game. They will win with frequency X and Y. What is the probability after N games that player A will win P more games than player B?

The first question is how to calculate this if they are independent games (ie, player A and B can both win or lose any given game).

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I'll assume that you mean exactly P more games, not at least P more games. I'll also assume that each player plays N games. Then the possibilities are (A wins P, B wins 0), (A wins P+1, B wins 1) ... (A wins N, B wins N-P). We must sum the probabilities of each of these as follows:

sum[k = P to N] {
C(N,k) * X^k * (1-X)^(N-k) *
C(N,k-P) * Y^(k-P) * (1-Y)^(N-k+P)
}


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The second question is how to calculate if they are dependent.

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Let A = number of games won by player A, and let B = number of games won by player B. Then we have:

A + B = N
A - B = P

So A = (N+P)/2, and B = (N-P)/2.

Note that A and B must both be odd or both be even. So your example of winning exactly 5 more games out of 30 is impossible.

The probability that A wins exactly this many games is just

C(N,A) * X^A * (1-X)^(1-A).