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Consider the scenario of two competing tennis players, player 1 and player 2. I have a function $g_i=f(p_i)$ which returns $g_1$, the probability of a player winning a game on serve, given a value of $p_1$, the probability of a winning a single point. The function has been built using the method described here.

I want to calculate the probability of a player winning a set with score $s_1$:$s_2$, given two inputs $p_1$, the probability of player 1 winning point on serve, and $p_2$, the probability of player 2 winning a point on serve.

For example, if $s_1=6$ and $s_2=0$, then I would want to calculate the probability of player 1 winning the set 6:0. Now this example is rather straightforward to calculate as it is simply $g_1^3*(1-g_2)^3$, where $g_1$ and $g_2$ represent the respective probabilities of each player winning a game. However I am getting lost when considering set scores that have many more combinations of game sequences, e.g. $6:4$

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  • $\begingroup$ You mean $p_2$ is the probability of Player $2$ winning a point on serve. $\endgroup$ – quasi May 24 '17 at 10:56
  • $\begingroup$ Also, how do you handle $6$-$6$? Is it decided by a tie-breaker, or do they keep playing until someone is ahead by two games? $\endgroup$ – quasi May 24 '17 at 10:57
  • $\begingroup$ Thank you, yes have corrected the definition of $p_2$ $\endgroup$ – user3725021 May 24 '17 at 11:03
  • $\begingroup$ Hadn't thought of $6:6$, but perhaps to simplify, assume that no tiebreakers are played, so a set must be won by two games, i.e. 9-7 would be a valid score. This is an edge case of minor practical importance though, so less if it's easier to assume tiebreakers, I'd go with that option. $\endgroup$ – user3725021 May 24 '17 at 11:05
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Here's a recursive solution, implemented as a Maple program: enter image description here

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