I am making a model in Matlab that calculates the winning probability of darts player A against darts player B with help of a paper (see bottom line). I have been able to implement the calculation of the probabilities of player A winning 1 set but have not been succesful at applying this to a multi-set match. Most importantly I want this model to take the advantage of beginning a set into account.
So, A = probability of player A to win a set he start, B = probability of player B to win a set B started, 1-A = probability of player B to win a set player A started and 1-B = probability of player A to win a set player B started.
So for example, according to the model Player A has a A = 0.5912 probability to win a set that he started and therefore a 1-A = 0.4088 probability to lose it, i.e. a 'break'. But after player A started the set it is player B's turn to start following set. According to the model, the probability of player B to win his 'own' set is B = 0.4474 and 1-B = 0.5526 for player A to win the set that player B started, i.e. again a break.
My model assumes player A starts the match, so the probability of player A to win 2:0 is 0.5912 * 0.5526 = 0.3267. The probability of winning 2:1 is 0.5912 * 0.4474 * 0.5912 + 0.4088 * 0.5526 * 0.5912 = 0.1563 + 0.1336 = 0.2899. So the probability to win a best of 3 match is 0.3267 + 0.2899 = 0.6166 For a best of 3 sets match this is easy but for up to a best of 13 sets match it gets complicated due to all combinations. I manually calculated the chances up to a best of 7 sets match but after this it got too complicated due to all different combinations. I will share my probabilities up to a best of 7 sets match (using the same aforementioned probabilities) so this can validate your code. Honestly I dont have a clue where to start. Here are the probabilities (and I'm sure they are correct):
Best of 5 sets P (3:0) = 0.1931 P (3:1) = 0.2336 p (3:2) = 0.2128 Best of 7 sets P (4:0) = 0.1067 P (4:1) = 0.1894 P (4:2) = 0.1900 P (4:3) = 0.1737
I hope I have provided enough information. Thank you in advance!
Paper used: Predicting the outcome of professional darts tournaments by Steffen Liebscher & Thomas Kirschstein
For $A$ to win a best-of-$(2n-1)$ match $n:k$, the first $n+k-1$ sets must contain $n-1$ wins for $A$ and $k$ wins for $B$, and then $A$ must win the $(n+k)$-th match.
If $n+k$ is odd, the probability for $A$ to win the $(n+k)$-th match is $A$, and $n+k-1$ is even, so each player starts $\frac{n+k-1}2$ of the first $n+k-1$ matches. Thus in this case the probability for $A$ to win $n:k$ is
$$ A\sum_{j=0}^k\binom{\frac{n+k-1}2}jB^j(1-B)^{\frac{n+k-1}2-j}\binom{\frac{n+k-1}2}{k-j}(1-A)^{k-j}A^{\frac{n+k-1}2-(k-j)}\;. $$
If $n+k$ is even, the probability for $A$ to win the $(n+k)$-th match is $1-B$, and $n+k-1$ is odd, so $A$ starts $\frac{n+k}2$ of the first $n+k-1$ matches and $B$ only starts $\frac{n+k}2-1$. Thus in this case the probability for $A$ to win $n:k$ is
$$ (1-B)\sum_{j=0}^k\binom{\frac{n+k}2-1}jB^j(1-B)^{\frac{n+k}2-1-j}\binom{\frac{n+k}2}{k-j}(1-A)^{k-j}A^{\frac{n+k}2-(k-j)}\;. $$