This could be any 'best of 15 games' contest between two players where the object is to be the first to win 8 games (called frames in snooker) and the probability $p$ of winning a single game is constant.

We have to find the probability of player A winning the snooker match by adding the probabilities for all the possible outcomes. Player A can win in 8 frames (by winning the first 8 frames outright), or by winning any 7 of the first 8 then winning the ninth (when the match lasts 9 frames), or by winning any 7 of the first 9 then winning the tenth (when the match last 10 frames), or similarly player A can win a match which lasts for 11, 12 13 14 or 15 frames. Note that the last game, which decides the contest, must be won by A. Let $P(x)$ denote the probability of A winning a match with $x$ games in total.

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{P(8)}& =\left(\genfrac{}{}{0ex}{}{7}{7}\right)p^{7}p\\ \multicolumn{1}{c}{P(9)}& =\left(\genfrac{}{}{0ex}{}{8}{7}\right)p^7(1-p)p\\ \multicolumn{1}{c}{P(10)}& =\left(\genfrac{}{}{0ex}{}{9}{7}\right)p^7(1-p{)}^{2}p\\ \multicolumn{1}{c}{\dots }\\ \multicolumn{1}{c}{P(15)}& =\left(\genfrac{}{}{0ex}{}{14}{7}\right)p^7(1-p{)}^{7}p\end{array}}$

Let $f(p)$ denote the probability of A winning a 'best of 15' match when the probability of winning each frame is $p$.

${\displaystyle p=P(8)+P(9)+P(10)+P(11)+P(12)+P(13)+P(14)+P(15).}$

When $p=0.4$ this gives the probability $f(0.4)$ of A winning a 'best of 15' match to be 0.2131.

When the probability of A winning each frame is $p=0.55$ then the probability of the opponent winning a frame is $1-p=0.45$ and the probability $f(0.55)$ of A winning the match is 0.6535.

When the probability of A winning each frame is $p=0.5$ then the probability of A winning the match is 0.5 as expected.