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'Snooker' printed from https://nrich.maths.org/
Thank you Sue Liu of Madras College, St Andrews for this solution.
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.
$$\begin{eqnarray} P(8) &= {7\choose 7}p^7p \\ P(9) &=
{8\choose 7}p^7(1-p)p\\ P(10)&= {9\choose 7}p^7(1-p)^2p \\
\ldots \\ P(15)&= {14\choose 7}p^7(1-p)^7p \end{eqnarray}$$
Let $f(p)$ denote the probability of A winning a 'best of 15' match
when the probability of winning each frame is $p$. \[p = P(8) +
P(9) + P(10) + P(11) + P(12) + P(13) + P(14) + P(15).\]
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.