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Thank you Sue Liu of Madras College, St Andrews for this solution. 
\par

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. 
\par

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*}

\par

Let $f(p)$ denote the probability of A winning a 'best of 15' match when 
the probability of winning each frame is $p$.
\[p = P(8) + P(9) + P(10) + P(11) + P(12) + P(13) + P(14) + P(15).\]

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

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.
\par

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.
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