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## 'Snooker Frames' printed from http://nrich.maths.org/

Ben has solved this Tough Nut. He did not
say which school he comes from.
The probability of winning a 15 frame match was shown in the
problem Snooker to be $0.2131$ for a weaker player who has a
consistent probability of $0.4$ of winning each single frame. We
use the same method here.

To win an 11 frame match the player must be the first one to win 6
frames. He may win 6 games outright or win any 5 of the first 6
games and lose one then win one, or any 5 of the first 7 games and
lose 2 then win one, or any 5 of the first 8 games and lose 3 then
wins one or any 5 of the first 9 games and lose 4 then win one or
any 5 of the first 10 games and lose 5 then win one. The
probability is $$p^6 + {6\choose 5}p^5(1-p)p + {7 \choose
5}p^5(1-p)^2p + {8 \choose 5}p^5(1-p)^3p+{9\choose 5}p^5(1-p)^4p+
{10 \choose 5}p^5(1-p)^5$$ For $p=0.4$ and $1-p=0.6$ this becomes
$$0.4^6[1 + 6 \times 0.6 + 21 \times 0.6^2 + 56 \times 0.6^3 + 126
\times 0.6^4 + 252 \times 0.6^5 = 0.2465018$$ As $0.2465 >
0.2131$ this result gives evidence that weaker players are more
likely to win 11 frame matches than they are to win 15 frame
matches.

James from Bay House school did a numerical
investigation and showed that the chances of winning the tournament
were very slight with a 0.4 chance of winning each
frame.
Even with a 40% chance of winning each frame, the chance of being
the first to 18 frames is less than 1 in 10 million!

We can also look at the chance of being the
first to 18 frames for different probabilities of winning each
frame.
If you plot a log-log graph then you can
see a highly linear relationship for values of probability less
than about $90$ percent.