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} + (\binom{6}{5}) p^{5} (1 - p) p + (\binom{7}{5}) p^{5} (1 - p)^{2} p + (\binom{8}{5}) p^{5} (1 - p)^{3} p + (\binom{9}{5}) p^{5} (1 - p)^{4} p + (\binom{10}{5}) p^{5} (1 - p)^{5}

For

p = 0.4

and

1 - p = 0.6

this becomes

0 . 4^{6} [1 + 6 * 0.6 + 21 * 0 . 6^{2} + 56 * 0 . 6^{3} + 126 * 0 . 6^{4} + 252 * 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.