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

p6 + æ ç è 6 5 ö ÷ ø p5(1-p)p + æ ç è 7 5 ö ÷ ø p5(1-p)2p + æ ç è 8 5 ö ÷ ø p5(1-p)3p+ æ ç è 9 5 ö ÷ ø p5(1-p)4p+ æ ç è 10 5 ö ÷ ø p5(1-p)5

For p=0.4 and 1-p=0.6 this becomes

0.46[1 + 6*0.6 + 21*0.62 + 56*0.63 + 126*0.64 + 252*0.65 = 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.