Suppose now that instead of taking turns to roll the die, they both roll together. The first to roll a 6 wins, but if they both roll a 6 at the same time, Sue wins by default (to reflect the ‘Sue rolled first’ condition). The Game is the same, as are the probabilities of each player winning. Now the maths…..

Both Sue and Bob have an equal chance of winning except for Sue’s extra 1/6 chance by having a ‘stronger’ 6 than Bob. Sue can win by the 50:50 chance of getting a 6 first, or the 1/6 chance of getting a 6 when Bob does (given Bob has a 6, Sue’s chance of a 6 in that round is 1/6, not 1/36)

So Sue’s chance of winning is (1/6) + (5/6)*(1/2) = 7/12
and Bob’s is (5/6)*(1/2) = 5/12

This shows the bias of Sue rolling first and their equal chance of winning – weighted by 5/6 to make the total probability = 1

As we all know, Bob’s chance of winning in the second round is (5^3)/(6^4) = 125/1296.

So now the conclusion.

The conditional probability for dependant probabilities is P(B|A)=P(A and B)/P(A)

P(Bob winning on 2nd roll | Bob wins) = P(Bob wins AND wins on the second roll) / P(Bob wins)

P(Bob wins AND wins on the 2nd roll) is the same as P(Bob wins on the 2nd roll) because he can’t ‘lose and win on 2nd roll’. So this boils down to:

P(Bob wins 2nd roll)/P(Bob wins)
=(125/1296)/(5/12)
=25/108
= approx 0.23

….I think