Electroholic wrote ...

you sure the "correct answer"is right? coz it can't be more than 1/6.

The probability rolling a 6 with a fair die is ALWAYS 1/6.
And every event is an independent event.

But in this case, there are a few other conditions.
1, Sue rolled a non-six (5/6)
2, Bob rolled a non-six (5/6)
3, Sue rolled a non-six (5/6)
4, Bob rolled a SIX (1/6)

multiply them all together and you should get the answer.

Steve McConner wrote ...

I see the logic behind it as follows:
In order for Bob to get to a second turn, the following conditions must apply:

1. Sue must not win on her first turn (probability of this condition = 5/6)
2. Bob must not win on his first turn (p=5/6)
3. Sue must not win on her second turn (p=5/6)

So the odds of getting to this point is (5/6) cubed.

4. Now it's Bob's second turn and he must roll a 6, the probability of which is 1/6.

So I make that (5*5*5*1)/(6*6*6*6)

If we want the odds counting only those games that Bob wins, as Chris just mentioned, then I guess we should multiply the above answer by two. The reasoning behind this is that if the die is fair, then Bob will win 50% of games. Unless Sue gets an advantage by rolling first, which I can't be bothered calculating. (Edit: she does, and I didn't know how to calculate it, hence the answer below is slightly wrong)

So I guess the answer is (5*5*5*2)/(6*6*6*6) = 0.192

neuralinhibitor wrote ...

Here's my solution:

The trick is to divide the probability of Bob winning on his second turn by the probability Bob wins in general.

p(Bob wins on his second turn) = 5^3/6^4

p(Bob wins in general) = p(Bob wins on his first turn + Bob wins on his second turn + Bob wins on his third turn + ...) = (5/6)(1/6) + (5/6)(5/6)(5/6)(1/6) + ... = sum( (1/6)(5/6)^(2k + 1), k=0, infinity) = (5/11). Note that p(Alice wins) is 6/11.

So p(it's Bob's second turn | Bob wins) = (5^3/6^4) / (6/11) = the answer.

I love probability problems like this because there's usually a hidden level of complexity and elegance.