The best way to look at this problem is not to search for "the probability that he rolled a 6", but instead find the probability that he did NOT roll a 6. Also ignore who is rolling, only look at what # turn it is. EX: Sue (Turn 1) then Bob (Turn 2) then Sue (Turn 3) etc.
The chance to not roll a 6 on turn 1 is 5/6. The chance to not roll a 6 on turn 2 (viewed independently)is 5/6. However the chance that a 6 was not rolled for two turns in a row is (5/6)(5/6) = 25/36 The chance to not roll a 6 on turn 3 (viewed independently) is 5/6. However the chance that a 6 was not rolled for 3 turns in a row is (5/6)(5/6)(5/6) = 125/216 Finally, for the 4th turn (which would be Bob's 2nd turn), the chance to not roll a 6 (independently) is 5/6, but the chance to not roll a 6 for the 4th time in a row is (5/6)(5/6)(5/6)(5/6) = 625/1296.
SO: on Bob's 2nd turn, the chance to NOT roll a 6 is 625/1296 That means that the chance for him to end the game on that turn is the remaining 671/1296 (which is about 51.8%).
Thanks for the riddle :) --Msstaffo 13:26, 12 February 2009 (UTC)
You are very close. Remember, we're looking for the probability in a particular universe where Bob wins. You've calculated the probablility of Bob winning on his second turn...not the probability of him winning on his second turn given that he has won. -Mark
