In Game 6 of a 6-team double-elimination tournament, Team 1, the top-ranked team, faces the loser of a previous game involving Team 2, the second-ranked team, against one of either Team 3, the third-ranked team or Team 6, the sixth-ranked team.
This exercise tests the limits of my expertise with Boolean algebra. I think I've managed to get a handle on the problem, but I would very much appreciate a confirmation of my calculations. Or if I have strayed from the path of Boolean truth, I would even more appreciate your letting me know.
According to pregame rankings, the probability of Team 1's losing to Team 2, Team 3 and Team 6, respectively is .48, .46 and .40. The corresponding probabilities that Team 1 will not lose to Team 2, Team 3 and Team 6 is .52, .54, and .60. What is the probability that Team 1 will lose the game?
Here's my solution:
The Attempt at a Solution
p(T1 loses to T2) = .48 p(T1 does not lose to T2) = .52
p(T1 loses to T3) = .46 p(T1 does not lose to T3) = .54
p(T1 loses to T6) = .40 p(T1 does not lose to T6) = .60
p(T1 loses to T3 and not to T6) = (.46) (.60) = .2760
p(T1 loses to T6 and not to T3) = (.40) (.54) = 2160
p(T1 loses to either T3 or T6 ) = .4920
p (T1 loses to neither T3 nor T6) = ..5028
p(T1 loses to T2 and not to (T3 or T6) = (.48) (.5028) = .2438
p(T1 loses to (T3 or T6) and not to T2) = (.4920) (.52) = .2558
p(T1 loses to either T2 or to (T3 or T6) = .4996
p(T1 loses neither to T2 nor to (T3 or T6) = .5004