Exercise on Boolean Algebra

  • #1

Homework Statement


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.

Homework Equations


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
 

Answers and Replies

  • #2
473
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Well, it's only one game (involving Team 1) that we're talking about here, right?

And we may not yet know who the opponent is, but it's one of the three teams mentioned; at worst Team 1 has a 0.48 chance of losing and at best only a 0.40 chance of losing. If our answer isn't between those two numbers there is probably something wrong.

It might help if we had some way of determining who the likely opponent is between the three possible teams.
 
  • #3
haruspex
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I believe that Joffan's point (inter alia) is that none of these calculations appear to be relevant:
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
since T1 will only face one opponent. I agree.
 
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