Two teams, team A and team B, are playing in a best 2 out of 3 tournament. Team A has a 60% chance of winning any game. What is the probability that team A wins the tournament?
+633 This probability problem is simple enough that it seems directed towards a brute force method, which is what I will use.
Let's first write out all the ways that team A could win the tournament, then sum the probabilities.
By games that would have been won:
A-B-A
B-A-A
A-A
You might be scratching your head at only having 2 games with the A-A permutation, but in this case, the third game doesn't matter, and might not be played in some tournament formats. Thus, we only need to calculate for until team A wins.
A-A = 60% * 60% = 36%
A-B-A = 60% * 60% * 40% = 36% * 40% = 14.4% (substitution from A-A chance calculation)
B-A-A = 14.4% by the commutative property of multiplication, it's the same as A-B-A
36%
+ 14.4%
+ 14.4%
=======
64.8%
Therefore, Team A has a 64.8% chance of winning the tournament.
