Brain Teasers
Chess Percents
Two
chess players compete in a best-of-five match. If Chekmatova has a 60% chance
of winning any particular game, what is the likelihood that she will win the
match?
This brainteaser was
written by Derrick Niederman.
Solution:
68.256%.
There are three possibilities:
- Chekmatova wins the first
three games, in which case it doesn’t matter what happens in the fourth or
fifth games.
- Chekmatova wins two of the
first three games and then wins the fourth game, in which case it doesn’t
matter what happens in the fifth game.
- Chekmatova wins two of the
first four games and then wins the fifth game.
In the first scenario, it doesn’t matter what happens in the
fourth or fifth games, and the probability of Chekmatova winning the first
three games is (0.6)3 = 0.216.
In the second scenario, it doesn’t matter what happens in
the fifth game. The probability of Chekmatova winning three out of four is 3 ×
(0.6)3 × (0.4) = 0.2592. (The 3 at the beginning indicates that her
opponent could win any of the first three games.)
In the third scenario, the probability of Chekmatova winning
three out of five is 6 × (0.6)3 × (0.4)2 = .20736. (The 6
at the beginning of the expression represents the 4C2 ways to choose which two
games her opponent wins.)
The combined probability is 0.216 + 0.2592 + 0.20736 =
0.68256, or 68.256%.
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