2001 AIME II Problem 11

Attempt Problem 11 of the 2001 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2001 AIME II solutions, or check the answer key.

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11.

Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 66 matches, the probabilities that Club Truncator will win, lose, or tie are each 13.\frac{1}{3}. The probability that Club Truncator will finish the season with more wins than losses is mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

Answer: 341
Concepts:basic probabilitymultiset permutationscomplementary countingsymmetry
Difficulty rating: 2560
Solution:

Swapping wins and losses is a probability-preserving symmetry, so the probability PP of more wins than losses equals the probability of more losses than wins, giving P=1p02,P = \frac{1 - p_0}{2}, where p0p_0 is the probability of equally many wins and losses.

An outcome with kk wins, kk losses, and 62k6 - 2k ties can be arranged in 6!k!k!(62k)!\frac{6!}{k! \, k! \, (6 - 2k)!} ways: 1,1, 30,30, 90,90, 2020 for k=0,1,2,3,k = 0, 1, 2, 3, totaling 141.141. Each of the 36=7293^6 = 729 outcome sequences is equally likely, so p0=141729=47243.p_0 = \frac{141}{729} = \frac{47}{243}.

Therefore P=12(147243)=98243,P = \frac{1}{2}\left(1 - \frac{47}{243}\right) = \frac{98}{243}, and m+n=98+243=341.m + n = 98 + 243 = 341.

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