2001 AIME II Problem 11

Question

Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its

6

matches, the probabilities that Club Truncator will win, lose, or tie are each

\frac{1}{3}.

The probability that Club Truncator will finish the season with more wins than losses is

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

Answer

By symmetry, more wins than losses and more losses than wins are equally likely

Find the probability of equally many wins and losses: count outcomes with

k

wins,

k

losses, and

6 - 2k

ties for each

k

Swapping wins and losses is a probability-preserving symmetry, so the probability

P

of more wins than losses equals the probability of more losses than wins, giving

P = \frac{1 - p_0}{2},

where

p_0

is the probability of equally many wins and losses.

An outcome with

k

wins,

k

losses, and

6 - 2k

ties can be arranged in

\frac{6!}{k! \, k! \, (6 - 2k)!}

ways:

1,

30,

90,

20

for

k = 0, 1, 2, 3,

totaling

141.

Each of the

3^6 = 729

outcome sequences is equally likely, so

p_0 = \frac{141}{729} = \frac{47}{243}.

Therefore

P = \frac{1}{2}\left(1 - \frac{47}{243}\right) = \frac{98}{243},

and

m + n = 98 + 243 = 341.

Problem 11 in other years