2006 AIME II Problems/Problem 10

Revision as of 15:34, 12 March 2008 by Jam (talk | contribs) (All the binomial coffieients were upside down.)

Problem

Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumilated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution

You can break this into cases based on how many rounds A wins out of the remaining 5 games.

If A wins 0 games, then B must win 0 games and the probability of this is $\frac{{5 \choose 0}}{2^5} \frac{{5 \choose 0}}{2^5} = \frac{1}{1024}$.

If A wins 1 games, then B must win 1 or less games and the probability of this is $\frac{{5 \choose 1}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}}{2^5} = \frac{5}{1024}$.

If A wins 2 games, then B must win 2 or less games and the probability of this is $\frac{{5 \choose 2}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}}{2^5} = \frac{160}{1024}$.

If A wins 3 games, then B must win 3 or less games and the probability of this is $\frac{{5 \choose 3}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}}{2^5} = \frac{260}{1024}$.

If A wins 4 games, then B must win 4 or less games and the probability of this is $\frac{{5 \choose 4}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}+{5 \choose 4}}{2^5} = \frac{155}{1024}$.

If A wins 5 games, then B must win 5 or less games and the probability of this is $\frac{{5 \choose 5}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}+{5 \choose 4}+{5 \choose 5}}{2^5} = \frac{32}{1024}$.

Summing these 6 cases, we get $\frac{638}{1024}$, which simplifies to $\frac{319}{512}$, so our answer is $319 + 512 = 831$.

See also

2006 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AIME Problems and Solutions
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