2006 AIME II Problems/Problem 10

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

Solution 1

The results of the five remaining games are independent of the first game, so by symmetry, the probability that $A$ scores higher than $B$ in these five games is equal to the probability that $B$ scores higher than $A$ . We let this probability be $p$ ; then the probability that $A$ and $B$ end with the same score in these give games is $1-2p$ .

Of these three cases ( $|A| > |B|, |A| < |B|, |A|=|B|$ ), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases).

There are ${6\choose k}$ ways to $A$ to have $k$ victories, and ${6\choose k}$ ways for $B$ to have $k$ victories. Summing for all values of $k$ ,

$1-2p = \frac{1}{2^{5} \times 2^{5}}\left(\sum_{k=0}^{5} {6\choose k}^2\right) = \frac{1^2+5^2+10^2+10^2+5^2+1^2}{1024} = \frac{126}{512}.$

Thus $p = \frac 12 \left(1-\frac{126}{512}\right) = \frac{193}{512}$ . The desired probability is the sum of the cases when $|A| \ge |B|$ , so the answer is $\frac{126}{512} + \frac{193}{512} = \frac{319}{512}$ , and $m+n = \boxed{831}$ .

Solution 2

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 $\frac{{5 \choose 1}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}}{2^5} = \frac{30}{1024}$ .

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 $319 + 512 = \boxed{831}$ .

2006 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2006 AIME II Problems/Problem 10

Problem

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Solution

Solution 1

Solution 2

See also