Difference between revisions of "2006 AIME II Problems/Problem 10"
m (→Solution) |
m (→Solution 4) |
||
(11 intermediate revisions by 10 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a <math> 50\% </math> 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 <math> A </math> beats team <math> B. </math> The probability that team <math> A </math> finishes with more points than team <math> B </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers. Find <math> m+n. </math> | + | Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a <math> 50\% </math> 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 <math> A </math> beats team <math> B. </math> The [[probability]] that team <math> A </math> finishes with more points than team <math> B </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers. Find <math> m+n. </math> |
+ | __TOC__ | ||
== Solution == | == Solution == | ||
− | + | === Solution 1 === | |
+ | The results of the five remaining games are independent of the first game, so by symmetry, the probability that <math>A</math> scores higher than <math>B</math> in these five games is equal to the probability that <math>B</math> scores higher than <math>A</math>. We let this probability be <math>p</math>; then the probability that <math>A</math> and <math>B</math> end with the same score in these five games is <math>1-2p</math>. | ||
− | + | Of these three cases (<math>|A| > |B|, |A| < |B|, |A|=|B|</math>), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases). | |
− | + | There are <math>{5\choose k}</math> ways to <math>A</math> to have <math>k</math> victories, and <math>{5\choose k}</math> ways for <math>B</math> to have <math>k</math> victories. Summing for all values of <math>k</math>, | |
+ | <center><math>1-2p = \frac{1}{2^{5} \times 2^{5}}\left(\sum_{k=0}^{5} {5\choose k}^2\right) = \frac{1^2+5^2+10^2+10^2+5^2+1^2}{1024} = \frac{126}{512}.</math></center> | ||
+ | Thus <math>p = \frac 12 \left(1-\frac{126}{512}\right) = \frac{193}{512}</math>. The desired probability is the sum of the cases when <math>|A| \ge |B|</math>, so the answer is <math>\frac{126}{512} + \frac{193}{512} = \frac{319}{512}</math>, and <math>m+n = \boxed{831}</math>. | ||
− | + | === Solution 2 === | |
+ | You can break this into cases based on how many rounds <math>A</math> wins out of the remaining <math>5</math> games. | ||
− | If A wins | + | *If <math>A</math> wins 0 games, then <math>B</math> must win 0 games and the probability of this is <math> \frac{{5 \choose 0}}{2^5} \frac{{5 \choose 0}}{2^5} = \frac{1}{1024} </math>. |
− | If A wins | + | *If <math>A</math> wins 1 games, then <math>B</math> must win 1 or less games and the probability of this is <math> \frac{{5 \choose 1}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}}{2^5} = \frac{30}{1024} </math>. |
− | If A wins | + | *If <math>A</math> wins 2 games, then <math>B</math> must win 2 or less games and the probability of this is <math> \frac{{5 \choose 2}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}}{2^5} = \frac{160}{1024} </math>. |
− | Summing these 6 cases, we get <math> \frac{638}{1024} </math>, which simplifies to <math> \frac{319}{512} </math>, so our answer is <math>319 + 512 = 831</math>. | + | *If <math>A</math> wins 3 games, then <math>B</math> must win 3 or less games and the probability of this is <math> \frac{{5 \choose 3}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}}{2^5} = \frac{260}{1024} </math>. |
+ | |||
+ | *If <math>A</math> wins 4 games, then <math>B</math> must win 4 or less games and the probability of this is <math> \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} </math>. | ||
+ | |||
+ | *If <math>A</math> wins 5 games, then <math>B</math> must win 5 or less games and the probability of this is <math> \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} </math>. | ||
+ | |||
+ | Summing these 6 cases, we get <math> \frac{638}{1024} </math>, which simplifies to <math> \frac{319}{512} </math>, so our answer is <math>319 + 512 = \boxed{831}</math>. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | |||
+ | We can apply the concept of generating functions here. | ||
+ | |||
+ | The generating function for <math>B</math> is <math> (1 + 0x^{1}) </math> for the first game where <math>x^{n}</math> is winning n games. Since <math>B</math> lost the first game, the coefficient for <math>x^{1}</math> is 0. The generating function for the next 5 games is <math>(1 + x)^{5}</math>. Thus, the total generating function for number of games he wins is | ||
+ | |||
+ | <math>{(1 + 0x)(1 + x)^{5}} = (1 + 5x^{1} + 10x^{2} + 10x^{3} + 5x^{4} + x^{5})</math>. | ||
+ | |||
+ | The generating function for <math>A</math> is the same except that it is multiplied by <math>x</math> instead of <math>(1+0x)</math>. | ||
+ | Thus, the generating function for <math>A</math> is | ||
+ | |||
+ | <math>1x + 5x^{2} + 10x^{3} + 10x^{4} + 5x^{5} + x^{6}</math>. | ||
+ | |||
+ | The probability that <math>B</math> wins 0 games is <math>\frac{1}{32}</math>. Since the coefficients for all <math>x^{n}</math> where | ||
+ | |||
+ | <math>n \geq 1</math> sums to 32, the probability that <math>A</math> wins more games is <math>\frac{32}{32}</math>. | ||
+ | |||
+ | Thus, the probability that <math>A</math> has more wins than <math>B</math> is <math>\frac{1}{32} \times \frac{32}{32} + \frac{5}{32} \times \frac{31}{32} + \frac{10}{32} \times \frac{26}{32} + \frac{10}{32} \times \frac{16}{32} + \frac{5}{32} \times \frac{6}{32} +\frac{1}{32} \times \frac{1}{32} = \frac{638}{1024} = \frac{319}{512}</math>. | ||
+ | |||
+ | Thus, <math>319 + 512 = \boxed{831} </math>. | ||
+ | |||
+ | === Solution 4 === | ||
+ | After the first game, there are <math>10</math> games we care about-- those involving <math>A</math> or <math>B</math>. There are <math>3</math> cases of these <math>10</math> games: <math>A</math> wins more than <math>B</math>, <math>B</math> wins more than <math>A</math>, or <math>A</math> and <math>B</math> win the same number of games. Also, there are <math>2^{10} = 1024</math> total outcomes. By symmetry, the first and second cases are equally likely, and the third case occurs <math>\binom{5}{0}^2+\binom{5}{1}^2+\binom{5}{2}^2+\binom{5}{3}^2+\binom{5}{4}^2+\binom{5}{5}^2 = \binom{10}{5} = 252</math> times, by [[Combinatorial identity#Another Identity|a special case of Vandermonde's Identity]]. There are therefore <math>\frac{1024-252}{2} = 386</math> possibilities for each of the other two cases. | ||
+ | |||
+ | If <math>B</math> has more wins than <math>A</math> in its <math>5</math> remaining games, then <math>A</math> cannot beat <math>B</math> overall. However, if <math>A</math> has more wins or if <math>A</math> and <math>B</math> are tied, <math>A</math> will beat <math>B</math> overall. Therefore, out of the <math>1024</math> possibilites, <math>386+252 = 638</math> ways where <math>A</math> wins, so the desired probability is <math>\frac{638}{1024} = \frac{319}{512}</math>, and <math>m+n=\boxed{831}</math>. | ||
== See also == | == See also == | ||
Line 23: | Line 60: | ||
[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:47, 22 March 2018
Problem
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a 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 beats team The probability that team finishes with more points than team is where and are relatively prime positive integers. Find
Contents
Solution
Solution 1
The results of the five remaining games are independent of the first game, so by symmetry, the probability that scores higher than in these five games is equal to the probability that scores higher than . We let this probability be ; then the probability that and end with the same score in these five games is .
Of these three cases (), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases).
There are ways to to have victories, and ways for to have victories. Summing for all values of ,
Thus . The desired probability is the sum of the cases when , so the answer is , and .
Solution 2
You can break this into cases based on how many rounds wins out of the remaining games.
- If wins 0 games, then must win 0 games and the probability of this is .
- If wins 1 games, then must win 1 or less games and the probability of this is .
- If wins 2 games, then must win 2 or less games and the probability of this is .
- If wins 3 games, then must win 3 or less games and the probability of this is .
- If wins 4 games, then must win 4 or less games and the probability of this is .
- If wins 5 games, then must win 5 or less games and the probability of this is .
Summing these 6 cases, we get , which simplifies to , so our answer is .
Solution 3
We can apply the concept of generating functions here.
The generating function for is for the first game where is winning n games. Since lost the first game, the coefficient for is 0. The generating function for the next 5 games is . Thus, the total generating function for number of games he wins is
.
The generating function for is the same except that it is multiplied by instead of . Thus, the generating function for is
.
The probability that wins 0 games is . Since the coefficients for all where
sums to 32, the probability that wins more games is .
Thus, the probability that has more wins than is .
Thus, .
Solution 4
After the first game, there are games we care about-- those involving or . There are cases of these games: wins more than , wins more than , or and win the same number of games. Also, there are total outcomes. By symmetry, the first and second cases are equally likely, and the third case occurs times, by a special case of Vandermonde's Identity. There are therefore possibilities for each of the other two cases.
If has more wins than in its remaining games, then cannot beat overall. However, if has more wins or if and are tied, will beat overall. Therefore, out of the possibilites, ways where wins, so the desired probability is , and .
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.