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Difference between revisions of "2001 AIME II Problems/Problem 11"

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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 <math>\frac {1}{3}</math>. The [[probability]] that Club Truncator will finish the season with more wins than losses is <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
 
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 <math>\frac {1}{3}</math>. The [[probability]] that Club Truncator will finish the season with more wins than losses is <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
  
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== Solution ==
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== Solution 1==
 
Note that the probability that Club Truncator will have more wins than losses is equal to the probability that it will have more losses than wins; the only other possibility is that they have the same number of wins and losses. Thus, by the [[complement principle]], the desired probability is half the probability that Club Truncator does not have the same number of wins and losses.
 
Note that the probability that Club Truncator will have more wins than losses is equal to the probability that it will have more losses than wins; the only other possibility is that they have the same number of wins and losses. Thus, by the [[complement principle]], the desired probability is half the probability that Club Truncator does not have the same number of wins and losses.
  
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The possible ways to achieve the same number of wins and losses are <math>0</math> ties, <math>3</math> wins and <math>3</math> losses; <math>2</math> ties, <math>2</math> wins, and <math>2</math> losses; <math>4</math> ties, <math>1</math> win, and <math>1</math> loss; or <math>6</math> ties. Since there are <math>6</math> games, there are <math>\frac{6!}{3!3!}</math> ways for the first, and <math>\frac{6!}{2!2!2!}</math>, <math>\frac{6!}{4!}</math>, and <math>1</math> ways for the rest, respectively, out of a total of <math>3^6</math>. This gives a probability of <math>141/729</math>. Then the desired answer is <math>\frac{1 - \frac{141}{729}}{2} = \frac{98}{243}</math>, so the answer is <math>m+n = \boxed{341}</math>.  
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The possible ways to achieve the same number of wins and losses are <math>0</math> ties, <math>3</math> wins and <math>3</math> losses; <math>2</math> ties, <math>2</math> wins, and <math>2</math> losses; <math>4</math> ties, <math>1</math> win, and <math>1</math> loss; or <math>6</math> ties. Since there are <math>6</math> games, there are <math>\frac{6!}{3!3!}</math> ways for the first, and <math>\frac{6!}{2!2!2!}</math>, <math>\frac{6!}{4!}</math>, and <math>1</math> ways for the rest, respectively, out of a total of <math>3^6</math>. This gives a probability of <math>141/729</math>. Then the desired answer is <math>\frac{1 - \frac{141}{729}}{2} = \frac{98}{243}</math>, so the answer is <math>m+n = \boxed{341}</math>.
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==Solution 2==
 
==Solution 2==

Latest revision as of 22:25, 20 February 2023

Problem

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 $m + n$.

Solution 1

Note that the probability that Club Truncator will have more wins than losses is equal to the probability that it will have more losses than wins; the only other possibility is that they have the same number of wins and losses. Thus, by the complement principle, the desired probability is half the probability that Club Truncator does not have the same number of wins and losses.

The possible ways to achieve the same number of wins and losses are $0$ ties, $3$ wins and $3$ losses; $2$ ties, $2$ wins, and $2$ losses; $4$ ties, $1$ win, and $1$ loss; or $6$ ties. Since there are $6$ games, there are $\frac{6!}{3!3!}$ ways for the first, and $\frac{6!}{2!2!2!}$, $\frac{6!}{4!}$, and $1$ ways for the rest, respectively, out of a total of $3^6$. This gives a probability of $141/729$. Then the desired answer is $\frac{1 - \frac{141}{729}}{2} = \frac{98}{243}$, so the answer is $m+n = \boxed{341}$.

Solution 2

At first, it wins $6$ games, only one way

Secondly, it wins $5$ games, the other game can be either win or loss, there are $\binom{6}{5}\cdot 2=12$ ways

Thirdly, it wins $4$ games, still the other two games can be either win or loss, there are $\binom{6}{4}\cdot 2^2=60$ ways

Fourthly, it wins $3$ games, this time, it can't lose $3$ games but other arrangements of the three non-winning games are fine, there are $\binom{6}{3}\cdot (2^3-1)=140$ ways

Fifth case, it wins $2$ games, only $0/1$ lose and $4/3$ draw is ok, so there are $\binom{6}{2}(1+\binom{4}{1})=75$ cases

Last case, it only wins $1$ game so the rest games must be all draw, $1$ game

The answer is $\frac{1+12+60+140+75+6}{3^6}=\frac{98}{243}$ leads to $\boxed{341}$

~bluesoul

See also

2001 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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