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Difference between revisions of "2006 AIME II Problems/Problem 10"

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== Problem ==
 
== Problem ==
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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 accumulated 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__
 
__TOC__
 +
 
== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== 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 give games is <math>1-2p</math>.
+
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).  
 
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>{6\choose k}</math> ways to <math>A</math> to have <math>k</math> victories, and <math>{6\choose k}</math> ways for <math>B</math> to have <math>k</math> victories. Summing for all values of <math>k</math>,
+
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} {6\choose k}^2\right) = \frac{1^2+5^2+10^2+10^2+5^2+1^2}{1024} = \frac{126}{512}.</math></center>
+
<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>.
 
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 ===
 
=== Solution 2 ===
−
You can break this into cases based on how many rounds A wins out of the remaining 5 games.
+
You can break this into cases based on how many rounds <math>A</math> wins out of the remaining <math>5</math> games.
  
−
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 <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 <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 <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 <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>.
+
*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>.
  
−
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 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 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>.
+
*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>.
 
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 34: Line 61:
  
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 12:42, 8 December 2021

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 accumulated 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 five 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 ${5\choose k}$ ways to $A$ to have $k$ victories, and ${5\choose k}$ ways for $B$ to have $k$ victories. Summing for all values of $k$,

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

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

Solution 3

We can apply the concept of generating functions here.

The generating function for $B$ is $(1 + 0x^{1})$ for the first game where $x^{n}$ is winning n games. Since $B$ lost the first game, the coefficient for $x^{1}$ is 0. The generating function for the next 5 games is $(1 + x)^{5}$. Thus, the total generating function for number of games he wins is 

${(1 + 0x)(1 + x)^{5}} = (1 + 5x^{1} + 10x^{2} + 10x^{3} + 5x^{4} + x^{5})$.

The generating function for $A$ is the same except that it is multiplied by $x$ instead of $(1+0x)$. Thus, the generating function for $A$ is

$1x + 5x^{2} + 10x^{3} + 10x^{4} + 5x^{5} + x^{6}$.

The probability that $B$ wins 0 games is $\frac{1}{32}$. Since the coefficients for all $x^{n}$ where

$n \geq 1$ sums to 32, the probability that $A$ wins more games is $\frac{32}{32}$.

Thus, the probability that $A$ has more wins than $B$ is $\frac{1}{32} \times \frac{32}{32} + \frac{5}{32} \times \frac{31}{32} + \frac{10}{32} \times \frac{26}{32} + \frac{10}{32} \times \frac{16}{32} + \frac{5}{32} \times \frac{6}{32} +\frac{1}{32} \times \frac{1}{32} = \frac{638}{1024} = \frac{319}{512}$.

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

Solution 4

After the first game, there are $10$ games we care about-- those involving $A$ or $B$. There are $3$ cases of these $10$ games: $A$ wins more than $B$, $B$ wins more than $A$, or $A$ and $B$ win the same number of games. Also, there are $2^{10} = 1024$ total outcomes. By symmetry, the first and second cases are equally likely, and the third case occurs $\binom{5}{0}^2+\binom{5}{1}^2+\binom{5}{2}^2+\binom{5}{3}^2+\binom{5}{4}^2+\binom{5}{5}^2 = \binom{10}{5} = 252$ times, by a special case of Vandermonde's Identity. There are therefore $\frac{1024-252}{2} = 386$ possibilities for each of the other two cases.

If $B$ has more wins than $A$ in its $5$ remaining games, then $A$ cannot beat $B$ overall. However, if $A$ has more wins or if $A$ and $B$ are tied, $A$ will beat $B$ overall. Therefore, out of the $1024$ possibilites, $386+252 = 638$ ways where $A$ wins, so the desired probability is $\frac{638}{1024} = \frac{319}{512}$, and $m+n=\boxed{831}$.

See also

2006 AIME II (Problems • Answer Key • Resources)
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