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Difference between revisions of "2007 iTest Problems/Problem 12"

m (Created page with "This problem is phrased oddly, leading to the non-standard result of <math>\frac{3}{2}</math> instead of the standard <math>\frac{9}{8}</math>. Label the teams A,B. If A wins th...")
 
(Solution to Problem 12 (note: 9/8 not answer choice))
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==Problem 12==
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My Frisbee group often calls "best of five" to finish our games when it's getting dark, since we don't keep score.
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The game ends after one of the two teams scores three points (total, not necessarily consecutive).
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If every possible sequence of scores is equally likely, what is the expected score of the losing team.
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<math>\text{(A) }\frac{2}{3}\qquad
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\text{(B) }1\qquad
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\text{(C) }\frac{3}{2}\qquad
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\text{(D) }\frac{8}{5}\qquad
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\text{(E) }\frac{5}{8}\qquad
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\text{(F) }2\qquad\\ \\
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\text{(G) }0\qquad
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\text{(H) }\frac{5}{2}\qquad
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\text{(I) }\frac{2}{5}\qquad
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\text{(J) }\frac{3}{4}\qquad
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\text{(K) }\frac{4}{3}\qquad
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\text{(L) }2007\qquad</math>
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[[2007 iTest Problems/Problem 12|Solution]]
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==Solution==
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First, count the number of outcomes for each possible score of the losing team.  Let <math>W</math> be a game where the winning team won and <math>L</math> be a game where the losing team lost.  Note that for all the possible outcomes, the winning team must win the last game.
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For <math>0</math> points for the losing team, there is one outcome — the winning team winning three in a row.  For <math>1</math> point for the losing team, there are three outcomes (losing team can win first, second, or third game).    For <math>2</math> points for the losing team, the losing team has to win exactly two points in the first four games.  There are <math>\binom{4}{2} = 6</math> outcomes in this case.
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In total, there are <math>10</math> outcomes.  Using expected value, the expected number of points for the losing team is <math>0 \cdot \frac{1}{10} + 1 \cdot \frac{3}{10} + 2 \cdot \frac{6}{10} = \boxed{\textbf{(C) } \frac{3}{2}}</math>.
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==Note==
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This problem is phrased oddly, leading to the non-standard result of <math>\frac{3}{2}</math> instead of the standard <math>\frac{9}{8}</math>.
 
This problem is phrased oddly, leading to the non-standard result of <math>\frac{3}{2}</math> instead of the standard <math>\frac{9}{8}</math>.
  
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<math>1*(\tfrac{1}{2})^3+3*(\tfrac{1}{2})^4+6*(\tfrac{1}{2})^5=\frac{1}{2}</math>,
 
<math>1*(\tfrac{1}{2})^3+3*(\tfrac{1}{2})^4+6*(\tfrac{1}{2})^5=\frac{1}{2}</math>,
 
and an expected value for the losing team of <math>\frac{(0*2+1*3+2*3)}{(2+3+3)}=\frac{9}{8}</math>.
 
and an expected value for the losing team of <math>\frac{(0*2+1*3+2*3)}{(2+3+3)}=\frac{9}{8}</math>.
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==See Also==
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{{iTest box|year=2007|num-b=11|num-a=13}}
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[[Category:Introductory Combinatronics Problems]]
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[[Category:Introductory Probability Problems]]

Revision as of 03:37, 10 June 2018

Problem 12

My Frisbee group often calls "best of five" to finish our games when it's getting dark, since we don't keep score. The game ends after one of the two teams scores three points (total, not necessarily consecutive). If every possible sequence of scores is equally likely, what is the expected score of the losing team.

$\text{(A) }\frac{2}{3}\qquad \text{(B) }1\qquad \text{(C) }\frac{3}{2}\qquad \text{(D) }\frac{8}{5}\qquad \text{(E) }\frac{5}{8}\qquad \text{(F) }2\qquad\\ \\ \text{(G) }0\qquad \text{(H) }\frac{5}{2}\qquad \text{(I) }\frac{2}{5}\qquad \text{(J) }\frac{3}{4}\qquad \text{(K) }\frac{4}{3}\qquad \text{(L) }2007\qquad$

Solution

Solution

First, count the number of outcomes for each possible score of the losing team. Let $W$ be a game where the winning team won and $L$ be a game where the losing team lost. Note that for all the possible outcomes, the winning team must win the last game.

For $0$ points for the losing team, there is one outcome — the winning team winning three in a row. For $1$ point for the losing team, there are three outcomes (losing team can win first, second, or third game). For $2$ points for the losing team, the losing team has to win exactly two points in the first four games. There are $\binom{4}{2} = 6$ outcomes in this case.

In total, there are $10$ outcomes. Using expected value, the expected number of points for the losing team is $0 \cdot \frac{1}{10} + 1 \cdot \frac{3}{10} + 2 \cdot \frac{6}{10} = \boxed{\textbf{(C) } \frac{3}{2}}$.

Note

This problem is phrased oddly, leading to the non-standard result of $\frac{3}{2}$ instead of the standard $\frac{9}{8}$.

Label the teams A,B. If A wins the series, it necessarily wins the $\underline{last}$ game. A three game series with A winning occurs 1 way (AAA), a four game series occurs ${3\choose 1}=3$ ways (AABA,ABAA,BAAA), and a five game tournament occurs ${4\choose 2}=6$ ways (AABBA,ABABA,BAABA,ABBAA,BABAA,BBAAA). These series occur in the proportions 1:3:6. The problem is phrased to imply that the actual probabilities occur in these proportions as well - leading to an expected value of $\frac{(0*1+1*3+2*6)}{(1+3+6)}=\frac{3}{2}$. This assumption leads to the events (AAA) or (BBB) occurring with $p=\tfrac{1}{10}$ Which requires each game to be $\underline{negatively}$ correlated with the previous one (i.e. A is less likely to win a game if it won the previous game).


A more standard approach would be to assume the games are uncorrelated and that each team is equally likely to win. Then the probabilities would occur in the proportions $1*(\tfrac{1}{2})^3:3*(\tfrac{1}{2})^4:6*(\tfrac{1}{2})^5$ or $2:3:3$. These probabilities would imply that A wins the series with probability $1*(\tfrac{1}{2})^3+3*(\tfrac{1}{2})^4+6*(\tfrac{1}{2})^5=\frac{1}{2}$, and an expected value for the losing team of $\frac{(0*2+1*3+2*3)}{(2+3+3)}=\frac{9}{8}$.

See Also

2007 iTest (Problems, Answer Key)
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