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Difference between revisions of "1981 AHSME Problems/Problem 26"

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Let <math>a</math>, <math>b</math>, and <math>c</math> be Carol's probabilities of winning, depending on whether it is Alice's, Bob's or Carol's turn to roll the die next, respectively.  Since it is initially Alice's turn, the value of <math>a</math> will be the answer.
 
Let <math>a</math>, <math>b</math>, and <math>c</math> be Carol's probabilities of winning, depending on whether it is Alice's, Bob's or Carol's turn to roll the die next, respectively.  Since it is initially Alice's turn, the value of <math>a</math> will be the answer.
  
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Then <math>a = \frac{5}{6}b</math>, <math>b = \frac{5}{6}c</math>, and <math>c = \frac{1}{6} + \frac{5}{6}a</math>.  Substituting into the first equation shows that <math>a = \frac{5}{6}b = \frac{25}{36}c = \frac{25}{36}\left(\frac{1}{6} + \frac{5}{6}a\right) = \frac{25}{216} + \frac{125}{216}a</math>. Solving the equation <math>a = \frac{25}{216} + \frac{125}{216}a</math> gives <math>a = \frac{25}{91}</math> <math>\fbox{(D)}</math>  
+
Then <math>a = \frac{5}{6}b</math>, <math>b = \frac{5}{6}c</math>, and <math>c = \frac{1}{6} + \frac{5}{6}a</math>.  Substituting into the first equation shows that <math>a = \frac{5}{6}b = \frac{25}{36}c = \frac{25}{36}\left(\frac{1}{6} + \frac{5}{6}a\right) = \frac{25}{216} + \frac{125}{216}a</math>. Solving the equation <math>a = \frac{25}{216} + \frac{125}{216}a</math> gives <math>a = \frac{25}{91}</math>. <math>\fbox{(D)}</math>  
  
 
-j314andrews
 
-j314andrews

Latest revision as of 13:48, 29 June 2025

Problem 26

Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob; and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is $\dfrac{1}{6}$, independent of the outcome of any other toss.)

$\textbf{(A)}\ \dfrac{1}{3}\qquad \textbf{(B)}\ \dfrac{2}{9}\qquad \textbf{(C)}\ \dfrac{5}{18}\qquad \textbf{(D)}\ \dfrac{25}{91}\qquad \textbf{(E)}\ \dfrac{36}{91}$

Solution 1 (Infinite Series)

For Carol to win on her $n^{th}$ roll, the first $3n-1$ rolls must not be $6$ and the last roll must be $6$. Therefore, the probability that Carol wins on her $n^{th}$ roll is $\frac{1}{6} \cdot \left(\frac{5}{6}\right)^{3n-1} = \frac{1}{5} \cdot \left(\frac{125}{216}\right)^{n}$. Taking the sum of this expression over all positive integers $n$, her total probability of winning must be $\sum_{n=1}^{\infty} \frac{1}{5} \cdot \left(\frac{125}{216}\right)^{n}$. Since this is a geometric series with first term $\frac{25}{216}$ and common ratio $\frac{125}{216}$, Carol's probability of winning is $\frac{\frac{25}{216}}{1-\frac{125}{216}}=\frac{\frac{25}{216}}{\frac{91}{216}}=\frac{25}{91}$. $\fbox{(D)}$

Solution 2 (Probability States)

Let $a$, $b$, and $c$ be Carol's probabilities of winning, depending on whether it is Alice's, Bob's or Carol's turn to roll the die next, respectively. Since it is initially Alice's turn, the value of $a$ will be the answer.

Then $a = \frac{5}{6}b$, $b = \frac{5}{6}c$, and $c = \frac{1}{6} + \frac{5}{6}a$. Substituting into the first equation shows that $a = \frac{5}{6}b = \frac{25}{36}c = \frac{25}{36}\left(\frac{1}{6} + \frac{5}{6}a\right) = \frac{25}{216} + \frac{125}{216}a$. Solving the equation $a = \frac{25}{216} + \frac{125}{216}a$ gives $a = \frac{25}{91}$. $\fbox{(D)}$

-j314andrews

See also

1981 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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