Difference between revisions of "1981 AHSME Problems/Problem 26"
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== Solution 2 (Probability States) == | == Solution 2 (Probability States) == | ||
| − | 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 Alice's turn, | + | 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 Alice's turn, the value of <math>a</math> will be the answer. |
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}(\frac{1}{6} + \frac{5}{6}a) = \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}(\frac{1}{6} + \frac{5}{6}a) = \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> | ||
Revision as of 11:07, 28 June 2025
Problem
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 
, independent of the outcome of any other toss.)
Solution 1
The probability that Carol wins during the first cycle through is 
, and the probability that Carol wins on the second cycle through is 
. It is clear that this is an infinite geometric sequence, and we must find the sum of it in order to find the answer to this question. Thus we set up the equation: 
, or 
, which simplifies into 
Solution 2 (Probability States)
Let 
, 
, and 
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 Alice's turn, the value of will be the answer.
Then 
, 
, and 
. Substituting into the first equation shows that 
. Solving the equation 
gives 
-j314andrews
See also
| 1981 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by 1980 AHSME |
Followed by 1982 AHSME | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
