1981 AHSME Problems/Problem 26
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 the Carol's probabilities of winning, depending on whether it is Alice's, Bob's or Carol's turn to roll the die next, respectively.
Then 
, 
, and 
. Substituting into the first equation shows that 
. Solving the equation 
gives 
See also
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