Multiple choice

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 1/6, independent of the outcome of any other toss.)

  1. $\dfrac{1}{3}$
  2. $\dfrac{2}{9}$
  3. $\dfrac{5}{18}$
  4. $\dfrac{25}{91}$
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D Correct answer
Explanation

Let p = 1/6 be the probability of rolling a 6. Carol wins if she rolls a 6 on her 1st turn (prob = (1-p)^2 * p), 2nd turn (prob = (1-p)^5 * p), etc. This is a geometric series with first term a = (5/6)^2 * (1/6) = 25/216 and common ratio r = (5/6)^3 = 125/216. Sum = a / (1-r) = (25/216) / (1 - 125/216) = (25/216) / (91/216) = 25/91.

AI explanation

Since Alice, Bob and Carol take turns rolling a die, the game cycles in rounds where each has a 5/6 chance of not rolling a six. The probability that all three fail to roll a six in a single round is (5/6) multiplied by (5/6) multiplied by (5/6), which equals 125/216. Carol's probability of winning is the infinite geometric series of her winning in the first round, second round, third round and so on, given by the formula a divided by (1 minus r), where a is 25/216 and r is 125/216. Calculating (25/216) divided by (1 minus 125/216) yields (25/216) divided by (91/216), which gives the final probability of 25/91.

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