Three players A, B, and C take turns in throwing a die. The first one to throw a six wins. If A starts, what is the probability that C wins?

Solution

1. Let $p=1/6$ be the probability of rolling a six and $q=5/6$ be the probability of not rolling a six.
2. C wins if the first two players fail to roll a six and C succeeds, which occurs on turns $3,6,9,\dots$ with probabilities $q^{2}p,q^{5}p,q^{8}p,\dots$.
3. This is an infinite geometric series with the first term $$\begin{array}{rl}a& =q^{2}p\\ & =(5/6{)}^2(1/6)\\ & =25/216\end{array}$$ and a common ratio $$\begin{array}{rl}r& =q^3\\ & =(5/6{)}^3\\ & =125/216\end{array}$$.
4. The sum of the series is $$\begin{array}{rl}S& =\frac{a}{1-r}\\ & =\frac{25/216}{1-125/216}\\ & =\frac{25/216}{91/216}\\ & =25/91\end{array}$$.
5. Hence the answer is (C).