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Alternating dice race

ProbabilityMedium

Players A and B take turns rolling one fair $6$ -sided die, with A rolling first. The first player to roll a $6$ wins. What is the probability that A wins?

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Hint

Write an equation for A's winning chance by looking at what happens after a full round of $2$ failed rolls.

Solution

Let $P$ be the probability that A wins. A wins at once with probability $\frac{1}{6}$ .
Otherwise both A and B fail, with probability $(\frac{5}{6}) \times (\frac{5}{6}) = \frac{25}{36}$ , and the game restarts in the same position.
This gives $P = \frac{1}{6} + (\frac{25}{36}) \times P$ .
Then $P \times (\frac{11}{36}) = \frac{1}{6}$ , so $P = \frac{36}{66}$ , which reduces to $\frac{6}{11}$ .
The probability that A wins is $\frac{6}{11}$ .

Answer: $\frac{6}{11}$

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