Question

A and B alternatively throw a pair of dice. A wins if he throws 6 before B throws 7 and B wins if he throws 7 before A throws 6. If A begins, show that his chance of winning is 30/61.

Answer 1

Answer on Question #44952 – Math - Statistics and Probability

Problem.

A and B alternatively throw a pair of dice. A wins if he throws 6 before B throws 7 and B wins if he throws 7 before A throws 6. If A begins, show that his chance of winning is 30/61.

Solution.

A could win the game when he rolls a 6 at the 1-st throw or when he rolls 6 at the $k$-th throw ($k > 1$) and in the first $k - 1$ throws A does not roll a 6 and in the first $k - 1$ throws B does not roll a 7.

There are $6 \cdot 6 = 36$ different combinations, that could be obtained in the each throw. There are 5 possibilities to obtain $6(1 + 5 = 2 + 4 = 3 + 3 = 4 + 2 = 5 + 1 = 6)$ and 6 possibilities to obtain $7(1 + 6 = 2 + 5 = 3 + 4 = 4 + 3 = 5 + 2 = 6 + 1 = 7)$. Therefore

$$P(A \text{ roll a 6 at some throw}) = \frac{5}{36},$$ $$P(A \text{ does not roll a 6 at some throw}) = 1 - \frac{5}{36} = \frac{31}{36},$$ $$P(B \text{ roll a 7 at some throw}) = \frac{7}{36},$$ $$P(B \text{ does not roll a 7 at some throw}) = 1 - \frac{6}{36} = \frac{30}{36} = \frac{5}{6},$$ $$\begin{array}{l} P(A \text{ does not roll a 6 at some throw and then } B \text{ does not roll a 7 at some throw}) = \frac{31}{36} \cdot \frac{5}{6} \\ = \frac{155}{216}. \end{array}$$

Hence

$$P(A \text{ wins at the } k\text{-th throw}) = \left(\frac{155}{216}\right)^{k-1} \cdot \frac{5}{36}$$

The total probability that A wins equals

$$P(A \text{ wins}) = \sum_{k=0}^{\infty} P(A \text{ wins at the } k\text{-th throw}) = \sum_{k=0}^{\infty} \left(\frac{155}{216}\right)^{k-1} \cdot \frac{5}{36} = \frac{\frac{5}{36}}{1 - \frac{155}{216}} = \frac{30}{61},$$

as the sum of infinite geometric progression.

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