A game is composed of the following rules:

i) In each round, a common unbiased die is rolled.

ii) If the number 3 comes up, then player $A$ wins.

iii) If one of the numbers in the set $\{4,5,6\}$ comes up, then player $B$ wins.

iv) If one of the numbers in the set $\{1,2\}$ comes up, then the die is rolled again until it results in 3 or 4 or 5 or 6.

What is the probability that player $B$ wins?
Solution

Let $P_{i}(B)$ be the probability of player $B$ winning in round $i$, with $i$ a positive integer, and $P_{i}(\overline{A+B})$ the probability that neither $A$ nor $B$ wins in round $i$. Therefore, we have

$$
\begin{aligned}
& P_{1}(B)=\frac{3}{6}=\frac{1}{2} \\
& P_{2}(B)=P_{1}(\overline{A+B}) \cdot P_{2}(B)=\frac{2}{6} \cdot \frac{1}{2}=\frac{1}{6} \\
& P_{3}(B)=P_{1}(\overline{A+B}) \cdot P_{2}(\overline{A+B}) \cdot P_{3}(B)=\frac{2}{6} \cdot \frac{2}{6} \cdot \frac{1}{2}=\frac{1}{18} \\
& \quad \vdots \\
& P_{n}(B)=P_{1}(\overline{A+B}) \cdot P_{2}(\overline{A+B}) \cdot \ldots \cdot P_{n-1}(\overline{A+B}) \cdot P_{n}(B)=\left(\frac{1}{3}\right)^{n-1} \cdot \frac{1}{2}
\end{aligned}
$$

Notice that the results form a geometric progression with a common ratio of 1/3, which we need to sum to find the probability $P(B)$ of $B$ winning. Therefore,

$$
P(B)=\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\cdots+\left(\frac{1}{3}\right)^{n-1} \cdot \frac{1}{2}=\frac{\frac{1}{2}}{1-\frac{1}{3}}=\frac{3}{4}
$$

Observation: If $|q|<1$ and $S=1+q+q^{2}+\ldots+q^{n-1}$, then

$$
S q=q+q^{2}+q^{3} \ldots+q^{n}
$$

Thus, $S q-S=q^{n}-1$ and, consequently, $S=\frac{q^{n}-1}{q-1}$. At the end of the previous solution, we applied this formula for the sum of the terms of a geometric progression with $q=1 / 3$.