## Task B-4.5.

Andro and Borna take turns rolling a fair die whose sides are marked with the numbers 1, 2, 3, 4, 5, and 6. The winner is the first one to roll a six. If Andro starts the game, what is the probability that Borna wins?
## Solution.

For Borna to win after his first roll, Andro must get a number different from 6 on his first roll, and Borna must get a 6. Since the probability of rolling a six in one roll is $\frac{1}{6}$, and a number different from six is $\frac{5}{6}$, it follows that the probability of Borna winning after his first roll is $\frac{5}{6} \cdot \frac{1}{6}$.

Similarly, for Borna to win after his second roll, Andro in his first two rolls and Borna in his first roll must not get a six, which Borna must get on his second roll. The mentioned event will occur with a probability of $\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}=\left(\frac{5}{6}\right)^{3} \cdot \frac{1}{6}$.

Analogously, we conclude that the probability of Borna winning on his $n$-th roll is

$$
\left(\frac{5}{6}\right)^{2 n-1} \cdot \frac{1}{6}
$$

so the chance of Borna winning $p$ is equal to the infinite sum

$$
\begin{aligned}
p & =\frac{5}{6} \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^{3} \cdot \frac{1}{6}+\cdots+\left(\frac{5}{6}\right)^{2 n-1} \cdot \frac{1}{6}+\cdots \\
& =\frac{1}{6} \cdot\left(\frac{5}{6}+\left(\frac{5}{6}\right)^{3}+\cdots+\left(\frac{5}{6}\right)^{2 n-1}+\cdots\right)
\end{aligned}
$$

The expression in parentheses is a geometric series with the first term $\frac{5}{6}$ and the common ratio $\left(\frac{5}{6}\right)^{2}<1$, so its sum exists and is equal to $\frac{\frac{5}{6}}{1-\left(\frac{5}{6}\right)^{2}}=\frac{30}{11}$.

Now the probability of Borna winning is $p=\frac{1}{6} \cdot \frac{30}{11}=\frac{5}{11}$.