We roll a die until a six comes up. What is the probability that we do not roll a five in the meantime?
I. solution. Let $B_{k}$ and $A_{k}$ denote the events that the first six is rolled on the $k$-th throw, and that no five is rolled up to the $k$-th throw, respectively. The event that no five is rolled until the first six is rolled occurs if and only if the event

$$
B_{1} A_{1}+B_{2} A_{2}+\ldots+B_{n} A_{n}+\ldots
$$

occurs. Therefore, the sought probability $p$ is equal to the probability $P\left(A_{1} B_{1}+A_{2} B_{2}+\ldots+\ldots+A_{k} B_{k}+\ldots\right)$. Since $\left(A_{i} B_{i}\right)\left(A_{j} B_{j}\right)=0$ for $i \neq j$, we have

$$
p=P\left(A_{1} B_{1}\right)+P\left(A_{2} B_{2}\right)+\ldots+P\left(A_{k} B_{k}\right)+\ldots
$$

Using the fact that

$$
\begin{gathered}
P\left(A_{k} B_{k}\right)=\left(\frac{4}{6}\right)^{k-1} \cdot \frac{1}{6} \\
p=\frac{1}{6}+\left(\frac{1}{6}\right)\left(\frac{4}{6}\right)+\left(\frac{1}{6}\right)\left(\frac{4}{6}\right)^{2}+\ldots+\left(\frac{1}{6}\right)\left(\frac{4}{6}\right)^{k-1}+\ldots=\frac{1}{6} \cdot \frac{1}{1-\frac{4}{6}}=\frac{1}{2}
\end{gathered}
$$

(György Pócsi, Debrecen, Fazekas M. Gymnasium)

II. solution. Let $A$ be the event that a five is rolled before a six, $B$ the event that a six is rolled before a five, and $C$ the event that neither of the above cases occurs. Clearly, $P(A)=P(B)$, $H=A+B+C$, and $A B=A C=B C=0$. Therefore,

$$
P(A)+P(B)+P(C)=2 \cdot P(A)+P(C)=1
$$

Let's determine $P(C)$!

$$
P(C)=1-\sum_{k=1}^{\infty}\left(\frac{4}{6}\right)^{k-1} \cdot \frac{2}{6}=1-\frac{2}{6} \cdot \frac{1}{1-\frac{2}{3}}=1-\frac{2}{6} \cdot 3=0
$$

Substituting this into (1):

$$
P(A)=1 / 2 \text { follows. }
$$

(Based on the work of Erzsébet Szécsi, Debrecen, Fazekas M. Gymnasium II. o. t.)