5. (1981 American High School Mathematics Exam Question) Alice, Bob, and Carl take turns rolling a die. Alice rolls first, Bob always rolls after Alice, and Carl always rolls after Bob. Find the probability that Carl is the first to roll a 6. (In each roll of the die, the probability of getting a 6 is $\frac{1}{6}$, and the rolls are independent of each other. )
A. $\frac{1}{3}$
B. $\frac{2}{9}$
C. $\frac{5}{18}$
D. $\frac{25}{91}$
E. $\frac{36}{91}$
5. D. Reason: The probability that Karl rolls a 6 first on his first roll is $\left(\frac{5}{6}\right)^{2} \cdot \frac{1}{6}$; the probability that he rolls a 6 first on his second roll is $\left(\frac{5}{6}\right)^{5} \cdot \frac{1}{6}$;
In general, the probability that Karl rolls a 6 first on his $n$-th roll is $\left(\frac{5}{6}\right)^{3 n-1} \cdot \frac{1}{6}$. Therefore, the probability that Karl rolls a 6 first is
$$
\left(\frac{5}{6}\right)^{2} \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^{5} \cdot \frac{1}{6}+\cdots+\left(\frac{5}{6}\right)^{3 n-1} \cdot \frac{1}{6}+\cdots
$$

This is the sum of an infinite geometric series with the first term $a=\left(\frac{5}{6}\right)^{2} \cdot \frac{1}{6}$ and common ratio $r=\left(\frac{5}{6}\right)^{3}$, which is
$$
\frac{a}{1-r}=\frac{25}{91} \text {. }
$$