$14 \cdot 26$ Alice, Bob, and Carl take turns rolling a die. Alice rolls first, Bob always follows Alice, and Carl always follows Bob. Find the probability that Carl is the first to roll a 6. (On each roll, 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) $\frac{36}{91}$.
(32nd American High School Mathematics Examination, 1981)
[Solution] The probability that Karl rolls a 6 for the first time on his first roll is
$$
\left(\frac{5}{6}\right)^{2} \cdot \frac{1}{6} ;
$$

The probability that he rolls a 6 for the first time on his second roll is
$$
\left(\frac{5}{6}\right)^{5} \cdot \frac{1}{6} \text {; }
$$

In general, the probability that Karl rolls a 6 for the first time on his $n$-th roll is
$$
\left(\frac{5}{6}\right)^{3 n-1} \cdot \frac{1}{6} \text {. }
$$

Therefore, the probability that Karl rolls a 6 for the first time 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
$$
\frac{a}{1-r}=\frac{25}{91} \text {. }
$$

Therefore, the answer is $(D)$.