9. Eleonora plays with a die and a (stopped) clock that initially shows 12. She rolls the die 2008 times and moves the hour hand forward by as many hours as the result. What is the probability that the hour hand will be horizontal at the end?
(A) 0
(B) $\frac{1}{2008}$
(C) $\frac{1}{1004}$
(D) $\frac{1}{12}$
(E) $\frac{1}{6}$.
9. The answer is $\mathbf{( E )}$. The hour hand is horizontal if the clock shows 3 or 9. Let $K$ be the hour indicated by the clock just before the last die roll. If $K$ is one of the 6 numbers between 3 and 8, there is exactly 1 outcome of the die out of 6 that would allow reaching 9, and there is none that would allow reaching 3. If $K$ is one of the 6 numbers between 9 and 12 or between 1 and 2, there is exactly 1 outcome of the die out of 6 that would allow reaching 3, and there is none that would allow reaching 9.

The total number of combinations of 2008 die rolls can be divided into 12 groups depending on the situation before the last roll. For $i=1,2, \ldots, n$, let $n_{i}$ be the number of combinations such that $K=i$; and let $a_{i}$ be the number of combinations such that $K=i$ and the hour hand is horizontal after the last roll.

From the above, for each of these groups, the probability that the hour hand will be brought to a horizontal position with the last roll is $\frac{1}{6}$, so for each $i$, we have that $\frac{a_{i}}{n_{i}}=\frac{1}{6}$, or that $n_{i}=6 a_{i}$. These relationships allow us to easily obtain the overall probability that the hour hand will be horizontal at the end, that is:

$$
\frac{\text { favorable cases }}{\text { possible cases }}=\frac{a_{1}+a_{2}+\cdots+a_{12}}{n_{1}+n_{2}+\cdots+n_{12}}=\frac{a_{1}+a_{2}+\cdots+a_{12}}{6 a_{1}+6 a_{2}+\cdots+6 a_{12}}=\frac{a_{1}+a_{2}+\cdots+a_{12}}{6\left(a_{1}+a_{2}+\cdots+a_{12}\right)}=\frac{1}{6} \text {. }
$$