10. (20 points) Rolling Dice (a uniform cube, with six faces marked with $1,2,3,4,5,6$) Game rules are as follows: First roll 9 dice, take out the dice showing 1 and set them aside; on the second roll, take out the dice showing 1 from the remaining dice; $\cdots \cdots \cdots$, until no dice show 1 or all dice are taken out, the game ends. It is known that the probability of the game ending exactly after 9 rolls is $\frac{a b^{u}}{c^{u} d^{u}}(a, b, c, d$ are different prime numbers, $u, v \in \mathbf{N}_{+}$). Find $u v+\overline{b c d}$.
10. According to the game rules, if the game ends exactly after 9 rounds, then in the first eight rounds, each time exactly 1 die shows a 1, and the ninth round ends the game regardless of whether it shows a 1 or not. Among these, the probability that exactly 1 die shows a 1 in the $k(k=1,2, \cdots, 8)$-th round, where $10-k$ dice are thrown, is $\frac{\mathrm{C}_{10-k}^{1} \times 1 \times 5^{9-k}}{6^{10-k}}$.
$$
\begin{array}{l}
\text { Then } \frac{a b^{u}}{c^{v} d^{u}}=\prod_{k=1}^{8} \frac{\mathrm{C}_{10-k}^{1} \times 5^{9-k}}{6^{10-k}}=\prod_{k=2}^{9} \frac{k \times 5^{k-1}}{6^{k}} \\
=\frac{9!\times 5^{36}}{6^{4}}=\frac{56 \times 5^{37}}{6^{40}}=\frac{7 \times 5^{37}}{3^{40} \times 2^{37}} \\
\Rightarrow a=7, b=5, c=3, d=2, u=37, v=40 . \\
\text { Hence } u v+\overline{b c d}=37 \times 40+532=2012 .
\end{array}
$$