9. In a uniform small cube, three

faces are marked with the number 0, two faces are marked with the number 1, and one face is marked with the number 2. If the small cube is thrown twice, then the mathematical expectation of the product of the numbers facing up is $\qquad$
9. $\frac{4}{9}$.

According to the problem, the probability of the number facing up on the small cube being 0 is $\frac{1}{2}$, the probability of it being 1 is $\frac{1}{3}$, and the probability of it being 2 is $\frac{1}{6}$, as shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|}
\hline Second Roll & 0 & 1 & 2 \\
\hline 0 & $\frac{1}{2} \times \frac{1}{2}$ & $\frac{1}{2} \times \frac{1}{3}$ & $\frac{1}{2} \times \frac{1}{6}$ \\
\hline 1 & $\frac{1}{3} \times \frac{1}{2}$ & $\frac{1}{3} \times \frac{1}{3}$ & $\frac{1}{3} \times \frac{1}{6}$ \\
\hline 2 & $\frac{1}{6} \times \frac{1}{2}$ & $\frac{1}{6} \times \frac{1}{3}$ & $\frac{1}{6} \times \frac{1}{6}$ \\
\hline
\end{tabular}

The distribution of the product of the numbers facing up is shown in Table 2.
Table 2
\begin{tabular}{|c|c|c|c|c|}
\hline$\xi$ & 0 & 1 & 2 & 4 \\
\hline$P$ & $\frac{3}{4}$ & $\frac{1}{9}$ & $\frac{1}{9}$ & $\frac{1}{36}$ \\
\hline
\end{tabular}

Therefore, $E \xi=0 \times \frac{3}{4}+1 \times \frac{1}{9}+2 \times \frac{1}{9}+4 \times \frac{1}{36}=\frac{4}{9}$.