8. There are five table tennis balls, three of which are new, and two are old (i.e., used at least once). Each time a match is played, two balls are taken out and used, and then all are put back. Let the number of new balls taken in the second match be $\xi$, then the mathematical expectation $E \xi=$ $\qquad$ .
8. $\frac{18}{25}$.

Let the probability of drawing $m$ new balls in the $i$-th game be $P_{i}(m)$. It is easy to see that
$$
\begin{array}{l}
P_{1}(0)=\frac{\mathrm{C}_{2}^{2}}{\mathrm{C}_{5}^{2}}=\frac{1}{10}, \\
P_{1}(1) \frac{\mathrm{C}_{3}^{1} \mathrm{C}_{2}^{1}}{\mathrm{C}_{5}^{2}}=\frac{6}{10}, \\
P_{1}(2) \frac{\mathrm{C}_{3}^{2}}{\mathrm{C}_{5}^{2}}=\frac{3}{10} . \\
\text { Therefore, } P_{2}(0)=\sum_{m=0}^{2}\left(P_{1}(m) \times \frac{\mathrm{C}_{2+m}^{2}}{\mathrm{C}_{5}^{2}}\right) \\
=\frac{1}{10} \times \frac{1}{10}+\frac{6}{10} \times \frac{3}{10}+\frac{3}{10} \times \frac{6}{10}=\frac{37}{100}, \\
P_{2}(1)=\sum_{m=0}^{3}\left(P_{1}(m) \times \frac{\mathrm{C}_{3-m}^{1} \mathrm{C}_{2+m}^{1}}{\mathrm{C}_{5}^{2}}\right) \\
=\frac{1}{10} \times \frac{6}{10}+\frac{6}{10} \times \frac{6}{10}+\frac{3}{10} \times \frac{4}{10}=\frac{54}{100}, \\
P_{2}(2)=\sum_{m=0}^{1}\left(P_{1}(m) \times \frac{\mathrm{C}_{3-m}^{2}}{\mathrm{C}_{5}^{2}}\right) \\
=\frac{1}{10} \times \frac{3}{10}+\frac{6}{10} \times \frac{1}{10}=\frac{9}{100} .
\end{array}
$$

Thus, $E \xi=0 \times \frac{37}{100}+1 \times \frac{54}{100}+2 \times \frac{9}{100}=\frac{18}{25}$.