Example 10 (2003 New Curriculum College Entrance Examination Question) Two representative teams, A and B, are having a table tennis match. Each team has three players. Team A's players are $A_{1}, A_{2}, A_{3}$, and Team B's players are $B_{1}, B_{2}, B_{3}$. According to the statistics from many previous matches, the probability of winning and losing between the opposing players is as follows:
\begin{tabular}{|c|c|c|}
\hline Opposing Players & Probability of A Team Player Winning & Probability of A Team Player Losing \\
\hline$A_{1}$ vs $B_{1}$ & $\frac{2}{3}$ & $\frac{1}{3}$ \\
\hline$A_{2}$ vs $B_{2}$ & $\frac{2}{5}$ & $\frac{3}{5}$ \\
\hline$A_{3}$ vs $B_{3}$ & $\frac{2}{5}$ & $\frac{3}{5}$ \\
\hline
\end{tabular}

Now, the players will compete according to the matchups in the table. Each winning team gets 1 point, and each losing team gets 0 points. Let $\xi$ and $\eta$ be the total points of Team A and Team B, respectively.
(1) Find the probability distribution of $\xi$ and $\eta$;
(2) Find $E(\xi), E(\eta)$.
(1) Find the possible values of $\xi$ and $\eta$, which are $3, 2, 1, 0$.
$$
\begin{array}{l}
P(\xi=3)=\frac{2}{3} \times \frac{2}{5} \times \frac{2}{5}=\frac{8}{75}, \\
P(\xi=2)=\frac{2}{3} \times \frac{2}{5} \times \frac{3}{5}+\frac{1}{3} \times \frac{2}{5} \times \frac{2}{5}+\frac{2}{3} \times \frac{3}{5} \times \frac{2}{5}=\frac{28}{75}, \\
P(\xi=1)=\frac{2}{3} \times \frac{3}{5} \times \frac{3}{5}+\frac{1}{3} \times \frac{2}{5} \times \frac{3}{5}+\frac{1}{3} \times \frac{3}{5} \times \frac{2}{5}=\frac{2}{5}, \\
P(\xi=0)=\frac{1}{3} \times \frac{3}{5} \times \frac{3}{5}=\frac{3}{25} .
\end{array}
$$

According to the problem, $\xi+\eta=3$, so
$$
\begin{array}{l}
P(\eta=0)=P(\xi=3)=\frac{8}{75}, \\
P(\eta=1)=P(\xi=2)=\frac{28}{75}, \\
P(\eta=2)=P(\xi=1)=\frac{2}{5}, \\
P(\eta=3)=P(\xi=0)=\frac{3}{25} .
\end{array}
$$
(2) $E(\xi)=3 \times \frac{8}{75}+2 \times \frac{28}{75}+1 \times \frac{2}{5}+0 \times \frac{3}{25}=\frac{22}{15}$.

Since $\xi+\eta=3$,
then $E(\eta)=3-E(\xi)=\frac{23}{15}$.
Note: If the problem only requires the expected values of $\xi$ and $\eta$, it can be simplified as follows:
Let $\xi_{i}$ be the score of player $A_{i}(i=1,2,3)$ of team A in each game, then $\xi=\xi_{1}+\xi_{2}+\xi_{3}$.
The distribution of $\xi_{1}$ is
\begin{tabular}{|c|c|c|}
\hline$\xi_{1}$ & 1 & 0 \\
\hline$P$ & $\frac{2}{3}$ & $\frac{1}{3}$ \\
\hline
\end{tabular}

Therefore, $E\left(\xi_{1}\right)=1 \times \frac{2}{3}+0 \times \frac{1}{3}=\frac{2}{3}$.
Similarly, $E\left(\xi_{2}\right)=1 \times \frac{2}{5}+0 \times \frac{3}{5}=\frac{2}{5}$,
$$
E\left(\xi_{3}\right)=1 \times \frac{2}{5}+0 \times \frac{3}{5}=\frac{2}{5} \text {. }
$$

Therefore, $E(\xi)=E\left(\xi_{1}+\xi_{2}+\xi_{3}\right)=E\left(\xi_{1}\right)+E\left(\xi_{2}\right)+E\left(\xi_{3}\right)=\frac{2}{3}+\frac{2}{5}+\frac{2}{5}=\frac{22}{15}$,
$$
E(\eta)=E(3-\xi)=3-E(\xi)=\frac{23}{15} .
$$