(3) 甲、乙两人进行乒兵球比赛, 约定每局胜者得 1 分, 负者得 0 分, 比赛进行到有一人比对方多 2 分或打满 6 局时停止. 设甲在每局中获胜的概率为 $\frac{2}{3}$, 乙在每局中获胜的概率为 $\frac{1}{3}$, 且各局胜负相互独立, 则比赛停止时已打局数 $\xi$ 的期望 $E \xi$ 为 ( ).
(A) $\frac{241}{81}$
(B) $\frac{266}{81}$
(C) $\frac{274}{81}$
(D) $\frac{670}{243}$

(3) Two players, A and B, are playing a table tennis match, where the winner of each game gets 1 point and the loser gets 0 points. The match stops when one player is ahead by 2 points or after 6 games have been played. Suppose the probability of A winning each game is $\frac{2}{3}$, and the probability of B winning each game is $\frac{1}{3}$, with the outcomes of each game being independent. Then, the expected number of games played, $E \xi$, when the match stops is ( ).
(A) $\frac{241}{81}$
(B) $\frac{266}{81}$
(C) $\frac{274}{81}$
(D) $\frac{670}{243}$
Method 1: According to the problem, all possible values of $\xi$ are $2, 4, 6$. Let each two games be a round, then the probability that the match stops at the end of this round is $\left(\frac{2}{3}\right)^{2}+\left(\frac{1}{3}\right)^{2}=\frac{5}{9}$. If the match continues after this round, then A and B must each have scored one point in this round, and the result of this round has no effect on whether the match stops in the next round. Therefore, we have
$$
\begin{array}{l}
P(\xi=2)=\frac{5}{9} \\
P(\xi=4)=\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)=\frac{20}{81} \\
P(\xi=6)=\left(\frac{4}{9}\right)^{2}=\frac{16}{81}
\end{array}
$$

Thus,
$$
E \xi=2 \times \frac{5}{9}+4 \times \frac{20}{81}+6 \times \frac{16}{81}=\frac{266}{81} .
$$

Therefore, the answer is B.
Method 2: According to the problem, all possible values of $\xi$ are $2, 4, 6$. Let $A_{k}$ represent A winning the $k$-th game, then $\bar{A}_{k}$ represents B winning the $k$-th game. By independence and mutual exclusivity, we get
$$
\begin{aligned}
P(\xi=2)= & P\left(A_{1} A_{2}\right)+P\left(\bar{A}_{1} \bar{A}_{2}\right)=\frac{5}{9} \\
P(\xi=4)= & P\left(A_{1} \bar{A}_{2} A_{3} A_{4}\right)+P\left(A_{1} \bar{A}_{2} \bar{A}_{3} \bar{A}_{4}\right) \\
& +P\left(\bar{A}_{1} A_{2} A_{3} A_{4}\right)+P\left(\bar{A}_{1} A_{2} \bar{A}_{3} \bar{A}_{4}\right) \\
= & 2\left[\left(\frac{2}{3}\right)^{3}\left(\frac{1}{3}\right)+\left(\frac{1}{3}\right)^{3}\left(\frac{2}{3}\right)\right] \\
P(\xi=6)= & P\left(A_{1} \bar{A}_{2} A_{3} \bar{A}_{4}\right)+P\left(A_{1} \bar{A}_{2} \bar{A}_{3} A_{4}\right) \\
& +P\left(\bar{A}_{1} A_{2} A_{3} \bar{A}_{4}\right)+P\left(\bar{A}_{1} A_{2} \bar{A}_{3} A_{4}\right) \\
= & \frac{20}{81}, \\
= & 4\left(\frac{2}{3}\right)^{2}\left(\frac{1}{3}\right)^{2} \\
= & \frac{16}{81}
\end{aligned}
$$

Therefore,
$$
E \xi=2 \times \frac{5}{9}+4 \times \frac{20}{81}+6 \times \frac{16}{81}=\frac{266}{81} .
$$

Therefore, the answer is B.