China Mathematical Competition

Solution I It is easy to see that $\xi$ can only be $2$, $4$ or $6$. We divide the six games into three rounds, each consisting of two consecutive games. If one of the players wins two games in the first round, the match ends and the probability is $${\left(\frac{2}{3}\right)}^2+{\left(\frac{1}{3}\right)}^2=\frac{5}{9}.$$ Otherwise the players tie with each other, earning one point each, and the match enters the second round; this probability is $1-\frac{5}{9}=\frac{4}{9}$. We have similar discussions for the second and third rounds. So we get $$\begin{array}{rl}P(\xi =2)& =\frac{5}{9},\\ P(\xi =4)& =\frac{4}{9}\times \frac{5}{9}=\frac{20}{81},\\ P(\xi =6)& ={\left(\frac{4}{9}\right)}^2=\frac{16}{81}.\end{array}$$ Then $$E\xi =2\times \frac{5}{9}+4\times \frac{20}{81}+6\times \frac{16}{81}=\frac{266}{81}.$$

Solution II Let $A_k$ denote the event that $A$ wins the $k$th game, while ${\overline{A}}_k$ means that $B$ wins the game. Since $A_k$ and ${\overline{A}}_k$ are incompatible, and are independent of the other events, we have $$P(\xi =2)=P(A_1{A}_2)+P({\overline{A}}_1{\overline{A}}_2)=\frac{5}{9},$$ $$\begin{array}{rl}P(\xi =4)& =P(A_1{\overline{A}}_2{A}_3{A}_4)+P(A_1{\overline{A}}_2{\overline{A}}_3{A}_4)+\\ & P({\overline{A}}_1{A}_2{A}_3{A}_4)+P({\overline{A}}_1{A}_2{\overline{A}}_3{A}_4)\\ & =2\left[{\left(\frac{2}{3}\right)}^3\left(\frac{1}{3}\right)+{\left(\frac{1}{3}\right)}^3\left(\frac{2}{3}\right)\right]=\frac{20}{81},\end{array}$$ $$\begin{array}{rl}P(\xi =6)& =P(A_1{\overline{A}}_2{A}_3{\overline{A}}_4)+P(A_1{\overline{A}}_2{\overline{A}}_3{A}_4)+\\ & P({\overline{A}}_1{A}_2{A}_3{\overline{A}}_4)+P({\overline{A}}_1{A}_2{\overline{A}}_3{A}_4)\\ & =4{\left(\frac{2}{3}\right)}^2{\left(\frac{1}{3}\right)}^2=\frac{16}{81}.\end{array}$$ Then $$E\xi =2\times \frac{5}{9}+4\times \frac{20}{81}+6\times \frac{16}{81}=\frac{266}{81}.$$