8. Team $A$ and Team $B$ are having a table tennis team competition. Each team has three players, and each player will play once. The three players of Team $A$ and Team $B$ are $A_{1}, A_{2}, A_{3}, B_{1}, B_{2}, B_{3}$, respectively, and the winning probability of $A_{i}$ against $B_{j}$ is $\frac{i}{i+j}(1 \leqslant i, j \leqslant 3)$. Then the maximum possible value of the expected score of Team $A$ is
8. $\frac{91}{60}$.

It can be discussed that when $A_{1}: B_{3}, A_{2}: B_{1}, A_{3}: B_{2}$, the maximum expected value is $\frac{91}{60}$.