8. Two Go teams, Team A and Team B, each have 5 players who compete in a relay match according to a predetermined order. The No. 1 players of both teams start the match, and the loser is eliminated; then the No. 2 player of the losing team competes against the winner, and the loser is eliminated. This continues until all players of one team are eliminated, at which point the other team is declared the winner. Assuming each player has equal strength, the expected number of Team A's players who have not played when the match ends, $E(X)=$ $\qquad$

---

I hope this translation meets your needs. If you have any further requests or need additional assistance, feel free to let me know!
8. $\frac{187}{256}$.

Let the event of Team A winning be denoted as $A$. Then
$$
P(A)=P(\bar{A})=\frac{1}{2} \text {. }
$$

When Team B wins, all members of Team A have participated, so the number of members who did not participate $X=0$.

When Team A wins and $X=k(k=0,1,36,4)$, $4-k$ members of Team A are eliminated, and the match needs to go through $9-k$ rounds, with the 5th member of Team A defeating the 5th member of Team B. The number of ways the first four members can be eliminated is $\mathrm{C}_{8-k}^{4-k}=\mathrm{C}_{8-k}^{4}$, and the probability of each situation is $\frac{1}{2^{9-k}}$. Therefore, the probability of $X=k(k=1,2,3,4)$ is
$$
\begin{array}{l}
P(X=k)=\mathrm{C}_{8-k}^{4} \times \frac{1}{2^{9-k}}, \\
P(X=0)=\frac{1}{2}+\mathrm{C}_{8}^{4} \times \frac{1}{2^{9}} .
\end{array}
$$

Substituting the values into the calculation yields Table 1.
Thus, the expected value of $X$ is
$$
\begin{array}{l}
E(X)=0 \times \frac{163}{256}+1 \times \frac{35}{256}+2 \times \frac{15}{128}+ \\
\quad 3 \times \frac{5}{64}+4 \times \frac{1}{32} \\
=\frac{187}{256} .
\end{array}
$$