8. Two Go teams, Team A and Team B, each have 5 players who compete in a broadcast match according to a predetermined order. The No. 1 players of both teams compete first, and the loser is eliminated; then the No. 2 player of the losing team competes with the winner, and the loser is eliminated again. 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, what is the expected number of Team A's players who have not yet played when the match ends, i.e., \( E X = \) \(\qquad\)

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The problem is translated as follows:

8. Two Go teams, Team A and Team B, each have 5 players who compete in a broadcast match according to a predetermined order. The No. 1 players of both teams compete first, and the loser is eliminated; then the No. 2 player of the losing team competes with the winner, and the loser is eliminated again. 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, what is the expected number of Team A's players who have not yet played when the match ends, i.e., \( E X = \) \(\qquad\)
8. $\frac{1}{6}$.

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

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

When Team A wins and $X=k(k=0,1, \cdots, 4)$, $4-k$ members of Team A are eliminated, and the game needs to be played for $9-k$ rounds, with the $(5-k)$-th member of Team A defeating the 5th member of Team B. Their order is fixed. Additionally, the order in which the 5 members of Team A play is $\mathrm{C}_{10}^{5}$, and the order in which the first four members are eliminated is $\mathrm{C}_{8-k}^{4-k}=\mathrm{C}_{8-k}^{4}$. Therefore, when Team A wins, the probability that $X=k(k=0,1, \cdots, 4)$ is
$$
P(X=k \mid A)=\frac{\mathrm{C}_{8-k}^{4}}{\mathrm{C}_{10}^{5}} \text {. }
$$

In summary, the probability distribution of the number of Team A members who did not play, $X$, is
$$
\begin{array}{l}
P(X=0)=P(\bar{A})+P(X=k \mid A) P(A), \\
P(X=k)=P(X=k \mid A) P(A)(k=1,2 ; 3,4) .
\end{array}
$$

Substituting the values, we get Table 1:
Table 1
\begin{tabular}{|c|c|c|c|c|c|}
\hline$X$ & 0 & 1 & 2 & 3 & 4 \\
\hline$P$ & $\frac{23}{36}$ & $\frac{35}{504}$ & $\frac{15}{504}$ & $\frac{5}{504}$ & $\frac{1}{504}$ \\
\hline
\end{tabular}

Thus, the expected value of $X$ is
$$
\begin{array}{c}
E X=0 \times \frac{23}{36}+1 \times \frac{35}{504}+2 \times \frac{15}{504}+ \\
3 \times \frac{5}{504}+4 \times \frac{1}{504}=\frac{1}{6} .
\end{array}
$$