8. Two players, A and B, are competing in a certain game. The rules state that the winner of each round gets 1 point, and the loser gets 0 points; the game ends when one player's score is 2 points higher than the other's. Additionally, the game is limited to a maximum of 20 rounds; if after 20 rounds, the player with the higher score wins, and if the scores are equal, it is a draw. It is known that the probability of A winning each round is $p$ $(0<p<1)$, and the probability of B winning is $q=1-p$. Assuming the results of each round are independent, the game ends after $\xi$ rounds. The range of the expected value $E \xi$ is $\qquad$
8. $2<E \xi \leq \frac{1023}{256}$

Analysis: Let $p(\xi=k)$ denote the probability that the game ends after $k$ rounds. If $k$ is odd, then the difference in scores between A and B is also odd, hence $p(\xi=k)=0$.
Consider the results of the first two rounds: (1) A wins both or B wins both, referred to as a decisive two rounds, with a probability of $p^{2}+q^{2}$;
(2) A and B each win one round, referred to as an indecisive two rounds, with two possible outcomes, hence a probability of $2 p q$. If the game ends after $k$ rounds, $k$ must be even, so the 1st and 2nd, 3rd and 4th, ..., $(k-3)$th and $(k-2)$th rounds are all indecisive.
If $k \neq 20$, then the $(k-1)$th and $k$th rounds are decisive, thus $p(\xi=k)=(2 p q)^{\frac{k}{2}-1}\left(p^{2}+q^{2}\right)$. If $k=20$, the game must end, so $p(\xi=20)=(2 p q)^{9}$. Therefore, $E \xi=\left(p^{2}+q^{2}\right) \sum_{i=1}^{9} 2 i(2 p q)^{i-1}+20(2 p q)^{9}$.
Given $p+q=1$, we know $p^{2}+q^{2}=1-2 p q$. Let $u=2 p q$, then $p^{2}+q^{2}=1-u$, so $E \xi=(1-u) \sum_{i=1}^{9} 2 i u^{i-1}+20 u^{9}$.
Let $s=\sum_{i=1}^{9} 2 i u^{i-1}$, then $u s=\sum_{i=1}^{9} 2 i u^{i}=\sum_{i=2}^{10} 2(i-1) u^{i-1}=\sum_{i=1}^{10} 2(i-1) u^{i-1}$,
$(1-u) s=\sum_{i=1}^{9} 2 u^{i-1}-18 u^{9}=\frac{2\left(1-u^{9}\right)}{1-u}-18 u^{9}$,
$E \xi=(1-u) s+20 u^{9}=\frac{2}{1-u}\left[1-u^{9}-9 u^{9}(1-u)+10 u^{9}(1-u)\right]=\frac{2\left(1-u^{10}\right)}{1-u}$.
Since $0<u \leq \frac{1}{2}$, we have $2<E \xi \leq \frac{1023}{256}$.