8. Two players, A and B, are playing a table tennis match, using the 11-point system, which means that for each ball, the winner gets 1 point and the loser gets 0 points. When one side accumulates 11 points, if the other side has less than 10 points, the one who reaches 11 points wins; if the other side has 10 points, the game continues until the score difference is 2 points, and the one with the higher score wins. It is known that A and B have similar skills, meaning that for each ball, the probability of A winning or B winning is $\frac{1}{2}$. Then, the probability that the winner scores 11 points at the end of a game is . $\qquad$
$8.1-\frac{\mathrm{C}_{20}^{10}}{2^{20}}$.
We only need to find the probability $P$ that one side reaches 11 points but cannot win yet. At this point, the score in the first 20 balls is 10:10. It can be seen that, in the first 20 balls, arranging 10 positions for A to win, there are $\mathrm{C}_{20}^{10}$ possibilities. And all possible win-loss scenarios have $2^{20}$ possibilities. Therefore, $P=\frac{\mathrm{C}_{20}^{10}}{2^{20}}$.