8. Two individuals, A and B, independently toss a fair coin. A tosses the coin 10 times, while B tosses it 11 times. The probability that the number of times the coin lands heads up for B is greater than that for A is $\qquad$ (Provided by Gan Zhiguo)
8. $\frac{1}{2}$.
(Method 1) Let the number of times the coin lands heads up for person A and B be $X, Y$, respectively, then $X \sim B\left(10, \frac{1}{2}\right), Y \sim B\left(11, \frac{1}{2}\right)$. Therefore,
$$
\begin{array}{l}
P(X=k)=\mathrm{C}_{10}^{k}\left(\frac{1}{2}\right)^{k}\left(\frac{1}{2}\right)^{10-k}=\frac{\mathrm{C}_{10}^{k}}{2^{10}} \quad(k=0,1, \cdots, 10) \\
P(Y=l)=\mathrm{C}_{11}^{l}\left(\frac{1}{2}\right)^{l}\left(\frac{1}{2}\right)^{11-l}=\frac{\mathrm{C}_{11}^{l}}{2^{11}} \quad(l=0,1, \cdots, 11)
\end{array}
$$

Thus, the required probability is
$$
\begin{array}{l} 
P(Y>X) \\
= P(Y=1, X=0)+P(Y=2, X \leqslant 1)+P(Y=3, X \leqslant 2)+\cdots+P(Y=11, X \leqslant 10) \\
= \frac{1}{2^{21}}\left(\mathrm{C}_{11}^{1} \mathrm{C}_{10}^{0}+\mathrm{C}_{11}^{2}\left(\mathrm{C}_{10}^{0}+\mathrm{C}_{10}^{1}\right)+\mathrm{C}_{11}^{3}\left(\mathrm{C}_{10}^{0}+\mathrm{C}_{10}^{1}+\mathrm{C}_{10}^{2}\right)+\cdots\right. \\
\left.+\mathrm{C}_{11}^{11}\left(\mathrm{C}_{10}^{0}+\mathrm{C}_{10}^{1}+\mathrm{C}_{10}^{2}+\cdots+\mathrm{C}_{10}^{10}\right)\right) \\
= \frac{1}{2^{21}}(11 \times 1+55 \times 11+165 \times 56+330 \times 176+462 \times 386+462 \times 638 \\
+330 \times 848+165 \times 968+55 \times 1013+11 \times 1023+1 \times 1024) \\
= \frac{1048576}{2097152}=\frac{1}{2} .
\end{array}
$$
(Method 2) Consider the two people each tossing the coin 10 times, there are three possible scenarios:
(a) The number of times the coin lands heads up for person A is more than for person B;
(b) The number of times the coin lands heads up for person A and B is the same;
(c) The number of times the coin lands heads up for person B is more than for person A.

Let the probability of scenario (a) occurring be $p_{1}$, and the probability of scenario (b) occurring be $p_{2}$, then the probability of scenario (c) occurring is also $p_{1}$, so $2 p_{1}+p_{2}=1$, i.e., $p_{1}+\frac{1}{2} p_{2}=\frac{1}{2}$.

At this point, since the total number of coin tosses is even, the difference in the number of times the coin lands heads up for the two people is even, i.e., the number of times the coin lands heads up for the two people is either the same, or one person has at least two more heads than the other. Therefore, when person A tosses the coin 10 times and person B tosses the coin 11 times, the event that the number of times the coin lands heads up for person B is more than for person A is equivalent to, after each person tosses the coin 10 times, scenario (c) occurring, or scenario (b) occurring and the 11th toss by person B results in heads, thus, the required probability is $p_{1}+\frac{1}{2} p_{2}=\frac{1}{2}$.