2. Two individuals, A and B, independently flip a fair coin. A flips the coin 10 times, and B flips it 11 times. The probability that B gets more heads than A is $\qquad$ .
2. $\frac{1}{2}$.

Let "the number of times the coin lands heads up for Yi is more than the number of times the coin lands heads up for Jia when each of them independently tosses the coin 10 times" be event $A$. Then the probability of the desired event is
$$
P=P(A)+\frac{1}{2}(1-2 P(A))=\frac{1}{2} .
$$