2. Person A and Person B each independently toss a fair coin, A tosses 10 times, B tosses 11 times, then the probability that the number of times B gets heads is more than the number of times A gets heads is


Assume that both A and B toss a coin 10 times. At this point, the probability that the number of times the coin lands heads up for B is more than for A is equal to the probability that A is more than B, which is set as $p$. The probability that both are equal is $1-2p$. Therefore, when B tosses the coin for the 11th time, if the number of times the coin lands heads up in the first 10 tosses is more for B than for A, the probability is $p$; if the number of times the coin lands heads up in the first 10 tosses is the same for both, the probability is $\frac{1}{2}(1-2 p)$. In summary, the probability that the number of times the coin lands heads up for B is more than for A is $P=p+\frac{1}{2}(1-2 p)=\frac{1}{2}$.