jmc

More generally, suppose coin $A$ is tossed $n+1$ times and coin $B$ is tossed $n$ times.

Consider the situation when coin $A$ has been tossed $n$ times and coin $B$ has been tossed $n$ times. At this point, in terms of heads, either $A$ is ahead of $B,$ $B$ is ahead of $A,$ or they are tied. Let $p$ be the probability that $A$ is ahead of $B.$ Then by symmetry, $p$ is the probability that $B$ is ahead of $A,$ so the probability of a tie is $1-2p.$

Now coin $A$ is tossed one more time. If coin $A$ ends up with more heads, then either it was ahead before the last flip (which occurs with probability $p$), or it was tied before the last flip and got heads for the last flip (which occurs with probability $(1-2p)\cdot \frac{1}{2}$). Therefore, the probability that coins $A$ ends up with more heads is $$p+(1-2p)\cdot \frac{1}{2}=\boxed{\frac{1}{2}}.$$