Let $X$ be the number of tails Jane counted, and $Y$ be the number of tails John counted. Then $X$ and $Y$ are independent, and $X \sim \operatorname{Bin}\left(n+1,\frac{1}{2}\right)$, $Y \sim \operatorname{Bin}\left(n,\frac{1}{2}\right)$. Note that $X$ is equal in distribution to $Z + U$, where $Z \sim \operatorname{Bin}\left(n,\frac{1}{2}\right)$ and $U \sim \operatorname{Bern}\left(\frac{1}{2}\right)$, where $U$, $Z$ and $Y$ are independent. Thus:
$$
\mathbb{P}\left(X > Y\right) = \mathbb{P}\left(Z + U > Y\right) = \mathbb{P}\left(Z + U > Y | U = 0 \right) \mathbb{P}\left(U = 0 \right) + \mathbb{P}\left(Z + U > Y | U = 1 \right) \mathbb{P}\left(U = 1 \right)
$$
Meaning
$$ \begin{eqnarray}
\mathbb{P}\left(X > Y\right) &=& \frac{1}{2} \left( \mathbb{P}\left(Z > Y\right) + \mathbb{P}\left(Z + 1 > Y\right) \right) \\ &=& \frac{1}{2} \left( \mathbb{P}\left(Z > Y\right) + \mathbb{P}\left(Z \geqslant Y\right) \right) = \frac{1}{2} \left( \mathbb{P}\left(Z > Y\right) + \left(1- \mathbb{P}\left(Z < Y\right) \right) \right) = \frac{1}{2}
\end{eqnarray}
$$
since by symmetry $\mathbb{P}\left(Z>Y\right) = \mathbb{P}\left(Z<Y\right)$.
