It is $50\%$.

Let's draw a picture and consider two stages. The first stage is for the first $n$ times. The second stage is for Joe to make his $(n+1)$ toss.

In the first stage, there are three possible outcomes:

1. Joe has more heads than Mary does.
2. Joe has less heads than Mary does.
3. They have the same number of heads.

By the principle of symmetry, the first two cases should occur with the same probability.

Note that if Joe already has more heads in the first $n$ flips, regardless of his last flip, he always gets more heads than Mary does. Meanwhile, if he gets less heads in the first $n$ flips, the best he can achieve after his last flip is to have the same heads as Mary does. This means he cannot have more heads in this case. It follows that case $1$ and $2$ above will lead to an average probability of $50\%$ that Joe has more heads in this game. 

Now, in the $3^{rd}$ case when they have the same heads during the first $n$ flips, it is obvious that whether Joe will end up with more heads depends on his last flip. There is $50\%$ chance that he will get a head.

Therefore, all in all, Joe has a half-half chance to have more heads than Mary does.

$\textbf{Note}$

This problem can be solved using mathematical calculation. However, applying the principle of symmetry can lead to a neat solution.