This question is surprisingly straightforward. Suppose with both toss a coin $N$ times. Clearly, the expected number of $H$s we throw will be the same. Consequently, you throwing more $H$s than me is contingent on your final throw. Half of the time you will throw a $H$ and end up with more than me, and half of the time you will throw a $T$ and the number of $H$s we throw will be the same. $$ $$ Therefore, the probability that you throw more $H$s than me is $0.5$.