Question: I have $6$ fair coins and you have $5$ fair coins. We both flip our own coins and observe the number of heads we each have. What is the probability that I have more heads than you?

Not sure how to start this, any help please?

Answer: 
Let $X$ be the number of heads in my 5 flips.

Let $Y$ be the number of heads in the first 5 of your 6 flips, and $Z$ be the number of heads from the flip of the sixth coin.

The probability that you have more heads than me is, by the Law of Total Probability

$$\begin{align}
\mathsf P(Y+Z > X) & = \mathsf P(Y>X\mid Z=0) \mathsf P(Z=0) + \mathsf P(Y\geq X\mid Z=1)\mathsf P(Z=1)
\end{align}$$

Now, use the fact that $X,Y,Z$ are mutually independent, and further that $X$ and $Y$ are identically and independently distributed. 

(**Hint:** use the law of complements to find symmetry)
Using a super simple approach, the only advantage the $6$ coin person has is the one "extra" coin. So it would be similar to both of them each giving up $5$ coins and one person just flipping the one remaining coin. Half the time that person would have more "heads" ($1$ vs. $0$) so the probability is $\frac 1 2$.