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: 
$$\sum\_{i=1}^{6} \binom{6}{i}p^i(1-p)^{6-i}(\sum\_{j=0}^{i-1}\binom{5}{j}p^j(1-p)^{5-j})$$

The term in the first sum gives $\Pr(i\text{ heads out of 6 flips})$. The second sum is $\Pr(\text{strictly less than $i$ heads out of 5 flips})$. Then we use the sum out in front to add up all this probability for each $i$.
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 $\frac 1 2$.