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 $i$.
An answer using generating functions:

$$G(x)=\left(\frac{1}{2}+\frac{x}{2}\right)^6 \left(\frac{1}{2}+\frac{1}{2 \
x}\right)^5$$

Expanding that I get:

$$\frac{231}{1024}+\frac{1}{2048 x^5}+\frac{11}{2048 \
x^4}+\frac{55}{2048 x^3}+\frac{165}{2048 x^2}+\frac{165}{1024 \
x}+\frac{231 x}{1024}+\frac{165 x^2}{1024}+\frac{165 \
x^3}{2048}+\frac{55 x^4}{2048}+\frac{11 x^5}{2048}+\frac{x^6}{2048}$$

Use only the terms that have positive powers of x:

$$\frac{231 x}{1024}+\frac{165 x^2}{1024}+\frac{165 x^3}{2048}+\frac{55 \
x^4}{2048}+\frac{11 x^5}{2048}+\frac{x^6}{2048}$$

Enter x = 1 to clear out the x variable and sum the terms.

$$ \frac{231}{1024}+\frac{165}{1024}+\frac{165}{2048}+\frac{55}{2048}+
\frac{11}{2048}+\frac{1}{2048}=\frac{1}{2} $$

The answer is $ \frac{1}{2}$