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: 
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}$
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 $\frac 1 2$.