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)
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}$