Flipping coins probability of $6$ flips having more heads than $5$ flips. 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?
 A: $$\sum_{i=1}^{6} \binom{6}{i}p^i(1-p)^{6-i}(\sum_{j=0}^{i-1}\binom{5}{j}p^j(1-p)^{5-j})$$
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 $i$.
A: 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)
A: 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}$
A: 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 $\frac 1 2$.
