M tosses $7$ fair coins and has $M$ heads. $A$ tosses $6$ fair coins and has $A$ heads. Find probability $P(M>A).$ $M$ tosses $7$ fair coins and has M heads. A tosses $6$ fair coins and has $A$ heads. Find probability $P(M>A)$.
I suppose that both distributions are binominal, but I don't know what to do next.
 A: The probability that $A=a,M=m$ is $\frac{1}{2^{13}}\binom{7}{m}\binom{6}{a}$. 
But that is the same as the probability that $A=6-a,M=7-m$. 
This map $(a,m)\to(6-a,7-m)$ is a bijection from $\{0,\dots,6\}\times\{0,\dots,7\}$ to itself, with the property that $a<m$ iff $7-m\leq 6-a$. So this means that the probability that $A<M$ is the same as the probability that $A\geq M$, and thus the probability is exactly $\frac{1}{2}$.
This will work more generally, if $M$ tosses $n+1$ counts and $A$ tosses $n$ coins, then $P(M>A)=\frac{1}{2}$.

Maybe an easier approach: Have them each flip six coins first, calling the results $M_6$ and $A$. Then with equal probability, $M_6<A$ and $M_6>A$.
If $M_6=A$, have $M$ flip another coin. Then $\frac{1}{2}$ the times when $M_6\neq A$, you get $M\geq M_6>A$, and $\frac{1}{2}$ the time when $M_6=A$, you get $M=1+M_6>A$. 
Basically, the seventh toss for $M$ "breaks the tie" case for $A=M_6$, and the non-tie cases are also equally likely.
A: The values $6$ and $7$ are sufficiently small that you could calculate this by hand.
The probabilty that $A$ gets zero heads is $ \frac{1}{64}$ and $M$ get one or more heads is $\frac{127}{128}$.
The probabilty that $A$ gets one head is $ \frac{6}{64}$ and $M$ get two or more heads is $\frac{120}{128}$.
...
The probabilty that $A$ gets six heads is $ \frac{1}{64}$ and $M$ gets seven heads is $\frac{1}{128}$.
We have
\begin{eqnarray*}
P= \frac{1}{64} \frac{127}{128}+\frac{6}{64} \frac{120}{128}+\frac{15}{64} \frac{99}{128}+\frac{20}{64} \frac{64}{128}+\frac{15}{64} \frac{29}{128}+\frac{6}{64} \frac{8}{128}+\frac{1}{64} \frac{1}{128}=\frac{4096}{8192}
\end{eqnarray*}
So the probability is $\color{red}{\frac{1}{2}}$.
