<s>[INST] Probability involving Binomial Summation The problem statement is :
$A$ has $n$ coins and $B$ has $n+1$ coins.  They toss their coins simultaneously. 
If $p$ be the probability that $B$ will have more heads than $A$ the find $p=?$
One way I did this is by this argument :
Since $B$ cannot have the  number of heads and tails both greater than $A$. 
But since one of those cases is always true. (either more heads($p$) or more tails(=$q$ (let)) 
So by symmetry  both should have equal probabilities. 
Since $p+q=1$.
Then $p=0.5$
One more way of doing it is by binomial summation i.e.  Lets say $A$ has $x$ heads and $B$ has $y$ heads then we add up 
The product  $$ \binom{n}{x} *  \binom{n+1}{y} $$for all $y>x$.
But I am not able to evaluate that... 
So how to do I go about summing this?
 [/INST] We have by inspection that the desired probability is
$$\frac{1}{2^{2n+1}} 
\sum_{k=1}^{n+1} {n+1\choose k}
\sum_{m=0}^{k-1} {n\choose m}.$$
Now introducing
$${n\choose m} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z^{m+1}} \; dz$$
we obtain for the inner sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z} 
\frac{1/z^{k}-1}{1/z-1}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
(1+z)^n
\frac{1/z^{k}-1}{1-z}
\; dz$$
The second term vanishes and we get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^k} \frac{1}{1-z} (1+z)^n
\; dz$$
which could have  been obtained by inspection. Note  that this is zero
when $k=0$ so we may extend the outer sum to include $k=0$, getting
$$\frac{1}{2^{2n+1}} 
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{1-z} (1+z)^n
\sum_{k=0}^{n+1} {n+1\choose k} \frac{1}{z^k}
\; dz
\\ = \frac{1}{2^{2n+1}} 
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{1-z} (1+z)^n
\left(1+\frac{1}{z}\right)^{n+1}
\; dz
\\ = \frac{1}{2^{2n+1}} 
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\frac{1}{1-z} (1+z)^{2n+1}
\; dz.$$
Extracting the residue now yields
$$\frac{1}{2^{2n+1}} 
\sum_{q=0}^n {2n+1\choose q}
= \frac{1}{2^{2n+1}} \times \frac{1}{2} 2^{2n+1}
= \frac{1}{2}.$$
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