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?
 A: 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
$$\frac{1}{2^{2n+1}} 
\sum_{q=0}^n {2n+1\choose q}
= \frac{1}{2^{2n+1}} \times \frac{1}{2} 2^{2n+1}
= \frac{1}{2}.$$
A: Instead of introducing two variables, let us just use $y$ to represent the number of heads Person $B$ gets. We wish to calculate the combinations such that Person $A$ gets less than $y$ heads, and sum these from $y = 1$ to $y = n + 1.$ This can be written as
$$\sum_{y = 1}^{n + 1}\sum_{k = 0}^{y - 1}\dbinom{n + 1}{y}\dbinom{n}{k}.$$
Notice that this is an extremely unwieldy count. Instead of counting that way, we can take advantage of that symmetry you noticed. In fact, this situation can be addressed with what is known as a $1-1$ correspondence. Say Person $A$ flips $x$ heads and Person $B$ gets $y > x$ heads. Then to make Person $B$ lose, all we have to do is switch all of each person's coins from $T$ to $H$ or from $H$ to $T.$ Since we have proven that the winning and losing counts are the same, the probability must be $\frac{1}{2}.$
A: $$\sum_{x=0}^n\sum_{y=x+1}^{n+1}{n\choose x}{n+1\choose y}=4^n$$
A: However, it was not stated that the coins were the same independent or identically distributed, I presume that the question in the textbook specified this:Or that both outcomes were possible on each coin for each person?
Person $B$ might have a deterministic coin which always lands heads,
and
Person $A$ may have a coin which always lands tails,
As this satisfies the hypothesis that person B cannot both have more heads and more tails then $A$ simultaneously, $p+q=1$ but $p=1 \neq q=0$
Buts its clearly that its not the case that the probability 'that person $B$ has more heads than A'=s $0.5$, rather its $1$
and its not the case that the probability that person $B$ has more tails then $A$ is $0.5$. In this counter-example its $0$. $p+q=1$  is satisfied nonetheless. Perhaps something further was specified in the question.?
So I presume that something else must have been specified. One generally cannot expect a probability distribution to fall out of the calculus alone.
I presume the question was phrased in terms of both outcomes being possible, and based on proportion counts?
But without a measure of some sort being specified its unclear what one deduce. Almost all binary sequences have relative frequency=$0.5$, but that does not mean that this always translates  into 'almost surely' and that 'necessarily  all two outcome processes have probability =$0.5$'
So long as you specify symmetry requirement further (I presume Rud-in specified the question further?); that presumably needs to be mentioned. Or that the trials are at least independently distributed with both outcomes possible (even if  identical distribution is not required, although it presumably is) .
