independent binomial variables $X$ and $Y$ be 2 independent bi variableswith parameters $n$ and $p=\frac12$. Is $P(X-Y=0)$ equal to this? Since ${X-Y=0}$ , then the events $X$ and $Y$ are equal, $X=Y$ then the event, $P(X-Y=0)$ equals to: $$P (\bigcup_{k=o}^n (X=k, Y=k) = \sum_{k=o}^n P(X=k, Y=k)=\sum_{k=o}^n P(X=k)P(Y=k)=\sum_{k=o}^n {n \choose k}\left(\frac12\right)^k \left(\frac12\right)^{n-k} {n \choose k}\left(\frac12\right)^k\left(\frac12\right)^{n-k}=\sum_{k=o}^n {n \choose k}\left(\frac12\right)^n {n \choose k}\left(\frac12\right)^n$$. How to simplify this one? Thanks I am stuck with this
 A: By symmetry the binomial coefficients part is $\sum_0^n \binom{n}{k}\binom{n}{n-k}$. This sum is $\binom{2n}{n}$.
To see that, suppose we have a box of $n$ distinct doughnuts and a box of $n$ distinct muffins, and want to choose a snack of $n$ items. This can be done in $\binom{2n}{n}$ ways. 
But it can also be done in $\sum_0^n\binom{n}{k}\binom{n}{n-k}$ ways, for that counts the ways we can pick, for any $k$, $k$ doughnuts and the rest muffins.
Another way: The random variable $X+Y$ has binomial distribution, for $X+Y$ is the number of "successes" in $2n$ independent trials, where the probability of success on any trial is $\frac{1}{2}$.
Note now that for any $k$, the probability that Alicia  gets $k$ heads and Beti gets $k$ heads is the same as the probability that Alicia gets $k$ heads and Beti gets $k$ tails, that is, the total number of heads between them is $n$.
We have just shown that $\Pr(X=Y)=\Pr(X+Y=n)$. This probability is $\binom{2n}{n}\frac{1}{2^{2n}}$.
A: Isn't that 
$\frac{1}{4^n}\sum_{k=0}^n  {n \choose k}^2$
