= Tie probability for independent binomial counts
{title2=$\Pr(X=Y)=\sum_{j=0}^n\binom nj^2p^{2j}(1-p)^{2n-2j}$}
For <independent random variables> $X,Y$ with the same <binomial distribution> $\operatorname{Bin}(n,p)$, partition the tie event into the mutually exclusive events $X=Y=j$. Independence makes each probability the square of its binomial probability, proving the formula. At $p=1/2$, <Vandermonde's identity> gives $\Pr(X=Y)=4^{-n}\binom{2n}{n}$. Symmetry under exchange of the two variables then gives $\Pr(X>Y)=\Pr(Y>X)=(1-\Pr(X=Y))/2$.
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