Tie probability for independent binomial counts ($\mathrm{Pr}(X=Y)={\sum }_{j=0}^n{\left(\genfrac{}{}{0ex}{}{n}{j}\right)}^2{p}^{2j}(1-p{)}^{2n-2j}$)

For independent random variables $X,Y$ with the same binomial distribution $\mathrm{Bin}(n,p)$, 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 $\mathrm{Pr}(X=Y)=4^{-n}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$. Symmetry under exchange of the two variables then gives $\mathrm{Pr}(X>Y)=\mathrm{Pr}(Y>X)=(1-\mathrm{Pr}(X=Y))/2$.