it turns out -- more computation than is necessary. As random variables, we could say that $A\sim\operatorname{Binom}\left(5,~\tfrac12\right)$ and $B\sim\operatorname{Binom}\left(4,~\tfrac12\right)$ are independent (but not identically distributed) binomial with $n=5,4$ respectivly and both with $p=\tfrac12$. Since they are independent, we can calculate the probability that $A>B$ thus: $$ \mathbb{P}\left(A > B\right) =\sum_{a=1}^5 {5\choose