How to lose as little as possible

Suppose Alice has a coin with heads probability $q$ and Bob has one with heads probability $p>q$. Now each of them will toss their coin $n$ times, and Alice will win iff she gets more heads than Bob does. Evidently the game favors Bob, but for the given $p,q$, what is the choice of $n$ that maximizes Alice's chances of winning? The problem of determining the optimal $N$ first appeared in \cite{wa}. We show that there is an essentially unique value $N(q,p)$ of $n$ that maximizes the probability $f(n)$ that the weak coin will win, and it satisfies $\frac{1}{2(p-q)}-\frac12\le N(q,p)\le \frac{\max{(1-p,q)}}{p-q}$. The analysis uses the multivariate form of Zeilberger's algorithm to find an indicator function $J_n(q,p)$ such that $J>0$ iff $n<N(q,p)$ followed by a close study of this function, which is a linear combination of two Legendre polynomials. An integration-based algorithm is given for computing $N(q,p)$.