Just for the record, here is a brief sketch of how to derive the general formula for $w_m$ (the probability of A winning, given that A started with $m$ points and B started with $n$ points).

The polynomials $\Delta_1 = \Delta_2 = 1$, $\Delta_3 = 1-pq$, $\Delta_4 = 1-2pq$, $\Delta_5 = 1-3pq + p^2q^2$, $\Delta_6 = 1-4pq + 3p^2q^2$, $\ldots$, satisfy the recurrence relation $\Delta_k = \Delta_{k-1} -pq\Delta_{k-2}\;(k\geqslant3)$, as you can check from the equation (*) in the previous comment. The auxiliary equation for this recurrence relation is $\lambda^2 - \lambda + pq = 0$, with solutions $\lambda = \frac12\bigl(1 \pm \sqrt{1-4pq}\bigr)$. This leads to the formula $$\Delta_k = \frac{\bigl( \frac12\bigl(1 + \sqrt{1-4pq}\bigr)\bigr)^k - \bigl( \frac12\bigl(1 - \sqrt{1-4pq}\bigr)\bigr)^k }{\sqrt{1-4pq}}.$$ Then $\boxed{w_m = \dfrac{p^n\Delta_m}{\Delta_{m+n}}}$. That looks very neat, but of course it disguises the complications in the definition of $\Delta_k$.