= Solution
This is a <gambler's ruin> hitting problem. For $0<p<1$, let $h_j$ be the <probability> of reaching $+3$ before $-3$ when the current difference is $j$. Conditioning on the next <independent> game gives
$$
h_j=ph_{j+1}+qh_{j-1}\quad(-2\leq j\leq2),\qquad h_{-3}=0,\quad h_3=1.
$$
Stopping occurs almost surely: conditional on being inside the barriers, three consecutive wins by either player force absorption within the next three games. This event has <probability> $p^3+q^3>0$, so the chance of surviving $m$ such blocks is at most $(1-p^3-q^3)^m$, which tends to zero.
If $p\ne q$, the characteristic roots of the interior recurrence are $1$ and $r=q/p$, so $h_j=A+Br^j$. Imposing the two boundary values yields
$$
h_j=\frac{1-r^{j+3}}{1-r^6}.
$$
In particular,
$$
h_0=\frac{1-r^3}{1-r^6}=\frac1{1+r^3}
=\frac{p^3}{p^3+q^3}.
$$
For $p=q=1/2$ the recurrence instead has affine solutions, and the boundary values give $h_j=(j+3)/6$, hence $h_0=1/2$. The deterministic cases $p=0,1$ give zero and one respectively. The formula covers all of them continuously:
$$
\boxed{\mathbb P(A\text{ wins overall})=\frac{p^3}{p^3+(1-p)^3},\qquad0\leq p\leq1.}
$$
The stopping argument and the boundary recurrence show why the answer concerns the first barrier reached rather than the win difference after a predetermined number of games.
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