= Solution
The equality must be a <first return time>: an order starting $GR$ or $RG$ would already have stopped at draw two. The only successful four-draw orders are therefore $GGRR$ and $RRGG$. Their <probabilities> under <sampling without replacement> add to
$$
\boxed{\frac gN\frac{g-1}{N-1}\frac r{N-2}\frac{r-1}{N-3}
+\frac rN\frac{r-1}{N-1}\frac g{N-2}\frac{g-1}{N-3}}
=\boxed{\frac{2g(g-1)r(r-1)}{N(N-1)(N-2)(N-3)}}.
$$
Here $N=g+r$. Counting all balanced four-draw orders would include games that have already stopped.
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