 First return to equal counts in sampling without replacement

Draw balls sequentially without replacement from $g$ green and $r$ red balls. Let $T$ be the first positive draw count at which the numbers of drawn green and red balls agree. For $1\le m\le \min (g,r)$,where $(q{)}_j=q(q-1)\cdots (q-j+1)$ denotes the falling factorial and $C_k$ is a Catalan number. Odd first return times are impossible because equality requires an even total number of draws.

To count the successful colour orders, encode green by a step $+1$ and red by a step $-1$. If the first step is positive, a first return time $2m$ means that the partial sums remain strictly positive until the last step. Delete the first positive step and the last negative step, and shift the remaining heights down by one. The result is a Dyck path of length $2m-2$, of which there are $C_{m-1}$. Interchanging the colours doubles this count. Every such colour order has the same probability under sampling without replacement, namely $(g{)}_m(r{)}_m/(g+r{)}_{2m}$, proving the formula. In particular, the numbers of successful colour orders at lengths two, four and six are two, two and four; balanced orders with an earlier return must not be counted.