Chaining the multiplication rule, drawing draw cards one at a time without replacement and having every one come from the special group gives:

$$\frac{g}{N} \times \frac{g-1}{N-1} \times \cdots$$

Return that probability. It must match the combination approach from "Drawing a Hand From One Group" exactly.

solve(52, 4, 2)  ->  0.004524886877828055     # 4/52 x 3/51