A bag contains black and white marbles $(X>0, Y>0)$. We are asked the probability of successfully drawing a black marbles and b white marbles $(a>0,b>0)$ in exactly $a+b$ draws. For what values $\{X,Y,a,b\}$ does drawing with or without replacement not change the probability of success? In other words, in what specific situations does replacement not matter? What is the smallest number of marbles in the bag to have a solution?
I've generalized the equations for the chances of success:
$$\text{w/o replacement} = \cfrac{\binom{X}{a}\binom{Y}{b}}{\binom{X+Y}{a+b}}\\ \text{w/ replacement} = \cfrac{\binom{a+b}{a}X^aY^b}{(X+Y)^{a+b}}$$
I've tested these equations with small values $\{1,1,1,1\}, \{2,2,1,1\},\{2,2,2,1\}$, etc. that are easily checked by hand, and they appear to hold true. What I have found is that drawing without replacement does not always give an advantage over drawing with, so I assume by IVT, as $n$ approaches infinity there will be some integer set where the probabilities intersect. The problem is I don't know how to proceed from here in finding a Magic Set other than with trial and error. A discovered answer would be easy to check, but I wouldn't know how to potentially find one other than brute force.