Given $N$ initial draws, a constant probability of success $P$ (with replacement), and the fact that each success adds $D$ additional draws, find the probability that our draws will be exhausted with $S$ total successes (ie. $SD$ additional draws were added)
My thoughts are to start with the binomial distribution to get the probability of $S$ successful draws in $N+S*D$ attempts. This will overestimate the probability as it will count cases where $1$ of the final $D$ draws was a success, or $2$ of the final $2D$ draws were successes, or....
I can subtract the combinations for each of these overestimates, but they have intersections (having 1 draw in the final $D$ and another in the penultimate $D$ fits both the examples above), so I will go from an overestimate to an underestimate, and I will need to add those combinations back in, but if $S$ is $3$ or higher, I think that adding those back in gets us back to an overestimate, causing us to oscillate around the answer I am looking for.
If I continue this way $S$ times I should land on the correct answer, but I am hoping for a formula or method of doing so that would not scale up so much in complexity with increased $S$. Is there a better way of going about this?