# Generalizing Combinations with added draws for success

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?

• I have fixed the error in the problem statement to show that $SD$ additional draws were added – Hoog Jan 9 at 15:45
• I haven't been able to come up with anything on this. I tried using the reflection principle (see e.g. the accepted answer to this question), but I don't think it works because the unequal step sizes break the symmetry. I also tried using generating functions to divide out multiple returns to $0$, but $\sum_n\binom{Sn}nx^n$ doesn't seem to have a closed form for $S\gt3$. (The result for $S=3$ is fascinating, though.) – joriki Jan 9 at 19:32
• Another approach would be to use inclusion–exclusion to systematically account for the over/underestimates that you describe, but that, too, doesn't really seem tractable, because if you have multiple zeros the number of paths depends in detail on their placement. – joriki Jan 9 at 19:34