# Expected number of attempts to get 2 out of 50 results at least once?

I looked at some material, such as markov chains and inclusion-exclusion, but was unable to wrap my head around them with my only cursory understanding of probability.

I'm trying to find the solution to a problem where there are 50 possible unique results of equal probability; you could imagine a deck of cards with the one-eyed jacks removed. How many times do I have to draw, with replacement, before I draw both two-eyed jacks (which are also replaced) at least once?

My initial intuition is that we can start with an expected 100 attempts, because it is $$1/p$$ or 50 expected attempts for the first jack and then you have the same deck you started with for +50 expected attempts for the other jack, but I suspect that I need to subtract some amount for the chance that the second jack was already drawn before the first.

I also tried to do what this answer suggests and treated the problem like a 3-sided die where... $$p(1) = 48/50\qquad p(2) = p(3) = 1/50$$ ...treating the 1-result as every other card in the deck, and the 2-result and 3-result as the desired draws. This, I guess, gives me an expected outcome of... $$4H_4-\frac{48}{50}\cdot4$$

...but I don't know how to account for the 0.04% chance that I get the 2-result and 3-result right away. That alone wouldn't be a problem as I doubt it changes the expected number of draws enough to matter, but I also don't know what $$H_4$$ means to do the calculation.

Any help you can offer to complete either of my two approaches (or to enlighten me on why these approaches don't work for my question) would be greatly appreciated.

I looked at these related questions, but the methods don't seem to work for this problem since they want each result at least once, and I only want two results (or perhaps I just don't understand the answers well enough to apply them to my question):

• For anyone wondering, the application of this question is related to D&D and trying to find the expected number of Wild Magic Surges before two desired outcomes are reached. Related to this answer on RPG StackExchange. Commented Apr 23, 2020 at 14:22

This is related to my answer here. For this question, at the start, the probability of drawing the first (as yet undetermined) two-eyed jack is $$\frac2{50}$$, so $$\frac{50}2$$ draws are expected to get it; once that has been drawn, there is a $$\frac1{50}$$ chance of drawing the other two-eyed jack, so $$\frac{50}1$$ draws are expected to get it. The expected total number of draws is then their sum: $$\frac{50}2+\frac{50}1=75$$.
In general, if you are waiting for $$k$$ specific results from repeating a draw on $$n$$ objects with replacement, the expected number of draws needed is $$n\sum_{i=1}^k\frac1i$$.