Expected number of trials before I get two out of three types

Question

I am a biologist trying to quantify the likelihood of observed genetic phenomena… I’ve tried my best to formalize my question in accepted terms as follows:

Given a stack of n cards, each colored red (x cards) , blue (y cards), or green (z cards) where x + y + z = n, what is the expected number of draws, without replacement, before you draw both a red and a blue card? It doesn’t matter whether the red or blue card comes first. Preferably, a solution would be defined in general terms, but for demonstration’s sake, one could assume a stack of 10 cards, where x = 3, y = 5, z = 2.

Intuitively, I figured this could be solved using multivariate hypergeometric distributions. However, because we are looking for the expected minimal number of draws before two successes occur (both a red and a blue card), this approach appeared to require some tricky combinatorics.

I have also looked, without success, into adopting solutions from other problems. My sense is that between the following two posts there is a solution, but one I have not been able to intuit.

Expected number of card draws to get all 4 suits

Expected number of trials before I get one of each type

Any direction would be greatly appreciated -- thanks so much!

Answer 1

This may help. Clearly the maximum number of draws is $z+x+1$ assuming with no loss of generality that $y\leq x.$ Then apply the inclusion exclusion approach in the first answer you linked.

Answer 2

If you just want at least one each of red and blue (and not something more complicated like at least 2 reds and at least 3 blues), then the combinatorics aren't too messy. Letting $T$ be the number of trials (draws) needed, the pmf of $T$ is$$\mathbb{P}\left(T=t\right)=\frac{{\displaystyle \binom{n-x}{t-1}\cdot x+\binom{n-y}{t-1}\cdot y-\binom{n-x-y}{t-1}\cdot\left(x+y\right)}}{{\displaystyle \binom{n}{t}}\cdot t}$$for $t=2,\ldots,n-\min\left\{x,y\right\}+1$. The expected value is$$\mathbb{E}\left(T\right)=\left(n+1\right)\left[\frac{1}{x+1}+\frac{1}{y+1}-\frac{1}{x+y+1}\right]\mbox{.}$$With $x=3,y=5,n=10$, this gives $\mathbb{E}\left(T\right)=\frac{121}{36}=3.36\overline{1}$.