Expected number of balls for an Urn problem with and without replacement

Question

Let's say an urn has two types of balls, $a$ and $b$. Balls $a$ are not put in the urn when pulled but balls $b$ get put back in the urn when drawn. What is the expected number of $a$ balls after $m$ draws?

Answer 1

This is an interesting problem, but I'm afraid that there will be no simple general solution.

Assume that initially there are $a$ balls of type A and $b$ balls of type B. Denote by $k$ the running number of A-balls, and let $p_m(k)$ $(0\leq k\leq a)$ be the probability that after $m$ drawings there are exactly $k$ A-balls left in the urn. Then $p_0(k)=\delta_{k\,a}$, and the $p_m(k)$ satisfy the recursion $$p_{m+1}(k)={b\over b+k}p_m(k)+{k+1\over b+k+1}p_m(k+1)\qquad(m\geq0)\ .$$ This recursion can be used to numerically compute the $p_m(k)$ for not all too large numerical data $a$, $b$, $m$.

Nevertheless there is one thing that we can compute explicitly, namely the expected number of draws for finding all A-balls. When there are $k$ A-balls left in the urn the probability of drawing such a ball in the next draw is ${k\over k+b}$. The expected number of draws necessary to obtain the next A-ball in this situation therefore is (exponential distribution!) ${k+b\over k}=1+{b\over k}$. It follows that the expected number of drawings necessary to find all A-balls is given by $$E_{\rm all}=\sum_{k=1}^a \left(1+{b\over k}\right)=a+ b H_a\approx a +b\,\log a\ ,$$ whereby $H_.$ denotes the harmonic numbers.