Suppose a red ball is selected with probability $p$, and a blue ball is selected with probability $q = 1 - p$ in each independent random draw. And one will stop sampling once he (or she) reaches his (or her) goal of $r$ red balls and $b$ blue balls, or once he (or she) has collected $m$ balls in total. Let $N_{r b m}$ be the number of samples drawn.
Now I want to know the expectation of $N_{r b m}$, i.e. $\mathrm{E} N_{r b m}$. Maybe a recurrence relation is enough.
Could anyone help me? Thanks in advance!
I suspect you can find an expression based on the binomial probability of reaching a position immediately before the last draw and then making a suitable final draw
This might produce something like this (unchecked): $$m-\sum_{k=r}^{m-d-1} m{m-1\choose k}p^kq^{m-k-1} \\+\sum_{j=r}^{m-d-1} (j+d){j+d-1\choose j}p^jq^{d} + \sum_{i=d}^{m-r-1} (i+r){i+r-1\choose i}p^rq^{i} $$
so if $m\le r+d$ then all the summations disappear and you just get $m$