Expected value of the longest run of red balls Suppose there's an urn containing $r$ red balls and $b$ blue balls. At each trial, I'm drawing a ball at random from the urn, without replacement. Let $R$ denote the event of drawing a red ball, and let $B$ denote the event of drawing a blue ball. I'm conducting $r+b$ trials of this experiment. So, there are a total of $r+b \choose r$ equally likely possibilities.
From this setting, it is clear that both $\Pr$[$R$ at $k^{th}$ trial] and $\Pr$[$B$ at $k^{th}$ trial] depend on the outcomes of the previous $k-1$ trials.
I'm interested in finding the distribution of the longest run of red balls, and thus its expected value. How do I approach this problem?

The problem can be solved by using a recursion. Let $p(r,b,L)$ be the probability that the longest run of red balls is larger than $L$. Then $p(r,b,L)=1-q(r,b,L)$, where $q(r,b,L)$  is the probability that there is no run of red balls of lenghth $L+1$. This probability can be obtained from the recursion
$q(r,b,L)=\frac{b}{r+b}q(r,b-1,L)+\sum_{k=1}^{L}\frac{r}{r+b}\times\cdots \times \frac{r-(k-1)}{r+b-(k-1)}\times\frac{b}{r+b-k}q(r-k,b-1,L)$.
The expected value of the longest run of red balls is $1+\sum_{L=1}^{r-1}p(r,b,L)$.
