Urn where black balls get swapped for red

An urn starts with $r$ red and $b$ black balls ($r,b>0$). Repeatedly, a ball is drawn at random: a red ball is put back, while a black ball is removed and a red ball is put in its place, so the total stays $r+b$. Let $f(r,b,n)$ be the probability that draw number $n$ is red. Compute $f(8,12,11)$ to two decimal places.

About $0.64$. Each draw removes a black ball with probability (blacks)/(total), so the expected number of black balls shrinks by the factor $1-\tfrac{1}{r+b}$ per draw. Hence $f=1-\tfrac{b}{r+b}\left(1-\tfrac1{r+b}\right)^{n-1}=1-0.6\,(0.95)^{10}\approx0.641$.