Red probability on draw 11 when black turns red

An urn starts with $r>0$ red and $b>0$ black balls. Balls are drawn one at a time: a red ball is put back, while a black ball is removed and a red ball is put in its place. Let $f(r,b,n)$ be the probability that the $n$th draw is red. Compute $f(8,12,11)$ to two decimal places.

About $0.64$. The urn always has $N=r+b$ balls, and each draw turns a black into red with probability (blacks)/$N$, so the expected black count shrinks by the factor $1-\tfrac1N$ per draw. Hence $f=1-\tfrac{b}{N}\left(1-\tfrac1N\right)^{n-1}=1-0.6\cdot0.95^{10}\approx0.64$.