Ans : $\frac{r}{r+b}$
Exp :
Assume that after $i_{th}$ iteration, we have $r$ red balls and $b$ black balls.
So, probability of choosing red ball in $(i+1)_{th}$ iteration will be :
$=\frac{r}{r+b}$
So, Probability of choosing red ball in $(i+2)_{th}$ iteration will be :
(note that, in $(i+1)_{th}$ iteration, we could either pick black or red ball. )
$\frac{r}{r+b} \times \frac{r+1}{r+b+1} + \frac{b}{r+b} \times \frac{r}{r+b+1} $
$=\frac{r}{r+b}$
So, in the beginning we have $r$ red balls, and $b$ green balls. So, in every iteration, the probability of choosing a red ball will be $\frac{r}{r+b}$ .
A beautiful question!! Whatever iteration we take, even in $100_{th}$ iteration, probability of picking a red ball will be same.
Refer following link. This GATE question can be found here as it is, along with many possible variations.
http://www.stat.yale.edu/~pollard/Courses/600.spring08/Handouts/Symmetry%5BPolyaUrn%5D.pdf