# Extension of polya's urn

An urn contains r > 0 red balls and b > 0 black balls. A ball is drawn at random from the urn, its color noted and returned to the urn. Further, d > 0 additional balls of the same color are added to the urn. This process of drawing a ball and adding d balls of the same color is continued. Define Xi = 1 if at the i-th draw the color of the ball drawn is red, and 0 otherwise. Compute E($$\sum_{i=1}^{\infty}{Xi}$$).

My attempt so far is~

Define the events $$\textit{R_n:="n-th ball drawn is red"}$$ and $$\textit{B_n:="n-th ball drawn is black"}$$. Then \begin{align*} \mathsf{P}(B_2)&=\mathsf{P}(B_2 \mid B_1) \mathsf{P}(B_1)+\mathsf{P}(B_2 \mid R_1) \mathsf{P}(R_1) \\ &= \frac{b+d}{r+b+d}\frac{b}{r+b}+\frac{b}{r+b+d}\frac{r}{r+b}=\frac{b}{r+b} \end{align*} and in general $$\mathsf{P}(B_n)=\frac{b}{r+b}$$.

Similarly for red ball,

$$\mathsf{P}(R_n)=\frac{r}{r+b}$$.

How do I find the expectation?

Intuitively it is essentially just counting the number of times you draw red, given you have an infinite number of draws. And the only way this kind of infinite sum converges is if it is a geometric sequence inside some radius of convergence. But you have shown the probability is constant for all $$n$$, so no geometric sequence exists.
The only other solution is trivially when $$r = 0$$. I could be missing something though because I do imagine if $$d$$ is incredibly large, intuitively it might converge. But your probability $$B_{n}$$ shows otherwise.