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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?

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Maybe I'm misunderstanding the question but I fail to see why that expectation doesn't just blow up to infinity.

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.

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  • $\begingroup$ Did you notice this? Define Xi=1 if at the ith draw, ball drawn is red and zero otherwise. $\endgroup$ Commented Jul 3, 2020 at 6:32
  • $\begingroup$ That means if by chance we draw a black ball at ith draw the expectation will be zero. How could I use this point to find the answer? $\endgroup$ Commented Jul 3, 2020 at 6:34
  • $\begingroup$ However what you addresed about expectation blowing up to infinity, i think it makes sense too. $\endgroup$ Commented Jul 3, 2020 at 6:38
  • $\begingroup$ You defined your expectation as the sum of the draws, not just the ith draw. If you mean the ith draw, then it is simply your probability since your value is 1. $\endgroup$
    – Dale C
    Commented Jul 4, 2020 at 6:50
  • $\begingroup$ So now, how do I solve this? $\endgroup$ Commented Jul 5, 2020 at 10:13

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