Finding probability that the $n$-th ball drawn was red in terms of the expected number of red balls in $n-1$ draws

Question

Assume that an urn contains $b$ blue ball and $r$ red balls. We pick balls from the urn one at a time, and inspect their color. After we inspect the ball, we put it back, and then we add $k$ balls of the same color as the ball picked to the urn.

For every $n$, let $p_n$ be the probability that the $n$-th drawn ball was red. How can we find a relation between $p_n$ the $\mathbb{E}[R_{n-1}]$, where $R_{n-1}$ is the number of red balls in the urn after $n-1$ draws?

I have no idea where to start. I tried finding a concrete expression for $p_n$ by conditioning on the $(n-1)$-th ball drawn, but it didn't help. I also tried to find an expression for $\mathbb{P}(R_{n-1} = s)$, but I couldn't.

Answer 1

Hint: $$P(\text{$n$th ball is red}) = E[P(\text{$n$th ball is red} \mid R_{n-1})]$$

$$=E\left[\frac{R_{n-1}}{b+r+(n-1)k}\right]$$