Consider the Polya urn model with $a$ black balls and $b$ red balls initially in an urn where $a$ and $b$ are arbitrary positive integers. At each turn you pick a ball at random and return it to the urn along with a ball with the same color. Let $X_n$ denote the number of black balls in the urn after the $n$th turn is completed.
I need to compute $P_b = $ {probability that the $n$th pick is black} and prove that $P_b = \frac{a}{a+b}$ for every n
I was able to prove that $$ E \left( \frac{X_n}{n+a+b} \mid X_{n-1} \right)= \frac{X_{n-1}}{n+a+b-1}. $$
And also computed $ E\left( \frac{X_n}{n+a+b} \right)= \frac{a}{a+b}$ But I'm still stuck on proving that $P_b = \frac{a}{a+b}$ for every n
edit : Is it ok to write $P_b = \frac{X_{n-1}}{n+a+b-1}$ which is the number of black balls after $n-1$ turns divided by the total number of balls after $n-1$ turns and now take the expected value from both sides and have $$E(P_b) = E( \frac{X_{n-1}}{n+a+b-1}) = E(X_{n-1}) \frac{1}{n+a+b-1} = \frac{ a(n+a+b-1) }{a+b} \frac{1} {(n+a+b-1) } = \frac{a}{a+b} \\$$ and since $P_b $ is a constant $E(P_b)= P_b$ and I got what I wanted ?
The expectation that you have computed is the expected number of black balls at time $n$, divided by the total number of balls at time $n$.
How would you define the probability of drawing a black ball?