Consider an urn, which initially has $w$ white balls and $b$ black balls. Make a sequence of draws from the urn as follows. Draw one of the balls in the urn at random, then put this ball back into the urn and add another ball to the urn of the same color of the one just drawn. Continue to draw and add balls in this manner indefinitely. Let $X_{n} = 1$ if the $n^{th}$ ball drawn is white and $X_{n} = 0$ if it is black.
Prove that Before starting the process, all of the draws have the same probability of being white.
$P(X_{n} = 1)$ = $\frac{w}{w+b}$ for $n\geq1$
I was able to prove it with induction (proved it for two). How can I prove it with Conditional probability (was able to prove it for two, need help with generalizing it)?
After exactly $m$ draws and corresponding addings the number of white balls in the urn equals $w+\sum_{n=1}^{m}X_{n}$ and the total number of balls equals $w+b+m$.
Our induction hypothese is:$$P(X_n=1)=\frac{w}{w+b}\text{ for }n=1,\dots,m$$
Then:
$$P\left(X_{m+1}=1\right)=\mathbb{E}X_{m+1}=\mathbb{E}\left[\mathbb{E}\left[X_{m+1}\mid\sum_{n=1}^{m}X_{n}\right]\right]=$$$$\mathbb{E}\frac{w+\sum_{n=1}^{m}X_{n}}{w+b+m}=\frac{w+m\left(\frac{w}{w+b}\right)}{w+b+m}=\frac{w}{w+b}$$