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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)?

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    $\begingroup$ @bof Yes, you are right. $\endgroup$ – almagest Jan 20 '20 at 10:14
  • $\begingroup$ Maybe you can condition on whether the ball drawn at the $n^{th}$ turh is one of the original $w+b$ balls (so the conditional probability is $w/(w+b)$), or the ball added at the $1^{st}$ turn, or the ball added at the $2^{nd}$, turn, etc. And if the ball drawn on the $n^{th}$ turn is the one added at the $i^{th}$ turn, then it has the same color as the ball drawn at the $i^{th}$ turn, so the conditional probability of a white ball is the same as the unconditional probability of drawing a white ball at the $i^{th}$ turn. Does that make sense? $\endgroup$ – bof Jan 20 '20 at 10:22
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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:

$$P\left(X_{m+1}=1\right)=\mathbb{E}X_{m+1}=\mathbb{E}\left[\mathbb{E}\left[X_{m+1}\mid\sum_{n=1}^{m}X_{n}\right]\right]=$$$$\mathbb{E}\frac{w+\sum_{n=1}^{m}X_{n}}{w+b+m}=\frac{w+m\left(\frac{w}{w+b}\right)}{w+b+m}=\frac{w}{w+b}$$

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