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An urn contains $1$ black and $2$ white balls. One ball is drawn at random and its color is noted. The ball is replaced in the urn, together with an additional ball of its color. There are now four balls int he urn. Again, one ball is drawn at random from the urn, then replaced along with an additional ball of its color. The process continues in this way. Let $B_n$ be the number of black balls in the urn just before the $n$th ball is drawn. I'm supposed to find some conditional expectations here, but I'm not sure how this would work.

Firstly, this is a homework question and I'm just looking for tips on maybe how to look at this in a different way than it's stated or just some different tid-bits of info on how to get started on it.

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Not sure if this helps you, but I think you can can write the distribution for $B_n$ in terms of the distribution for $B_{n-1}$. –  Itai Zukerman Dec 12 '12 at 6:06
    
If this is homework and you hand back the accepted solution below... well, you will see. –  Did Dec 13 '12 at 6:57
    
So, Kyle, did you hand in the answer below as a solution to your homework, in the end? And if you did, what happened? –  Did May 4 '13 at 13:23
    
@Did Considering it was only a tip and that the solution below wasn't the soluton I would be looking for, I don't think I would have put the answer below as a solution for my homework. –  TheHopefulActuary May 4 '13 at 19:21
    
Good for you--since there is a serious problem with this answer, as explained in the comments a loooong time ago. –  Did May 4 '13 at 20:46

1 Answer 1

up vote 1 down vote accepted

If you have $b$ black balls and $w$ white balls then the proportion of black balls is $\dfrac{b}{b+w}$.

You then add a black ball with probability $\frac{b}{b+w}$ and a white ball with probability $\frac{w}{b+w}$. So the expected proportion of black balls becomes $$\frac{b}{b+w} \times \frac{b+1}{b+w+1} + \frac{w}{b+w} \times \frac{b}{b+w+1}.$$ Multiply this out and simplify and you will find this is still $\dfrac{b}{b+w}$. So the expectation stays the same as you move forward through the process and so by induction remains the same in future.

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That the expectation stays the same as you move forward through the process is true but a proof requires more work since after one step (and afterwards) the number of black balls is random. –  Did Dec 12 '12 at 7:14
    
@did: "by induction" may be a little brief, but for the following step the expectations become $\frac{b+1}{b+w+1}$ and $\frac{b}{b+w+1}$ weighted by $\frac{b}{b+w}$ and $\frac{w}{b+w}$, so the same. And so on . . . –  Henry Dec 12 '12 at 7:23
    
Not unless your b and w are in fact $B_n$ and $n+2-B_n$... Note that the support of the distribution of $B_n$ is $\{1,2,\ldots,n\}$ hence your argument, in its present form, breaks down as soon as $n\geqslant2$.. –  Did Dec 12 '12 at 7:28

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