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There are $n$ urns each having $a$ white and $b$ black balls. One ball is taken from urn 1 and is transferred to urn 2. Then one ball is taken from urn 2 and transferred to urn 3 and so on. Find the probability that the ball drawn from $n$th urn is white.

I get the intuition that the answer should be $\frac{a}{a+b}$, but I'm unable to prove it.

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    $\begingroup$ Please edit your first sentence. As is, it is unclear and ungrammatical. Also, you need to show your work. What have you tried? Where did you get stuck? $\endgroup$ Commented Jan 1, 2018 at 17:19
  • $\begingroup$ How could the probability be $a(a+b)$? $\endgroup$
    – lulu
    Commented Jan 1, 2018 at 17:21
  • $\begingroup$ @lulu sorry I ve edited the question $\endgroup$
    – user481779
    Commented Jan 1, 2018 at 17:23
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    $\begingroup$ Please state the problem completely. Presumably you don't mean to stop after you have put a ball into $\#3$. Do I stop once I get add one to $\#n$ or do we move a ball from $\#n$ to $\#1$? Have you solved the problem for $n=2$? $n=3$? $\endgroup$
    – lulu
    Commented Jan 1, 2018 at 17:24
  • $\begingroup$ You stop once add one to nth urn $\endgroup$
    – user481779
    Commented Jan 1, 2018 at 17:26

3 Answers 3

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When you go from urn 1 to urn 2, the white probability in urn 2 is now $\frac{a+\frac{a}{a+b}}{a+b+1} = \frac{(a+b)a+a}{(a+b+1)(a+b)}=\frac{a}{a+b}$ Therefore as you continue, the probabilities in each urn remain the same after each transfer, leading to your final answer.

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Wrap the balls in a paper and put the number of the original urn to the outside of the ball. When you draw the ball from the last urn, first look at the number on the ball, and then open the package and look at the color.

The color of the ball doesn't have any effect in what happens, so after we have looked at the number, no matter what the number was, the probability that the ball is white is $\frac{a}{a+b}$. Therefore, the probability is $\frac{a}{a+b}$ even if we don't look at the number.


Mathematically, let $X$ be the number of the urn where the last ball originally comes from, and let $Y$ be the color. We know that for all $x$, $\mathcal{P}(Y=\text{white}|X=x)=\frac{a}{a+b}$, so we know that $\mathcal{P}(Y=\text{white})=\frac{a}{a+b}$.

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In the average of many trials, you can consider that, when you draw a ball from the first urn and move it to the second one, it is part white and part black, in fractions $\frac{a}{a+b}$ and $\frac{b}{a+b}$ respectively. This doesn't make sense in terms of one trial, but as an average, it does.

Thus, adding such a fractional ball to urn #2 does not change the probability of what is drawn from that urn. This argument extends to any number of urns.

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    $\begingroup$ I don't find the symmetry argument persuasive, but maybe I am missing the point. If $a=b$ then sure. But if, say, $a<b$ then I don't see how to invoke symmetry. Am I missing something? $\endgroup$
    – lulu
    Commented Jan 1, 2018 at 17:45
  • $\begingroup$ @lulu , eh, you may be right. I wonder if I should delete that part.... Hmmm. Is there a reason to think that filtering the selection through multiple urns would amplify the probability of the more numerous color, or that it would dampen the advantage? $\endgroup$ Commented Jan 1, 2018 at 17:47
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    $\begingroup$ Well, it's easy enough to show that the conclusion is correct...the probability doesn't change. But I haven't seen a way to intuit that. Of course, the fact that the answer is so simple suggests that there is some guiding principle... $\endgroup$
    – lulu
    Commented Jan 1, 2018 at 17:49
  • $\begingroup$ I think the top part of my answer is closer to that principle than the second part. $\endgroup$ Commented Jan 1, 2018 at 17:50
  • $\begingroup$ @GTonyJacobs Say white balls are more than black balls so chances That you transfer a white ball are more and hence it seems that effectively filtering the selection through many urns would amplify the probability of white ball $\endgroup$
    – user481779
    Commented Jan 1, 2018 at 17:51

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