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I am thinking on how to solve the following problem, from Rozanov's book on Probability:

Consider $n$ urns, each containing $w$ white balls and $b$ black balls. A ball is drawn at random from the first urn and put into the second urn, then a ball is drawn at random from the second urn and put into the third urn, and so on, until finally a ball is drawn from the last urn and examined. What is the probability of this ball being white?

To fix the ideas, take $n=2$. Then by conditioning on the first extraction, we have $$ P = \frac{w}{w+b}\cdot\frac{w+1}{(w+1)+b} + \frac{b}{w+b}\cdot\frac{w}{w+(b+1)}=\frac{w(w+b+1)}{(w+b)(w+b+1)}=\frac{w}{w+b}. $$ I guess that now it would be possible to keep going more or less in this way. If the probability of drawing a white ball from the second urn is $w/(w+b)$, then the probability of drawing a black ball will be $1-P=b/(w+b)$.

So, when I compute the probability of drawing a white ball from the third urn in the chain, it should give exactly the same computation as above. By induction, it should be possible to conclude the somewhat counterintuitive result that the required probability is simply: $$ P(\text{draw a white ball from the last urn of the chain})=\frac{w}{w+b}. $$ What do you think? Is there something wrong with this way of thinking about the problem?

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  • $\begingroup$ LGTM $\endgroup$ – sds Apr 13 '18 at 19:45
  • $\begingroup$ I would not say that the result is counerintuitive. $\endgroup$ – user Apr 13 '18 at 20:07
  • $\begingroup$ @user: at first sight, I would have expected something much more complicated, e.g. a polynomial of degree $n$ in $w$ $\endgroup$ – J. D. Apr 13 '18 at 20:10
  • $\begingroup$ I do not agree. It would be completely counterintuitive to have any answer deviating from $w/(w+b) $. But you can try another problem with every urn containing white and black balls in different proportions. $\endgroup$ – user Apr 13 '18 at 20:19
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Because the proportion of white balls is the same in each urn, moving a random ball from one to another does not change the expected ratio. No matter how you move the balls you will still have probability of $\frac w{w+b}$ for the ball you look at.

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