0
$\begingroup$

I tried to figure out following problem, but failed :/ My best guess is that it somehow is a mixture of a binomial and a hypergeometric distribution:

I have an urn with N white balls and 0 black balls. Every time I draw one white ball, I replace it with a black ball. A drawn black ball stays black (and is put back in the urn). I'm particularly interested in the mean of white balls in the urn after n draws.

Cheers, Andreas

$\endgroup$
  • $\begingroup$ Why did you accept an answer that doesn't answer the question, despite commenting under the answer that does answer the question that it does? $\endgroup$ – joriki Jun 18 '16 at 1:06
1
$\begingroup$

This is Coupon Collector's problem, which could also be viewed as sum of $n$ iid Geometric rvs. Mean time until the box has only white balls is exactly $n H_n = O(n \log n)$.

$\endgroup$
  • $\begingroup$ Perfect, thank you very much! $\endgroup$ – andreas_ Feb 23 '15 at 0:17
1
$\begingroup$
  • The probability that a particular ball is not drawn in one draw is $\dfrac{N-1}{N}.$

  • The probability that a particular ball is not drawn in $n$ draws (i.e. that a particular ball is white after $n$ draws) is $\left(\dfrac{N-1}{N}\right)^n.$

  • The expected number of white balls after $n$ draws is $$N\left(\dfrac{N-1}{N}\right)^n.$$

$\endgroup$
  • $\begingroup$ Thank you very much, that's what I was looking for. $\endgroup$ – andreas_ Feb 23 '15 at 12:10
  • $\begingroup$ One more question: So the probability to draw a white ball after n draws would be $\frac{N(\frac{N-1}{N})^n}{N} = \frac{N-1}{N}^n}$. But this would be the probability that a particular ball is not drawn in n draws again. $\endgroup$ – andreas_ Feb 23 '15 at 17:30
  • $\begingroup$ Yes: the probability that the $n+1$th draw is white is $\left(\frac{N-1}{N}\right)^n.$ You can either regard it as my final result divided by $N$, or more directly as my second result (you draw a particular ball: what is the probability that it has not already been drawn in $n$ draws?) $\endgroup$ – Henry Feb 23 '15 at 21:23

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.