# Formula for urn model with replacement and “color switching”

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

• 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? – joriki Jun 18 '16 at 1:06

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 $n H_n = O(n \log n)$.
• 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 $$N\left(\dfrac{N-1}{N}\right)^n.$$
• 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. – andreas_ Feb 23 '15 at 17:30
• 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?) – Henry Feb 23 '15 at 21:23