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There are b blue balls and r red balls in the urn. All balls currently in the urn have an equal probability of being selected each.
When drawing a red ball, put it back in the urn, but a blue ball doesn't. Ask the expected value of times to draw the red ball after n times drawn.
I'm really confused since if put both red and blue balls back in the urn, it will be binomial, if put neither of them back, it will be hypergeometric.
But only put back red ball but not blue is very hard to figure out.
I cannot see a simple closed form, but there will be a recurrence.
Suppose the expectation is $f(r,b,n)$. Then $$f(r,b,n)= \frac{r}{r+b}(1+f(r,b,n-1)) + \frac{b}{r+b}f(r,b-1,n-1)$$ starting with $f(r,b,0)=0$.
For example, if $r=10,b=5,n=3$ then you get $f(10,5,3)=\frac{35216}{17199} \approx 2.0475609$ as your expected number of red balls drawn, slightly more that the expectation of $2$ red balls you would get with the binomial or hypergeometric approach.
