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I have a box with $A$ red balls and $B$ blue balls, from which I sample with replacement with uniform probability for each ball. This would imply that the probability of selecting a blue ball would simply be $\frac{B}{A+B}$. I keep sampling with replacement until I select a blue ball exactly $k$ times, and as such, the mean for the total number of times I need to sample from the box for this to occur, $M$, should be $Mean[M] = \frac{k*(A+B)}{B}$.

However, how does the variance for $M$, the total number of samplings required to observe a blue ball $k$ times, change as a function of $k$?

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What you are sampling is sum of $k$ independent variables with geometric distribution. So, for $p=\frac{B}{A+B}$, the variance of this sum will be $k\frac{1-p}{p^2}$.

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