Mean and Variance of the Negative Binomial Distribution

Let $X \sim \text{NB}(r,p)$ count the number of independent Bernoulli$(p)$ trials needed to observe $r$ successes. We can decompose $X$ as a sum of waiting times between successes:

$$X = Y_1 + Y_2 + \cdots + Y_r,$$

where $Y_i$ is the number of trials starting just after the $(i-1)$-th success and ending with the $i$-th success. Each $Y_i \sim \text{Geom}(p)$ and the $Y_i$ are independent.

Recall that $E[Y_i] = \dfrac{1}{p}$. By linearity of expectation,

$$E[X] = E[Y_1] + E[Y_2] + \cdots + E[Y_r] = r \cdot \dfrac{1}{p} = \dfrac{r}{p}.$$

For instance, if $r = 3$ and $p = \dfrac{1}{4}$, then

$$E[X] = \dfrac{3}{1/4} = 12.$$

On average, it takes $12$ trials to collect $3$ successes when each trial succeeds with probability $\dfrac{1}{4}$.