Suppose that independent trials, each having probability $p$, $0 < p < 1$, of being a success are performed until a total of r successes is accumulated. If we let X equal the number of trials required, then
Above equation follows because, in order for the rth success to occur at the nth trial, there must be r − 1 successes in the first n − 1 trials and the nth trial must be a success.
where Y is a negative binomial random variable with parameters r + 1,
$\large E[X] = \frac{r}{p}$
$r = 4$
$p =$$\large \frac{1}{6}$
$\large E[X] = \frac{4}{\frac{1}{6}} = 24$
$Var[X] =$$\Large \frac{4(\frac{5}{6})}{(\frac{1}{6})^2} $$\large= 120$
Note: Above content is referred from Sheldon Ross.