Mean of the geometric distribution

Statement

Let $p \in (0, 1]$ and let $X$ have the geometric distribution with parameter $p$, so $P(X = k) = (1 - p)^k p$ for $k = 0, 1, 2, \dots$. Then $$E[X] = \frac{1 - p}{p}.$$ For the number $T = X + 1$ of trials up to and including the first success, $E[T] = 1/p$.

Remarks

A proof uses the tail sum formula: $P(X \ge n) = (1 - p)^n$, so $E[X] = \sum_{n \ge 1} (1 - p)^n = (1 - p)/p$. The variance of $X$ and of $T$ is $(1 - p)/p^2$.

Explanations

L1 human

If each trial succeeds with probability $p$, you need on average $1/p$ trials to get the first success. For a fair die, you wait on average $6$ rolls for the first six.