---
tag: 0DB0R4N5XX
kind: statement
role: proposition
name: Mean of the geometric distribution
msc: ["60E05"]
uses: [0XCVV5Q500, 0DCT0QZVQ1, 0Y6QNF1J49]
sources:
  - url: https://en.wikipedia.org/wiki/Geometric_distribution
    locator: Proof of expected value
---

## Statement

Let $p \in (0, 1]$ and let $X$ have the [[geometric distribution|0XCVV5Q500]] 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|0Y6QNF1J49]]: $P(X \ge n) = (1 - p)^n$, 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.