Geometric Distribution¶

Definition¶

Let $0<p\le1$ and $q=1-p$ $p$ . A discrete random variable $X$ has a geometric distribution with parameter $p$ , written $X\sim\operatorname{Geo}(p)$ $Geo (p)$ , if $X$ counts the failures before the first success in independent trials with success probability $p$ :

\mathbb{P}(X=k)=q^kp,\qquad k=0,1,2,\ldots.

At $p=1$ the first trial is a success, so $X=0$ with probability $1$ .

Interpretation¶

The geometric distribution counts how many failures happen before the first success in independent trials with success probability $p$ . A value of zero means that the first trial succeeds.

Probability bars for the number of failures before the first success with success probability 0.3, starting at 0.3 at 0 and shrinking by a factor 0.7 at each step.

In the figure $p=0.3$ $0.3$ : the bar at $0$ is $\mathbb{P}(X=0)=p=0.3$ $0.3$ , and each next bar is $0.7$ times the previous one, so the bars decrease like the terms of a geometric series, which gives the distribution its name. The most likely value is $0$ , while the mean $q/p=0.7/0.3\approx2.33$ given in the figure lies further to the right because of the long tail.

Each bar is $q$ times the one before, starting at $\mathbb{P}(X=0)=p$ $p$ . A smaller $p$ means a longer wait: $\mathbb{E}X=q/p$ $q / p$ .

Each trial is a Bernoulli variable, and $X=k$ means $k$ failures followed by a success, with probability $q^kp$ by independence. The masses sum to $1$ because $\displaystyle p\sum_{k\ge0}q^k=p/(1-q)=1$ $1$ . The tail formula in the Key Results, $\mathbb{P}(X\ge m)=q^m$ $q^{m}$ , has an equally direct reading: at least $m$ failures occur exactly when the first $m$ trials all fail. It also shows that waiting does not help: given $m$ failures so far, the conditional probability of at least $n$ more is $q^{m+n}/q^m=q^n$ $q^{n}$ , the memoryless property.

Books and software use the name "geometric" for two versions. This page counts failures, with values $0,1,2,\ldots$ ; the first success distribution counts trials including the success, $T=X+1$ $1$ , with values $1,2,\ldots$ . Before using a formula, check which values it starts from.

Key Results¶

$\mathbb{E}X=(1-p)/p$ and $\operatorname{Var}(X)=(1-p)/p^2$ $p) / p^{2}$ .
For integers $m\ge0$ $0$ , $\mathbb{P}(X\ge m)=(1-p)^m$ $p)^{m}$ .

Examples¶

For $p=1/2$ $1/2$ , $\mathbb{P}(X=2)=1/8$ $1/8$ .
For $p=1/4$ $1/4$ , the expected failure count is $(3/4)/(1/4)=3$ $3$ .

Prerequisites¶

Sources¶

Probability Theory for EOR · Week 3
Probability Theory for EOR · Week 4
Nicky van Foreest · Probability Distributions · Refresher

Aliases: Geom(p), geometric random variable, zero-based geometric distribution, Geo(p) Keywords: failures before first success Symbols: sum

Last updated: 20 October 2026 03:02