Geometric distribution

The geometric distribution counts the number of independent trials up to and including the first success.

P(X = k) = (1-p)^{k-1} p, \quad k = 1, 2, 3, \dots

pmf (the engine counts TRIALS, so the support starts at 1)

$p$: probability of success on each trial
$k$: the trial on which the first success happens

P(X > k) = (1-p)^k, \qquad \E(X) = \frac{1}{p}, \qquad \Var(X) = \frac{1-p}{p^2}

Survival function, mean and variance

$P(X > k)$: probability that the first $k$ trials are all failures

When to use

Waiting for the first success in repeated independent trials with constant $p$ : calls until the first sale, items inspected until the first defect.

Assumptions

Independent trials, same $p$ each time, no upper limit on the number of trials.

How to interpret

The most likely value is always 1, yet the mean is $1/p$ : the distribution is strongly right-skewed.
Memoryless: $P(X > s + t \given X > s) = P(X > t)$ . Past failures do not make a success "due".

Common mistakes

Mixing conventions: some books and software count failures before the first success (support 0, 1, 2, …, mean $(1-p)/p$ ). Statistics Quest counts trials (support 1, 2, …, mean $1/p$ ).
Believing a long run of failures raises the chance of success next time (gambler's fallacy).

Example

Each call converts with $p = 0.2$ . $P(X = 3) = 0.8^2 \times 0.2 = 0.128$ ; $P(X \le 3) = 0.488$ ; $P(X > 3) = 0.8^3 = 0.512$ . Expected number of calls until the first sale: $1/0.2 = 5$ .