Geometric Distribution Calculator

Background.

Roll a die until a six comes up. Call strangers until one agrees to a survey. Take free throws until one goes in. Each is a run of identical, independent attempts that stops at the first success, and the geometric distribution describes how long such a run lasts.

The reasoning is short. For the first success to arrive on attempt k, the k − 1 attempts before it must all fail, each with probability 1 − p, and then attempt k must succeed, with probability p. Multiply those together and you have the chance of that exact outcome. The chance that success has arrived by attempt k is easier still to see from the other side: it is one minus the chance that all k attempts failed.

The shape that results surprises people. The single most likely trial for the first success is always the very first one, however small p is, and the probabilities then shrink by the same factor 1 − p with every further trial. Yet the average wait is 1 ÷ p trials, and the tail is long: with a one-in-four chance per attempt, about 4% of runs still have no success after eleven tries.

The distribution is also memoryless. A run of bad luck changes nothing about the next attempt: after ten failures the chance of succeeding on the eleventh is still p, and the expected number of further attempts is still 1 ÷ p. That is exactly what the gambler's fallacy denies, and why the model only fits when the attempts are genuinely independent and p does not drift — a salesperson who improves with practice, or a machine wearing out, breaks it.

What is geometric distribution calculator?

The geometric distribution gives the probability that the first success in a sequence of independent trials, each with the same success probability p, occurs on trial k. This page uses the first-success convention, where k counts the successful trial itself and starts at 1. Some textbooks instead count only the failures before the first success, starting at 0; that version shifts every k down by one.

How to use this calculator.

1. 01Enter the success probability p for a single attempt as a decimal greater than 0 and at most 1 — 0.25 for one chance in four, about 0.1667 for rolling a six.
2. 02Enter the trial number k as a whole number from 1 upward, counting the successful attempt itself: k = 3 means two failures followed by a success.
3. 03Read P(X = k) for the chance that the first success falls on exactly that trial, and P(X ≤ k) for the chance it has happened by then.
4. 04To find how many attempts give you a target level of confidence, raise k until P(X ≤ k) passes the level you need.

The formula.

P(X = k) = (1 − p)^(k − 1) × p and P(X ≤ k) = 1 − (1 − p)^k. In the first, (1 − p)^(k − 1) is the probability that the first k − 1 independent trials all fail, and the final p is the success on trial k. In the second, (1 − p)^k is the probability that all of the first k trials fail, so subtracting it from 1 leaves the probability that at least one succeeded. Each extra trial multiplies the exact probability by 1 − p, so the values fall geometrically — hence the name. The mean number of trials is 1 ÷ p and the variance (1 − p) ÷ p², neither of which is shown here. The page refuses p of 0 or above 1, and any k that is not a positive whole number.

A worked example.

Frequently asked questions.

References& sources.

1. [1]NIST/SEMATECH e-Handbook, Geometric Distribution.
2. [2]OpenStax, Introductory Statistics 2e, section 4.4, Geometric Distribution.
3. [3]BIPM, The International System of Units (SI Brochure), 9th ed., version 3.01, coherent derived units and quantity equations.