Geometric Distribution Calculator (with Graph)
Find the probability that the first success comes on a given trial.
What the geometric distribution models
The geometric distribution describes how long you wait for the first success in a series of independent trials with the same probability of success p. For example, how many rolls of a die until the first six, how many calls until a sale, or how many items are inspected before finding a defect. There are two common conventions, and the calculator supports both: counting the trials until the first success (1, 2, 3, …) or counting the failures before the first success (0, 1, 2, …).
The formula
- Trials: P(X = k) = (1 − p)ᵏ⁻¹ × p, with mean 1 ÷ p.
- Failures: P(X = k) = (1 − p)ᵏ × p, with mean (1 − p) ÷ p.
- Both have variance (1 − p) ÷ p².
- The chance of waiting more than k trials is (1 − p)ᵏ.
Worked examples (p = 0.2)
| Question | Answer |
|---|---|
| First success on exactly the 5th trial | 0.0819 |
| First success within 5 trials | 0.6723 |
| Need more than 5 trials | 0.3277 |
| Expected number of trials | 5 |
The graph under the result highlights the event you chose, and has Download PNG and Download SVG buttons for reports and slides. A table lists P(X = k), P(X ≤ k) and P(X ≥ k) for the values around the mean.
The memoryless property
The geometric distribution has no memory: after 10 failures, the chance of success on the next trial is still p, and the expected further wait is unchanged. This is why a streak of failures does not make a success “due”. For several successes, the negative binomial distribution is the generalisation.
How to cite this page
CalcPulse. (2026). Geometric Distribution Calculator. https://calcpulse.com/geometric-distribution-calculator/
Frequently asked questions
What is a geometric distribution?
The distribution of the number of trials needed to get the first success, when each trial has the same probability of success.
What is the mean of a geometric distribution?
1 ÷ p when counting trials, and (1 − p) ÷ p when counting failures before the first success.
What is the difference between the two definitions?
One counts the trial on which the success happens (starting at 1). The other counts only the failures before it (starting at 0). The probabilities are the same, shifted by one.
What does memoryless mean?
Past failures do not change the chances of the next trial. The distribution of the remaining wait is the same at every point.
How do I find the probability of waiting more than k trials?
It is (1 − p)ᵏ, the chance that the first k trials all fail.
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