Interactive module guide

Geometric Distribution Calculator & Guide

This geometric distribution calculator lets you calculate probabilities for the number of trials until the first success and explore the memoryless tail.

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The geometric distribution models the number of independent Bernoulli trials needed to obtain the first success, so its support starts at one and its mass declines by the constant factor 1−p with every additional trial.

It is the only discrete distribution with the memoryless property: a run of failures carries no information about the remaining wait.

This page uses the trials convention, P(X=k)=p(1−p)^(k−1) for k=1,2,3,…, with mean 1/p; some textbooks count only the failures before the first success, which shifts every value down by one, so always confirm the convention before comparing formulas.

Is this the right module for my question?

Use it when

Use a geometric model when a well-defined attempt repeats independently with the same success probability and the question is how long until the first success: calls until the first sale, inspections until the first defect, requests until the first accepted connection.
Use it when the memoryless structure is itself the modeling claim. If a process genuinely restarts fresh after every failure, then P(X>m+n | X>m)=P(X>n), and the geometric distribution is the only discrete model consistent with that statement.
Use the quantile lookup for planning questions such as how many attempts must be budgeted to reach a target assurance level; the smallest k with cumulative probability at least 0.95 is a direct planning number that the closed-form tail (1−p)^k makes easy to verify.

Choose another method when

Avoid it when the success probability changes across attempts—learning effects, fatigue, escalating retry backoff, or a depleting pool of prospects all violate the constant-p assumption, usually in a direction that makes the geometric tail misleading.
Avoid it when attempts are not independent. If a failed attempt makes the next one more likely to fail, as with a server that degrades under retries, the observed wait will be more dispersed than any geometric model predicts.
Avoid using it for the time until the r-th success or for total successes in a fixed number of trials; those questions belong to the negative binomial and binomial pages respectively, and mixing the three families is a common source of formula errors.

Interactive tool

The calculator loads as you approach this section so the guide remains fast on mobile connections.

Example preview

Calls until the first sale

Inputs: X ~ Geometric(p=0.3), trials until first success; evaluate needing more than 4 calls.
Representative result: P(X>4) = 0.7⁴ ≈ 0.2401 by the memoryless tail formula.

Illustrative only. Load the interactive tool to enter your own values and review assumptions.

Watch the explanation

2:56 min

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Read the complete transcript

The Geometric Distribution. How can repeated yes-or-no trials turn into a complete probability model for waiting? This page counts every attempted trial, including the first success, so support begins at one. Waiting exactly k trials means k minus one failures followed by success on trial k. Its point mass multiplies p by one minus p raised to k minus one. Every bar is one minus p times the previous bar, creating geometric decay. Cumulative probability through k is one minus the probability that all k trials failed.

Survival beyond k equals one minus p to k, directly describing an unlucky failure run. Quantiles choose the smallest whole-number budget whose cumulative probability reaches the requested assurance. Mean wait is one over p, while variance is one minus p over p squared. The mode stays at one because each successive point mass is strictly smaller. Median can be far below the mean because rare long waits pull the average rightward. Skewness and excess kurtosis are positive, reflecting a discrete distribution with a long right tail. As p rises waits collapse toward one; as p falls the tail and moments expand.

Suppose each cold call closes independently with probability point three, and X counts calls through sale. The first five masses are point three, point two one, point one four seven, and smaller. More than four calls means four consecutive failures, giving point seven to the fourth. That is about twenty-four percent, leaving cumulative probability about seventy-six percent through four. The mean is three point three three calls, but the median is only two calls. A ninety-five-percent assurance requires nine calls, much farther right than the mean.

Memorylessness says surviving m failures does not lengthen or shorten the distribution of remaining trials. After three failed calls, two more failures still have conditional probability point seven squared. This never means success is due; the next-call success probability remains point three. Counting failures instead shifts X down by one, changing support, mean, and every quantile. Changing p across attempts destroys constant hazard, so learning or depletion needs another model. Dependence between attempts also breaks the geometric tail and its memoryless interpretation entirely. The explorer evaluates a specified p; estimating p requires observed waiting-time data and uncertainty analysis. Report the convention, success definition, p source, exact event, and requested probability output. Explore geometric waits, exact tails, and planning quantiles free at Distri Scope dot com.

How to read the result

The bar chart always peaks at k=1 and declines geometrically; there is no interior peak, which surprises readers who expect the most likely wait to sit near the mean. The mean 1/p is pulled well right of the mode by the long tail.

The gap between the median and the mean is a direct display of skewness: for p=0.3 the median is 2 while the mean is 3.33, so typical waits are shorter than the average wait.

The survival value at k is exactly (1−p)^k, the probability that the first k trials all fail; reading it from the point calculator is often more natural than reading the CDF.

Inputs and parameter meaning

The single parameter is the per-trial success probability p between 0.01 and 0.99. Small p produces a long, slowly decaying tail; the chart range adapts to cover the bulk of the mass, extending to the 99.9th percentile.

This page counts trials, so the support begins at k=1 and the mass at one is exactly p. If your source counts failures before the first success, subtract one from every quantile and shift the mass function accordingly.

Interval probabilities include both integer endpoints, so the probability of needing between 2 and 4 trials is the sum of the masses at 2, 3, and 4.

Formula and available parameters

Density or probability mass

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

Cumulative distribution

F(k) = 1 - (1-p)^{\lfloor k \rfloor}

Mean

\mu = \frac{1}{p}

Variance

\sigma^2 = \frac{1-p}{p^2}

Probability of success (p): interactive range 0.01 to 0.99; default 0.3.

What the method is doing

The CDF has the closed form F(k)=1−(1−p)^⌊k⌋, so tail probabilities are computed directly as powers of 1−p rather than by summation, which keeps extreme tails accurate.

Quantiles are the smallest integer k whose cumulative probability reaches the requested level, found by exact search against the closed-form CDF; the summary median is that quantile at one half.

The theoretical moments are closed-form: mean 1/p, variance (1−p)/p², skewness (2−p)/√(1−p), and excess kurtosis 6+p²/(1−p). The mode is always 1 regardless of p, because each successive mass is a fixed fraction of the previous one.

Worked example: calls until the first sale

A sales representative closes on 30% of independent cold calls, so the number of calls needed for the first sale is modeled as X~Geometric(p=0.3) on the trials convention. The manager wants the probability that more than 4 calls are needed, the typical and average number of calls, and a call budget that covers the first sale with 95% assurance.

1Compute the tail directly: P(X>4)=(1−0.3)⁴=0.7⁴≈0.2401, so about one campaign in four needs a fifth call or more.
2Read the summary: the mean is 1/0.3≈3.33 calls and the median is 2, because F(2)=1−0.7²=0.51 already exceeds one half while F(1)=0.3 does not.
3Use the quantile lookup at 0.95: the smallest k with 1−0.7^k≥0.95 is k=9, so budgeting nine calls covers the first sale with 95% assurance under the model.
4Check memorylessness: after three failed calls, the chance the next call still fails is 0.7, unchanged—the conditional distribution of the remaining wait is again Geometric(0.3).

Interpretation

Under the stated model the first sale typically arrives by the second call, on average by the third or fourth, yet a 95% guarantee requires nine. The spread between those three numbers is the practical meaning of a geometric tail, and it is why planning on the average alone understates the resources needed for unlucky runs.

Common mistakes

Do not confuse the two support conventions. The trials version used here starts at 1 with mean 1/p; the failures version starts at 0 with mean (1−p)/p. Every quantile differs by exactly one between them, which is a silent off-by-one error in comparisons.
Do not read the memoryless property as a claim about luck evening out. After ten failures the expected remaining wait is still 1/p; the model never owes a success, and treating past failures as raising the next success chance is the gambler’s fallacy.
Do not fit a geometric model to waits measured in continuous time. The exponential distribution is the continuous analogue, and discretizing continuous waits into trials changes both parameter meaning and tail behavior.

Limits and independent validation

The page assumes a known constant p; it does not estimate p from observed waits. Estimation from data, with its own uncertainty, belongs to a fitting workflow rather than this theoretical calculator.

Extremely small p values produce very long supports; the chart truncates the display at the 99.9th percentile, so the visible bars intentionally omit a 0.1% tail that the closed-form CDF still accounts for.

The model has no mechanism for dependent attempts, changing success rates, or a maximum number of attempts; real retry systems with backoff or cutoffs need a modified model.

Before using the result

Justify the constant-p assumption from the mechanism: attempts should be operationally identical and independent. If a log of historical attempts exists, split it into early and late halves and compare observed success rates before pooling them into one p.
Recompute one tail by hand: the survival at k must equal (1−p)^k exactly, so p=0.3 must give P(X>4)=0.2401. A mismatch indicates a convention error rather than a numerical one.
Compare the observed mean wait with 1/p when data are available. A sample mean far above 1/p with excess dispersion suggests heterogeneity in p, which a negative binomial mixture represents better.

exponential distribution →See the continuous-time analogue that shares the memoryless property.

← Bernoulli Distribution Negative Binomial Distribution →

See every option in the Probability Distribution Explorer or review the DistriScope methodology. Educational information; last reviewed 2026-08-09. Verify consequential calculations independently.