Geometric Distribution Calculator

Why the failures behind you do not count

The geometric distribution is memoryless, and it is one of very few that are. The property has an exact statement:

P(X > m + n | X > m) = P(X > n), for every m.

In words: given that you have already failed m times, the chance of waiting at least n more is the same as the chance of waiting at least n from a standing start. The past contributes nothing.

The panel in the tool computes both sides on your own numbers. At p = 0.2 with n = 5, the answer is 0.3276800000 whether you have failed zero times or two hundred. Identical to the last digit, because it is an identity rather than an approximation: the survival function is (1−p)ᵏ, and (1−p)ᵐ⁺ⁿ divided by (1−p)ᵐ is (1−p)ⁿ exactly.

This is the gambler’s fallacy, formalised. “Red has come up eight times, so black is due” is precisely the belief that memorylessness denies. The wheel has no record of the previous eight spins, and neither does the arithmetic.

Worth being careful about what it does not say. It does not say long streaks are common — they are rare, which is why they feel meaningful. It says that having had one changes nothing about what comes next. Rare beforehand, irrelevant afterwards.

And it only holds when the trials really are independent with a constant p. If your applications are improving, or the machine is wearing, the model does not apply and the intuition that the past matters is correct — just not for this distribution.

Two definitions share the name

This is the commonest source of wrong answers with the geometric distribution, and it is not a subtlety — it is two different random variables.

Number of trials, including the success. Starts at 1. P(X = k) = (1−p)ᵏ⁻¹p. Mean 1/p. This is what the tool takes as its input, and what “how many attempts until it works” usually means.

Number of failures before the success. Starts at 0. P(X = k) = (1−p)ᵏp. Mean (1−p)/p. This is R’s dgeom and the convention most textbooks use when they generalise to the negative binomial.

The means differ by exactly 1, which is the tell. At p = 0.2 the mean is 5 trials or 4 failures. Both are correct; quoting one while computing the other is not.

The variance is the same for both, (1−p)/p², because shifting a distribution does not change its spread. That is a useful check: if two sources agree on the variance and differ by 1 in the mean, this is why.

The tool prints both means side by side on every calculation, so the convention is never inferred.

The most likely wait is always one trial

A property that surprises people and is worth sitting with, because it says something real about waiting times.

The mode is always 1, whatever p is. P(X = 1) = p is the largest single probability, and every subsequent trial is strictly less likely than the one before, because each requires an extra failure first.

So the single most likely outcome is succeeding immediately — even when p is small. At p = 0.05 the chance of success on the first attempt is 5%, and that is still higher than the chance on any other specific attempt.

Meanwhile the mean is 1/p = 20. Mode 1, mean 20: an enormous gap, and the mark of a heavily right-skewed distribution. The gap between centres is doing exactly what that page describes.

The median is smaller than the mean too — roughly ln(2)/p, about 0.69/p. At p = 0.05 the median wait is about 14 trials while the mean is 20, so more than half of all waits are shorter than average.

The practical reading: most waits are shorter than the average, and a few are much longer. Planning around the mean will leave you surprised by the tail, and the tail is where the cost is.

How long until you are probably done

A more useful question than the mean, and one the survival function answers in a line.

P(X > k) = (1−p)ᵏ. The chance of still waiting after k trials, with no sum required.

So the number of trials needed to be 95% confident of at least one success is ln(0.05)/ln(1−p). At p = 0.1 that is 29 trials; at p = 0.01 it is 299; at p = 0.001, 2,995.

Notice the pattern: roughly 3/p. Three times the mean gets you to about 95%, for any p. That is a genuinely useful planning heuristic, and it is the same constant behind the rule of three, arriving from the other direction.

It also quantifies how bad the mean is as a plan. If you budget 1/p attempts, your chance of having succeeded is 1 − (1−p)^(1/p), which converges to about 63% as p gets small. Budgeting the mean leaves you failing more than a third of the time.

For anything where running out of attempts is expensive — retries, test batches, drilling programmes — plan on 3/p, not 1/p.

What it becomes when you change one thing

The geometric sits at the corner of a family, and each neighbour relaxes exactly one of its assumptions.

Waiting for the r-th success: negative binomial. The direct generalisation — the geometric is the case r = 1. Its other use is as the fix for overdispersed counts, which is a different face of the same distribution.

Fixed number of trials instead: binomial. The mirror image. Binomial fixes n and lets the successes vary; geometric fixes the successes at one and lets n vary. Same trials, opposite question.

Continuous time instead of discrete trials: exponential. The exponential is the geometric’s continuous analogue, and it is the only continuous distribution that is memoryless — just as the geometric is the only discrete one. That uniqueness is not a coincidence; memorylessness forces the functional form.

Without replacement: negative hypergeometric. Drawing from a finite pool changes p as you go, which breaks the constant-probability assumption, exactly as it does for the binomial.

Where it applies, and where it quietly does not

Anywhere you repeat an independent attempt at constant odds until it works.

Retries and reliability. Packets until one gets through, attempts until a lock acquires, restarts until a flaky test passes. The memorylessness here is real and is why exponential backoff exists — if retries were memoryless in the useful sense, backoff would be pointless.

Quality inspection. Items examined before the first defect, when the batch is large enough that removing one does not change the rate.

Games of chance. Rolls until a six, spins until a number comes up. The canonical teaching case, and the one where the fallacy is most tempting.

And two where it looks right and is not.

Job applications. Tempting, and wrong if you are learning. Each application changes your CV, your interview technique and your network, so p is not constant. The model assumes you are no better on attempt fifty than on attempt one.

Anything with wear or fatigue. A machine that degrades has a rising failure probability, so the trials are not exchangeable. Reliability engineering uses the Weibull distribution precisely because it allows the hazard rate to change, which memorylessness forbids.

The fallacy, and its mirror image

Memorylessness is denied in two opposite directions, and both are common.

The gambler’s fallacy says a long run of failures makes success more likely — you are “due”. The arithmetic above refutes it directly: the conditional and unconditional probabilities are identical.

The hot-hand fallacy says the opposite — that a run predicts more of the same. Under a genuinely geometric process that is equally wrong, and for the same reason.

What makes this genuinely hard is that both can be correct outside the model. If a factory is drifting out of calibration, a run of defects really does predict more defects. If a player is warmed up, a run of hits may really predict more hits. The mistake is not believing the past matters — it is believing it matters in a process where p is constant, or believing it does not in a process where p is drifting.

So the useful question is never “am I due?” but “is p actually constant here?”. If it is, the streak is information about nothing. If it is not, the geometric distribution was the wrong model from the start and its probabilities were never the ones to use.

Sources and methodology

References for the distribution and the memoryless property.

• NIST/SEMATECH e-Handbook — the geometric distribution and its two parameterisations
• R documentation for dgeom — which uses the failures-before-success convention, one lower than this page's default
• Exponential Distribution Calculator — the continuous analogue, and the only other memoryless distribution
• Binomial Distribution Calculator — the mirror question: fixed trials, variable successes
• How we calculate — methodology

Method. The memorylessness panel computes both sides of P(X > m+n | X > m) = P(X > n) from the survival function rather than displaying a rearranged formula, so the identity is demonstrated on the reader’s own numbers instead of asserted. The suite checks it holds to within 1e-12 across a grid of p values and prior-failure counts including m = 200, where a naive implementation that accumulated error would visibly drift. Both conventions are computed and printed on every result, and the suite asserts that their means differ by exactly 1 while their variances are identical — which is the property that lets a reader tell which convention a source is using. That engine is verified on every change against 0 assertions, including that the geometric PMF sums to 1 over four thousand terms at p = 0.02. The count and the per-case breakdown are published on the formula verification page.

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