This geometric distribution calculator prices the wait for a first success: the probability that success first arrives on trial k is P(X = k) = (1 − p)^(k−1) × p — with the cumulative chance it has arrived by trial k, and the expected wait 1/p.

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The Mathematics of “How Long Until It Happens”

Every other discrete distribution counts successes in a fixed number of trials; the geometric asks the inverse question — how many trials until the first success — and its formula is the purest in the subject: fail (k − 1) times, then succeed once, (1 − p)^(k−1) p. The defaults play it out with a die: the first six appearing exactly on roll 4 costs (5/6)³ × (1/6) = 9.65%; appearing by roll 4 is 51.77%; and the mean wait is the reciprocal 1/p = 6 rolls — a mean that is also the anchor of sales-call planning (a 5% close rate prices the first sale at ~20 calls), A/B and reliability work, and the expected length of every dry spell anyone has ever complained about. The distribution is front-loaded by construction: trial 1 is always the single most likely success trial (probability p itself), and every later k is strictly less likely — long waits are possible but never probable.

Its famous property, memorylessness, deserves precision because it’s the only discrete distribution that has it: P(X > m + n | X > m) = P(X > n) — conditional on having waited m trials fruitlessly, the distribution of the REMAINING wait is the original distribution, untouched. (The exponential distribution is its continuous twin with the same property, and together they’re the mathematics of “no wear, no learning, no debt” processes: radioactive decays, component lifetimes under the constant-hazard model, arrivals at a queue.) The practical consequence cuts against deep intuition: a slot machine that hasn’t paid in 100 pulls is exactly as likely to pay on pull 101 as a fresh one — the (1−p)⁹⁹ factor is the price of the PATH, already paid, not an accumulating credit. Casinos’ “due” mythology is this formula misread as a promise; the formula makes no promises.

Worked Examples (Verified)

  • Defaults: die, k = 4 → 9.65% exactly on roll 4; 51.77% by roll 4; mean 6.
  • Coin: p = 0.5, first heads on flip 3 → (0.5)²×0.5 = 12.5%.
  • Sales calls: p = 0.05 → mean wait 20 calls; P(no sale in the first 10) = 0.95¹⁰ = 59.87% — dry spells are the median experience, not bad luck.
  • By-when planning: p = 0.1 → P(success within 7 trials) = 1 − 0.9⁷ = 52.17%; the median wait for p is ln 0.5 / ln(1 − p) ≈ 6.6 trials.

Geometric Discipline

  • Two conventions exist: trials-until-success (support 1, 2, 3…, this page) vs failures-before-success (support 0, 1, 2…, mean (1−p)/p) — check which your textbook/software uses before comparing numbers.
  • Constant p and independence are the price of admission: processes that learn, tire or deplete aren’t geometric — they’re the interesting ones.
  • Mean 1/p is an average over many repetitions, not a deadline: at k = mean, cumulative probability is only ~63% (1 − e⁻¹ in the limit) — planning to the mean fails a third of the time.
  • Memorylessness is a modeling assumption, not a physical law — mechanical systems age (rising hazard), so geometric lifetimes flatter old equipment.
  • Negative binomial is the elder sibling: trials until the r-th success (r = 1 collapses to this page).

Related Tools

Frequently Asked Questions

How do you calculate geometric probability?

P(X = k) = (1 − p)^(k−1) × p: raise the failure probability to the number of failures before the success, multiply by the success probability once. For the first six on roll 4 of a die: (5/6)³ × (1/6) = 9.65%. The probability of success BY trial k is 1 − (1−p)^k (no failures-only run of length k). This calculator returns both, plus the mean wait 1/p and the standard deviation √(1−p)/p.

What is the expected value of a geometric distribution?

E[X] = 1/p — the reciprocal of the per-trial success chance. A die’s first six averages 6 rolls; a 2% cold-call conversion averages 50 calls to the first sale; a p = 0.25 process averages 4 trials. The mean is famously unintuitive company: the median is smaller (ln 0.5/ln(1−p) — for the die, ~3.8 rolls), because the distribution’s long right tail (the occasional 20-roll drought) drags the average above the typical experience. Budget time by the mean, expect the median, and don’t be surprised by the tail.

What does memoryless mean in the geometric distribution?

That past failures change nothing: P(X > m + n | X > m) = P(X > n). After 10 failed rolls, the probability of needing more than 3 additional rolls is exactly what it was at the start — (5/6)³ ≈ 57.9% for the die. The geometric is the ONLY discrete distribution with this property (its continuous sibling, the exponential, is the only continuous one). Where memorylessness holds, “I’m due” is always false; where real processes age, learn or fatigue, the model — not the universe — is what fails.

What is the difference between geometric and binomial?

The random variable differs. Binomial fixes the number of trials n and counts successes (X ~ Bin(n, p)); geometric fixes the success count at 1 and counts trials (X ~ Geo(p)). Same Bernoulli trials, opposite question. “How many heads in 10 flips” is binomial; “how many flips until the first head” is geometric. The negative binomial generalizes the geometric (trials until the r-th success), and the binomial’s waiting-time shadow is exactly that family — the two distributions are one process read from opposite ends.

What is the probability of success within k trials?

1 − (1 − p)^k — one minus the probability of failing every trial. At p = 0.1, success within 7 trials is 1 − 0.9⁷ = 52.17% (so the median wait at p = 0.1 is 7 trials). Useful inversion: to reach a target confidence C you need k = ln(1 − C)/ln(1 − p) trials — 90% confidence at p = 0.05 demands 45 calls, which is the honest arithmetic behind sales managers’ activity quotas and recruiters’ pipeline math.

Is the geometric distribution’s support 0 or 1 based?

Both conventions live in the wild, and confusing them shifts every answer by one trial. This calculator uses the trials version: X counts the trial OF the first success, support {1, 2, 3, …}, mean 1/p. The failures version counts failures BEFORE the first success: support {0, 1, 2, …}, mean (1−p)/p, formula P(Y = y) = (1−p)^y · p. R’s dgeom and many software packages use the failures version; most probability textbooks use trials. Convert between them by ±1 — and always check which one a formula sheet means.

How is the geometric distribution used in real life?

Anywhere “time to first event” with constant per-period risk appears: sales and fundraising (calls to first conversion), reliability (cycles to first failure under constant hazard — the discrete shadow of the exponential), quality (items inspected to first defective), gaming (pulls to first jackpot, attempts to first rare drop — published “pity timer” debates are arguments about replacing geometric waits with capped ones), and epidemiology’s generation models. If the per-trial chance truly doesn’t change, this page’s three outputs (exact, cumulative, mean) are the whole story of the wait.

Is this geometric distribution calculator free?

Yes — free, instant, no sign-up. Enter p and k for the exact probability, the by-trial-k cumulative, mean wait and standard deviation.