Geometric Distribution Calculator
Answer the question “when does the first success arrive?” for repeated independent yes/no trials. Get exactly, at most, at least and between probabilities in either textbook convention, plus the mean, variance, a bar chart and a table.
Trials count the success itself, failures do not
A decimal from 0 to 1: a 20% chance per trial is 0.2
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Compare the binomial, Poisson, normal, uniform, and exponential distributions and learn how to choose the right model for counts, measurements, and waiting times.
Before you calculate
- Pick the convention your course or textbook uses: X counts trials including the success (k = 1, 2, …), or failures before it (k = 0, 1, …). The two describe the same experiment shifted by one.
- Enter the success probability as a decimal: a 20% chance per trial is p = 0.2.
- Trials must be independent with the same p every time, the same conditions as the binomial distribution, but with the number of trials left open.
Formulas in both conventions
Trials (k = 1, 2, 3, …): P(X = k) = (1 − p)^(k − 1) × p, P(X ≤ k) = 1 − (1 − p)^k, mean 1/p
Failures (k = 0, 1, 2, …): P(X = k) = (1 − p)^k × p, P(X ≤ k) = 1 − (1 − p)^(k + 1), mean (1 − p)/p
Both conventions: variance (1 − p)/p²
The logic is direct: for the first success to land on trial k, the first k − 1 trials must all fail, with probability (1 − p)k−1, and then trial k must succeed. The cumulative form is simpler through the complement: the first success arrives within k trials unless all k trials fail, so P(X ≤ k) = 1 − (1 − p)k. The two conventions differ only in whether the final success is counted, so every probability matches after shifting k by one, and the variance is identical because shifting does not change spread.
The “more than” card is computed from the upper tail directly, so it keeps a tiny probability that 1 minus the cumulative value would round away: with p = 0.5 and k = 60 in the trials convention it shows 8.6736e-19, where 1 − P(X ≤ 60) evaluates to exactly 0 in double precision.
Which convention does your software use?
| Software | P(X = k) | Counts |
|---|---|---|
| TI-84 | geometpdf(p, k) and geometcdf(p, k) | trials, k ≥ 1 |
| Python (SciPy) | scipy.stats.geom.pmf(k, p) | trials, k ≥ 1 |
| R | dgeom(k, p) and pgeom(k, p) | failures, k ≥ 0 |
| Excel | NEGBINOM.DIST(k, 1, p, FALSE) | failures, k ≥ 0 |
R and Excel count failures while SciPy and the TI-84 count trials, so dgeom(3, 0.2) in R and geom.pmf(4, 0.2) in SciPy return the same 0.1024. Choose the matching convention above to compare with any of them. The geometpdf and geometcdf calculator follows the TI-84 (trials).
Where it sits among the distributions
The binomial distribution fixes the number of trials and asks how many successes occur; the geometric fixes the number of successes at one and asks how many trials it takes. It is the discrete cousin of the exponential distribution and shares its defining quirk, memorylessness: given k failures so far, the distribution of the remaining wait is exactly the original distribution. With p = 0.2, P(X > 10 | X > 5) = 0.810 / 0.85 = 0.85 = 0.32768, the same as starting from scratch. It is also the r = 1 case of the negative binomial distribution, which waits for the r-th success instead of the first.
Worked example: cold calls
A salesperson closes 20% of cold calls (p = 0.2). What is the probability that the first sale comes exactly on the 4th call?
- Three failures first: 0.8³ = 0.512.
- Then one success: 0.512 × 0.2 = 0.1024.
- Within four calls: P(X ≤ 4) = 1 − 0.8⁴ = 1 − 0.4096 = 0.5904.
- Longer than four calls: P(X > 4) = 0.8⁴ = 0.4096.
The summary: mean 1/0.2 = 5 calls until the first sale, variance 0.8/0.04 = 20 and standard deviation ≈ 4.4721. The large spread relative to the mean is typical of geometric waiting times: first sales on call 1 and on call 15 are both entirely plausible. P(X ≤ 3) = 0.488 and P(X ≤ 4) = 0.5904 straddle one half, so the median wait is 4 calls, and the chance of waiting more than 10 calls is 0.107374. In the failures convention the same experiment is entered as k = 3 with p = 0.2 and gives the identical 0.1024. Load example reproduces the first four results.
Related guides and calculators
For the chance of at least one success in n tries use the at least one probability calculator, for a fixed number of tries the binomial distribution, and for coins the coin flip probability calculator. The continuous counterpart, for waiting times, is the exponential distribution, and probability distributions compares the whole family.
Frequently Asked Questions
What is the difference between the two geometric distribution conventions?
One convention counts the trial on which the first success occurs (1, 2, 3, ...), the other counts the failures before it (0, 1, 2, ...). They describe the same experiment with failures = trials - 1, so any probability in one convention can be read in the other by shifting k by one. Textbooks split roughly evenly, which is why this calculator supports both explicitly.
What is the formula for the geometric distribution?
In the trials convention, P(X = k) = (1 - p)^(k-1) x p: the first k - 1 trials fail, then trial k succeeds. The cumulative version is P(X <= k) = 1 - (1 - p)^k, since the only way the first success is not within k trials is for all k to fail. In the failures convention the exponent shifts: P(X = k) = (1 - p)^k x p.
What does the mean 1/p actually tell me?
It is the long-run average number of trials until the first success: with p = 0.2, sales come every 5 calls on average. It is not a guarantee or even the most likely single value; the most probable trial for the first success is always the first one (probability p), and the long tail means waits far beyond 1/p are common.
How is the geometric distribution related to the binomial?
Both assume independent trials with constant success probability p. The binomial fixes the number of trials n and counts successes; the geometric fixes the target at one success and counts trials. They answer complementary questions, and the identity P(X > k) = (1 - p)^k is just the binomial statement that k trials produced zero successes.
Is the geometric distribution memoryless?
Yes, it is the only discrete distribution with that property. Given that the first k trials all failed, the number of additional trials needed has exactly the original geometric distribution: P(X > s + t | X > s) = P(X > t). A losing streak does not make the next trial more likely to succeed; the process has no memory.
What if I am waiting for more than one success?
The waiting time for the r-th success follows the negative binomial distribution, of which the geometric is the r = 1 case. For 'how many successes in a fixed number of trials' use the binomial distribution calculator, and for 'at least one success in n trials' the at least one probability calculator gives the direct complement-rule answer.
How small can the success probability be?
The average number of failures before the first success, (1 - p)/p, must not exceed 1,000,000, which allows p down to about 0.000001. Within that range the probabilities have been checked against high-precision arithmetic to at least nine significant digits.
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