Geometric Distribution Calculator
Answer the question “when does the first success arrive?” for repeated independent yes/no trials. Get P(X = k), P(X ≤ k), and P(X > k) in either textbook convention, plus the mean, variance, and standard deviation.
Before You Calculate
- Pick the convention your course or textbook uses: X counts trials including the success (k = 1, 2, …), or Y counts 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.
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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.
Binomial vs Poisson vs Hypergeometric
Three questions decide the right counting distribution — fixed trials, independence, replacement — with one inspection scenario computed all three ways.
Formulas in Both Conventions
Trials (X = 1, 2, 3, …):
P(X = k) = (1 − p)^(k−1) × p
P(X ≤ k) = 1 − (1 − p)ᵇ, E[X] = 1/p
Failures (Y = 0, 1, 2, …):
P(Y = k) = (1 − p)ᵇ × p
P(Y ≤ k) = 1 − (1 − p)^(k+1), E[Y] = (1 − p)/p
Both: Var = (1 − p)/p²
The logic is direct: for the first success to land on trial k, the first k − 1 trials must all fail — probability (1 − p)^(k−1) — and then trial k must succeed. The cumulative form is even simpler through the complement: the first success arrives within k trials unless all k trials fail, so P(X ≤ k) = 1 − (1 − p)ᵇ.
The two conventions differ only by whether the final success is counted, so Y = X − 1: every probability matches after shifting k by one, and the variance is identical because shifting does not change spread.
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 — the process does not remember the losing streak. It is also the r = 1 special 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 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: 0.8⁴ = 0.4096.
The distribution summary: mean 1/0.2 = 5 calls until the first sale, variance 0.8/0.04 = 20, standard deviation ≈ 4.47. The large spread relative to the mean is typical of geometric waiting times: first sales on call 1 and call 15 are both entirely plausible. In the failures convention the same experiment is entered as k = 3 with p = 0.2, giving the identical 0.1024.
Frequently Asked Questions
What is the difference between the two geometric distribution conventions?
One convention counts the trial on which the first success occurs (X = 1, 2, 3, ...), the other counts the failures before it (Y = 0, 1, 2, ...). They describe the same experiment with Y = X - 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(Y = 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 distribution's 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). Practically, 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 instead, and for 'at least one success in n trials' the probability-of-at-least-one calculator gives the direct complement-rule shortcut.