How this instrument works
The geometric distribution models the number of independent, identical trials needed to get the very first success, when each trial has the same success probability p. Rolling a die until the first six comes up, dialing sales calls until the first yes, or testing components until the first defect all follow this pattern: repeated trials, constant probability p, stop as soon as one succeeds.
This instrument uses the trial-of-first-success convention, where k counts trials starting from 1 — k = 1 means the very first trial was already a success, k = 2 means the first trial failed and the second succeeded, and so on. (A related convention counts failures-before-the-first-success starting from 0 instead; this site's Negative Binomial Distribution calculator, generalized to r successes, uses that failures-counted convention for its own k input — worth noting if you compare the two side by side.)
The PMF, P(X = k) = (1−p)^(k−1) × p, multiplies the probability of k−1 consecutive failures by the probability of the final success. The CDF, P(X ≤ k) = 1 − (1−p)^k, gives the probability that the first success happens by trial k or earlier — the complement of the event that the first k trials are all failures. Like the exponential distribution, its continuous-time cousin, the geometric distribution is memoryless: how many failures you've already seen tells you nothing about how many more are coming.
- Enter the per-trial success chance into Probability of success on each trial (p), a value greater than 0, up to and including 1.
- Enter the trial number you're asking about into Trial number of the first success (k ≥ 1, not failures-before-success) — k = 1 for 'succeeds on the very first try'.
- Read P(X = k) — probability the first success is on trial k — the probability of that exact trial being the first success.
- Read P(X ≤ k) — probability the first success occurs by trial k — the probability the first success has happened by trial k or sooner.
- k must be 1 or greater: this instrument counts trials, not failures, so there is no k = 0 case here.
Worked example — first success on the 3rd trial, p = 0.2
Enter p = 0.2 into Probability of success and k = 3 into Trial number of the first success — a 20% success rate, asking about the first success landing exactly on the third attempt. The instrument computes (1−p)^(k−1) = 0.8^2 = 0.64, the probability of failing on trials 1 and 2, then multiplies by p = 0.2.
P(X = k) reads 0.128 — that's 0.64 × 0.2 exactly. P(X ≤ k) reads 0.488, computed as 1 − 0.8^3 = 1 − 0.512. So there's a 12.8% chance the first success happens on precisely the third trial, and a broader 48.8% chance it has happened by the third trial or sooner — meaning there's still a 51.2% chance the first three trials all fail.
Questions
What's the difference between P(X = k) and P(X ≤ k) here?
P(X = k) is the probability that the first success lands on exactly trial k — trials 1 through k−1 all fail, and trial k succeeds. P(X ≤ k) is the probability the first success has happened somewhere in trials 1 through k, i.e. it's not the case that all of the first k trials failed. P(X ≤ k) is always at least as large as P(X = k), since it accumulates every k value from 1 up to k.
How is the geometric distribution related to the negative binomial distribution?
The geometric distribution is the special case of the negative binomial distribution where the target number of successes r = 1. This site's Negative Binomial Distribution calculator confirms it directly: with r = 1, p = 0.5, k = 0 failures before the 1st success, its PMF gives exactly 0.5 — matching this geometric calculator's p = 0.5, k = 1 result of 0.5 (trial-of-first-success = 1 failure + 1 success, in the negative binomial's failures-counted convention, is the same event as 'first trial succeeds' here).
What is the mean number of trials until the first success?
The mean is 1/p. At p = 0.2, as in the worked example, the mean number of trials to the first success is 1/0.2 = 5 — even though the single most probable outcome, k = 1, has a lower individual probability (0.2) than the mean would suggest, because the distribution has a long tail of increasingly unlikely late successes that pull the average up.
Why is the geometric distribution called 'memoryless'?
Because past failures don't change future odds: if the first 10 trials have all failed, the probability of needing exactly 5 more trials to succeed is identical to the probability of a fresh sequence needing exactly 5 trials from trial 1. Each trial is independent and identically distributed, so the process genuinely has no memory of how many failures came before.
Can p be very small, like 0.001?
Yes — small p values are common and simply push the distribution's mass toward larger k, since the mean 1/p grows as p shrinks. At p = 0.001, the mean number of trials to first success is 1,000, and P(X = k) for small k (like k = 1) becomes tiny, while the distribution spreads out over a much wider range of plausible trial counts.
What real-world situations follow a geometric distribution?
Anything with repeated, independent, identical attempts stopped at the first success: the number of sales calls until the first sale, the number of lottery tickets bought until the first win, the number of manufactured parts inspected until the first defect is found, or the number of coin flips until the first heads. The requirement is always the same — constant success probability p and independence between trials.