Waiting for the first success
The geometric distribution answers the simplest waiting question in probability: how many independent trials until the first success, when each trial succeeds with probability p? Failures carry no progress and no penalty; the process simply continues. The probability that the wait ends exactly on trial k is a string of failures followed by one success: P(X = k) = (1 − p)k−1 p. The chart above always slopes downhill: the first trial is the single most likely ending point, however small p is, because every later trial must first survive all the failures before it.
An archer hits the gold on 1 shot in 4 (p = 0.25). First hit on the third shot: 0.75² × 0.25 = 0.1406. A hit within the first three shots: 0.25 + 0.1875 + 0.1406 = 0.5781. Still waiting after five shots: 0.755 = 0.2373. The average wait is 1/p = 4 shots, with a standard deviation of 3.464: the spread is nearly as large as the mean, which is characteristic of waiting-time distributions.
The memoryless property
The geometric distribution is the only discrete distribution with no memory: P(X > s + t | X > s) = P(X > t). Ten failures in a row do not make success due, and they do not make it less likely either; the remaining wait restarts statistically fresh. This is the gambler's fallacy stated as a theorem. The exponential distribution, its continuous cousin, shares the property, and both model the same physics: events arriving at a constant rate with no ageing.
Two conventions exist for what X counts: trials until the first success (this page, support 1, 2, 3...) or failures before it (support 0, 1, 2...). Software differs: R's dgeom uses the failures convention, so its probabilities shift by one trial relative to this page. Check the support before comparing answers.
Frequently asked questions
What is the mean of a geometric distribution?
1/p trials. At p = 0.25 the average wait is 4 trials; the variance is (1 - p)/p^2, here 12, so waits vary a lot around that mean.
Is the geometric distribution memoryless?
Yes, the only discrete distribution that is. Past failures change nothing about the remaining wait.
When should I use geometric instead of binomial?
The binomial fixes the number of trials and counts successes. The geometric fixes the number of successes (one) and counts trials. The question being asked picks the distribution.