How this calculation works
T counts all trials up to and including the first success. To succeed for the first time on trial k, the previous k−1 trials must all fail. The cumulative result includes success on any of trials 1 through k.
P(T = k) = (1−p)ᵏ⁻¹ p; P(T ≤ k) = 1−(1−p)ᵏA worked example
With a 25% chance each trial, first success on trial 3 has probability 0.75² × 0.25 = 14.0625%.
Good to know
A mathematical probability model, not a fit test or an empirical guarantee. Inputs must match the model assumptions. Floating-point arithmetic is used; extremely small probabilities may underflow or round to zero.
Reference: OpenStax: geometric distribution (opens in a new tab) ↗
Be clear about what the count includes
One common geometric convention counts trials up to and including the first success. Another counts only the preceding failures. This calculator uses the first convention, so immediate success is trial 1, with zero preceding failures.
With a 25% chance each trial, reaching the first success on trial 3 requires failure, failure, success. Multiplying 0.75 × 0.75 × 0.25 gives 14.0625%. A success on trial 1 or 2 belongs in “by trial 3” but not in “on trial 3”.
The expected count is 1/p. At p = 25%, that average is four trials. It does not mean the fourth trial is certain to succeed or that four failures contradict the model.
Past failures do not improve an independent trial
Under the model, each fresh trial has the same success probability even after a long run of failures. Feeling that a success is now “due” adds a dependence that is absent from the stated assumptions.
Some real tasks do become easier with practice, or harder as resources are depleted. Others select items without replacement. A constant independent-trial model should not be used merely because those tasks also repeat attempts.
At zero success probability, no finite trial produces a success and the expected waiting count is not finite. At 100%, the first trial always succeeds. The tool handles both boundaries directly rather than applying a logarithm to an invalid endpoint.
A few questions, answered
Does it count successes after the first one?
No. The random variable stops at the first success. Counting total successes in a fixed number of independent trials is a binomial question.
Does increasing k increase the exact probability?
Not generally. At a fixed positive p, the probability of first success on trial k decreases with k, while the cumulative probability by trial k increases.
Can k be a decimal?
No. Trials are counted in whole numbers, starting at 1. The supported maximum is one billion.