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Geometric Distribution (First Success)

P(X=k)=(1−p) k−1pP(X = k) = (1 - p)^{\,k-1} p

Worked example: p = 0.2 with P = 0.08192 → k = 5 — press Try an example to run it live, then adjust anything.

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Geometric Distribution (First Success) explained

kp

To have your first success on trial k you must fail k − 1 times and then succeed, which is (1−p)k−1×p(1-p)^{k-1} \times p. Rolling a fair die, the chance the first six comes on the third roll is (5/6)² × (1/6) = 25/216 ≈ 0.116. Enter p as a decimal from 0 to 1 or switch that input to %; k is a whole trial number, 1 or more.

The geometric distribution is memoryless — the only discrete distribution that is — so after twenty failed rolls the chance the next roll is a six is still 1/6, exactly as it was at the start. That is the gambler's fallacy in its purest form, and the "hot hand" is its mirror image. Notice that the single most likely value of k is always 1, however small p is: each additional wait multiplies by (1 − p) < 1. Solving for p has no closed form, so this calculator solves for P or for k: with p = 0.2 and P = 0.08192, k = 1 + ln(0.4096)/ln(0.8) = 5.

Geometric Distribution (First Success) formula

P(X=k)=(1−p) k−1pP(X = k) = (1 - p)^{\,k-1} p
Where
  • P(X=k)P(X = k)= Probability the first success is on trial k
  • pp= Probability of success per trial
  • kk= Trial of the first success

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