Updated 12 September 2026

Geometric distribution

Flip a biased coin with success probability p until the first head. The number of flips needed follows a Geometric distribution. One parameter does all the work.

What to notice

• Small p produces a long, geometrically decaying right tail — most runs take many trials, but occasionally the first flip succeeds.
• Large p pushes almost all probability onto k=1 and k=2 — success comes quickly.
• Mean $E[X]=1/p$ and variance $\text{Var}(X)=(1-p)/p^2$ both grow as p shrinks.

Memorylessness

For independent trials with a fixed known p, past failures do not change the success probability on the next trial.

If you’ve already failed m times, the distribution of remaining trials looks exactly like starting fresh. Among distributions on the positive integers, this property characterizes the geometric law. The exponential is its nondegenerate continuous counterpart on nonnegative waiting times.

Connection to the exponential

Split time into intervals of length δ. In a homogeneous Poisson process, an interval contains at least one event with probability p = 1 − exp(−λδ). The number G of intervals until the first nonempty one is geometric. As δ tends to zero, the waiting time δG converges in distribution to Exponential(λ). They are the same memorylessness expressed in discrete vs continuous time.

Count the successful trial consistently

Suppose the outcomes are failure, failure, success. The trial count G is three; the failure count K=G−1 is two. At p=0.3, both descriptions assign this sequence probability 0.7²×0.3=0.147. They describe the same experiment, with means 1/p≈3.333 and (1−p)/p≈2.333. The variance is unchanged by subtracting one.

For G≤5, the complement is five consecutive failures: P(G>5)=0.7⁵=0.16807, so P(G≤5)=0.83193. For K≤5, six trials are allowed; the probability is 1−0.7⁶=0.882351. Switch the convention without changing the numeric threshold to see why software parameter conventions matter.

Decompose the remaining wait

At p=0.3, P(G>3)=0.343 and P(G>5)=0.16807. Among runs surviving three failures, the chance of two more failures is 0.16807/0.343=0.49. A fresh run also has probability 0.7²=0.49 of failing twice. The control uses the surviving group as denominator, including waits beyond the chart; an empty simulated group has no empirical conditional probability.

A first-step calculation also explains the mean: one trial is always used. With probability 1−p it fails and the remaining wait has the original distribution. Thus E[G]=1+(1−p)E[G], giving 1/p. Adding r independent waits gives the total-trial version of a negative binomial.

Reference

Random Services: geometric distribution develops both counting conventions, the survival identity, and the first-step expectation argument. The exponential experiment compares memorylessness with changing-hazard alternatives.