Lesson · Discrete Probability · Distributions

The Geometric Distribution and Waiting Times

P(X = k) = (1−p)^{k−1}p for the trial of the first success. Its mean is 1/p, its distribution is heavily skewed, and it is memoryless — which is the property people find hardest to accept.

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P(X = k) = (1−p)^{k−1} p first success on trial k E[X] = 1/p Var(X) = (1−p)/p² MEMORYLESS: P(X > m+n | X > m) = P(X > n)

Waiting for a success

Core ideas

What this lesson establishes

The last distribution on this course, and the one whose mean is least representative of a typical outcome.

The number of trials is not fixed

That is what separates it from the binomial. The sample space is countably infinite, which is why the axioms had to be stated in general.

The mean is 1/p, and it is not typical

A 1-in-6 event takes 6 attempts on average, and yet the single most likely outcome is the very first attempt.

It is memoryless

Having failed 20 times does not improve the next attempt. The distribution of the remaining wait is unchanged.

The material

Waiting times

The distribution, its mean, memorylessness, and the coupon collector.

Theorem · Geometric distribution

In independent trials with success probability p > 0, let X be the number of the trial on which the first success occurs. Then P(X = k) = (1−p)^{k−1}p for k = 1, 2, 3, …

The first success on trial k means k − 1 failures followed by a success. By independence that probability is (1−p)^{k−1}p.

These sum to 1 by the geometric series of “Induction with Sums and Products” in Induction and Recursion: Σ_{k≥1} (1−p)^{k−1}p = p/(1 − (1−p)) = 1.

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Theorem · Mean and variance

E[X] = 1/p and Var(X) = (1−p)/p².

Condition on the first trial. With probability p it succeeds and X = 1; with probability 1 − p it fails and the remaining wait has the same distribution as X. So E[X] = p·1 + (1−p)(1 + E[X]).

That sentence is a computation from the definition, not a new rule. For k ≥ 2, P(X = k) = (1−p)·P(X = k−1) — the first trial failed, then the wait started again — so E[X] = 1·p + Σ_{k≥2} k(1−p)P(X = k−1) = p + (1−p) Σ_{j≥1} (j+1)P(X = j) = p + (1−p)(E[X] + 1), the same equation.

Expanding: E[X] = 1 + (1−p)E[X], hence pE[X] = 1 and E[X] = 1/p.

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That conditioning argument is worth more than the formula. It avoids summing an infinite series entirely, and the same trick handles many waiting-time problems where no series is available.

Theorem · Memorylessness

P(X > m + n | X > m) = P(X > n) for all m, n ≥ 0. The geometric is the only discrete distribution with this property.

P(X > k) = (1−p)^k, since it means the first k trials all failed. Then P(X > m+n | X > m) = (1−p)^{m+n} / (1−p)^m = (1−p)^n = P(X > n).

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This is the gambler's fallacy stated correctly. A roulette wheel that has come up red twenty times is exactly as likely to come up red next spin; the past failures are already accounted for and change nothing. The distribution has no memory because the trials are independent, and independence is what the model assumed.

Why the mean misleads

p = 1/6: E[X] = 6 P(X = 1) = 0.167 the single most likely value P(X ≤ 3) = 0.421 P(X ≤ 6) = 0.665 so 2/3 of the time you finish by the mean P(X > 12) = 0.112 and 1 in 9 waits more than twice as long P(X > 20) = 0.026

The distribution is strictly decreasing: the most likely single outcome is always k = 1, whatever p is. "On average 6" describes a long right tail balanced against a mode at 1, and quoting the mean alone hides both facts.

The coupon collector

Theorem · Collecting all n coupons

With n equally likely coupons, the expected number of purchases to collect all of them is n(1 + 1/2 + ⋯ + 1/n) ≈ n ln n + 0.577n.

Split the process into stages: stage i runs while you hold exactly i − 1 distinct coupons. The chance a purchase is new during that stage is (n − i + 1)/n, so the wait is geometric with that p and mean n/(n − i + 1).

By linearity of expectation the total is Σ_{i=1}^{n} n/(n−i+1) = n Σ_{j=1}^{n} 1/j.

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For n = 50 that is about 225 purchases, and the last coupon alone accounts for 50 of them on average. The final stages dominate, which is the structural reason collecting the last few of anything takes so long.

Interactive

Distribution, expectation, variance

Definition first, formula second

The shape behind the mean

The lab opens on the worked example, p = 2/12 = 1/6. The first row reads 0.166667, the largest in the table, and k = 6 reads 0.066980; the bars fall from the first one and the dashed mean sits at 6, out where they are already less than half their starting height. That gap is the whole caution of this lesson. E[X] summed reads 6.0000 against 1/p and Var(X) reads 30.0000. The sums run far past the thirty bars drawn — the status line says how far — because a sum stopped at 30 gives 5.848, and a lab that printed that beside 6 would be showing a disagreement that is not there.

Distribution

n10

p (in twelfths)6/12

E[X] summed—

E[X] formula—

Var(X)—

Method

Working with waiting times

Identify what is fixed: the trials, or the successes.

Check whether n is fixed

Fixed trials, count successes: binomial. Fixed target, count trials: geometric.

Use E[X] = 1/p

And derive it by conditioning on the first trial if you want to check it. The series is unnecessary.

Report the spread as well as the mean

The distribution is skewed enough that the mean alone misinforms. P(X > k) = (1−p)^k is the clean way to describe the tail.

Decompose staged waits by linearity

The coupon collector is a sum of geometric waits, and linearity handles it without any joint distribution.

Geometric

Score 0 / 0

Choose the answer you can justify.

Rolling for a six

p = 1/6. How long until the first six?

E[X] = 1/p = 6 rolls Var(X) = (1−p)/p² = (5/6)/(1/36) = 30, σ ≈ 5.48 P(X = 1) = 1/6 ≈ 0.1667 ← the most likely value P(X = 6) = (5/6)⁵(1/6) ≈ 0.0670 P(X ≤ 6) = 1 − (5/6)⁶ ≈ 0.6651 P(X > 12) = (5/6)¹² ≈ 0.1122 The standard deviation (5.48) is nearly as large as the mean (6), which is a distribution nobody should summarise with one number.

Two thirds of the time you finish within the mean, and one time in nine you wait more than twice it. Both facts follow from P(X > k) = (5/6)^k, which is a better summary of this distribution than its expectation.

Common mistakes

Confusing it with the binomialFixed trials counting successes is binomial; fixed target counting trials is geometric. The question decides, not the setting.

Believing a long run of failures improves the oddsMemorylessness says otherwise, and it is a theorem about the model rather than a claim about luck.

Reporting only the meanWith σ nearly equal to the mean, a single number conceals both the mode at 1 and the long tail.

Check your understanding

✓

Finish when you can solve a waiting time no series gives you.

How many flips of a fair coin, on average, until two heads in a row? Let E be the answer and condition on how the sequence starts: a tail (1/2) wastes one flip, head-then-tail (1/4) wastes two, and head-head (1/4) finishes in two. Solve E = ½(E + 1) + ¼(E + 2) + ¼·2 and check that you get 6. No series was available, and the argument did not need one.

Note: This course ends where several others begin. The geometric distribution is the discrete analogue of the exponential, memorylessness characterises both, and the coupon collector is the first example most people meet of a process whose expected time is dominated by its final stages.