Poisson and Geometric Distributions

Some of the most useful probability models answer questions that sound almost casual. "How many customer complaints should the call centre expect in an hour?" "How many times will I have to roll before I finally get a six?" The first is a counting question about rare events spread over time or space; the second is a waiting question about how long until something finally happens. The Poisson distribution and the geometric distribution are the two workhorses for exactly these situations, and once you can tell them apart you will start seeing them everywhere — in queues, in insurance, in biology, in network traffic.

This page teaches both from the ground up: what problem each was built to solve, the formulas and where they come from, and enough worked arithmetic that you can trust yourself to use them.

Learning Objectives

• State the Poisson probability formula and identify the conditions under which it applies.
• State the geometric probability formula and distinguish "trials until success" from "counts per interval."
• Compute probabilities, means, and variances for both distributions.
• Recognise the Poisson as the limit of the binomial for rare events.
• Choose the correct distribution for a described real-world scenario and avoid the classic mix-ups.

Quick Answer

The Poisson distribution models the number of times a rare, independent event happens in a fixed interval of time, area, or volume, given an average rate $\lambda$. Its probability mass function is $P(X=k)=\dfrac{\lambda^k e^{-\lambda}}{k!}$, and both its mean and variance equal $\lambda$. The geometric distribution models the number of independent trials, each with success probability $p$, needed to get the first success. Its pmf is $P(X=k)=(1-p)^{k-1}p$, with mean $1/p$ and variance $(1-p)/p^2$. Use Poisson when you are counting how many events occur; use geometric when you are counting how many attempts until one occurs.

Where It Came From

The Poisson distribution is named for the French mathematician Siméon Denis Poisson, who published it in 1837 in a work on the probability of judgments in criminal and civil matters. Poisson was studying how many jurors out of many would render a particular verdict — a rare-outcome counting problem — and derived his distribution as an approximation to the binomial when the number of trials is huge but the chance of the event on any single trial is tiny. For decades it was a mathematical curiosity.

What made it famous was a wonderfully grim dataset. In 1898 the Russian statistician Ladislaus Bortkiewicz published Das Gesetz der kleinen Zahlen ("The Law of Small Numbers"). He examined records from 14 corps of the Prussian army over 20 years — 200 corps-years in total — counting how many soldiers were killed each year in each corps by being kicked by a horse. Death by horse-kick is a textbook rare event: each soldier has a minuscule chance of it, but there are many soldiers, and the outcomes are essentially independent. Bortkiewicz found that the yearly counts (0, 1, 2, 3, 4 deaths) matched the Poisson distribution almost perfectly with an average of about $0.61$ deaths per corps per year. The point he made — that even seemingly chaotic rare events follow a precise statistical law — cemented the Poisson distribution as the model for counting rare events.

The geometric distribution has no single celebrated origin story; it grew naturally out of the study of repeated independent trials (Bernoulli trials), formalised by Jacob Bernoulli around 1700. The need it answers is intuitive and ancient: gamblers and insurers alike wanted to know how long they should expect to wait for a first success or first failure. The name comes from the fact that its probabilities $p, (1-p)p, (1-p)^2 p, \dots$ form a geometric sequence.

The Poisson Distribution: Counting Rare Events

The Poisson distribution applies when four conditions hold:

1. Events occur in a fixed interval (time, length, area, volume).
2. Events happen at a constant average rate $\lambda$.
3. Events are independent of one another.
4. Two events cannot occur at exactly the same instant.

The probability of observing exactly $k$ events is

$P(X=k)=\frac{\lambda^k e^{-\lambda}}{k!}, \qquad k=0,1,2,3,\dots$

Here $\lambda$ (lambda) is the expected number of events in the interval. A remarkable feature is that the mean and variance are both equal to $\lambda$:

$E(X)=\lambda, \qquad \operatorname{Var}(X)=\lambda.$

Worked example. A hospital emergency department receives on average $5$ ambulances per hour. Assuming arrivals are Poisson, what is the probability of exactly $3$ ambulances in a given hour?

Here $\lambda=5$ and $k=3$:

$P(X=3)=\frac{5^3 e^{-5}}{3!}=\frac{125 \times e^{-5}}{6}.$

Since $e^{-5}\approx 0.006738$,

$P(X=3)=\frac{125 \times 0.006738}{6}=\frac{0.84225}{6}\approx 0.1404.$

So there is about a 14% chance of exactly three ambulances. Notice the variance is also $5$, so the standard deviation is $\sqrt{5}\approx 2.24$ ambulances.

Scaling the interval. Because the rate is constant, if you change the length of the interval you simply scale $\lambda$. If ambulances arrive at 5 per hour, then over a 30-minute window $\lambda = 2.5$, and the probability of no ambulances in that half hour is $P(X=0)=e^{-2.5}\approx 0.082$.

Poisson as the Limit of the Binomial

The Poisson distribution is not arbitrary — it is what the binomial distribution becomes when trials are many and successes are rare. If $X\sim \text{Binomial}(n,p)$ and you let $n\to\infty$ while keeping $np=\lambda$ fixed (so $p\to 0$), the binomial pmf converges to the Poisson pmf.

Worked check. Suppose a factory produces $n=1000$ items and each is defective with probability $p=0.002$, independently. The exact expected number of defectives is $np=2$. The binomial probability of exactly 2 defectives is $\binom{1000}{2}(0.002)^2(0.998)^{998}\approx 0.2709$. The Poisson approximation with $\lambda=2$ gives

$P(X=2)=\frac{2^2 e^{-2}}{2!}=\frac{4 \times 0.13534}{2}=0.2707.$

The two agree to three decimal places — which is why Poisson is the go-to model whenever $n$ is large and $p$ is small.

The Geometric Distribution: Waiting for Success

The geometric distribution answers a different question: in a sequence of independent trials each with success probability $p$, how many trials until the first success? The number of trials $X$ (counting the successful one) satisfies

$P(X=k)=(1-p)^{k-1}\,p, \qquad k=1,2,3,\dots$

The logic is direct: you need $k-1$ failures (each probability $1-p$) followed by one success (probability $p$). The mean and variance are

$E(X)=\frac{1}{p}, \qquad \operatorname{Var}(X)=\frac{1-p}{p^2}.$

Worked example. You roll a fair die until you get a six. Here $p=\tfrac{1}{6}$. What is the probability the first six appears on the fourth roll?

$P(X=4)=\left(\frac{5}{6}\right)^{3}\times \frac{1}{6}=\frac{125}{216}\times\frac{1}{6}=\frac{125}{1296}\approx 0.0965.$

On average you expect to wait $E(X)=1/p = 6$ rolls for your first six — which matches most people's intuition.

The memoryless property. The geometric distribution has a striking feature: it has no memory. If you have already rolled five times without a six, the probability that you need at least three more rolls is exactly the same as if you were starting fresh. Formally, $P(X > m+n \mid X > m)=P(X > n)$. The die does not "owe" you a six. This is the discrete analogue of the exponential distribution's memorylessness.

A cumulative example. The probability of getting the first success within the first $n$ trials is $P(X \le n)=1-(1-p)^n$. For the die, the probability of at least one six in four rolls is $1-(5/6)^4 = 1 - 0.4823 = 0.5177$, just over half.

Real-World Applications

• Insurance and risk (Poisson): number of claims filed per month, number of accidents at an intersection per year — Bortkiewicz's horse-kicks are the ancestor of modern actuarial claim-count models.
• Telecommunications and computing (Poisson): packets arriving at a router, requests hitting a web server, radioactive decay counts on a Geiger counter.
• Quality control (Poisson): flaws per square metre of fabric, typos per page, contamination particles per litre.
• Reliability and testing (Geometric): number of test attempts until a defective unit is found; number of login attempts until success.
• Marketing and sales (Geometric): number of sales calls until the first "yes"; number of ad impressions until the first click.
• Games and genetics (Geometric): trials until a rare card is drawn, or the number of offspring until a particular trait appears.

Common Mistakes

Mistake 1: Confusing "counting events" with "counting trials." Students often reach for the wrong model. Why it's wrong: Poisson counts how many events occur in an interval; geometric counts how many trials until the first success. Correction: Ask "am I counting occurrences within a fixed window (Poisson) or waiting for the first hit (geometric)?" "How many emails in an hour?" is Poisson; "How many emails until the first spam?" is geometric.

Mistake 2: Forgetting to scale $\lambda$ to the interval. If the rate is 5 per hour and you're asked about 20 minutes, you cannot use $\lambda=5$. Why it's wrong: the Poisson parameter is the expected count for the interval in question. Correction: scale it — $\lambda = 5 \times \tfrac{20}{60} = 1.667$ for a 20-minute window.

Mistake 3: Using $p$ where you need $1-p$ in the geometric formula. Writing $P(X=k)=p^{k-1}(1-p)$ inverts the roles. Why it's wrong: you need $k-1$ failures before the success, and failures carry probability $1-p$. Correction: the exponent $k-1$ always attaches to the failure probability $(1-p)$, and the lone factor of $p$ is the final success.

Comparison and Connections

Both distributions are built from independent Bernoulli-type events, but they answer opposite questions. Poisson relates to the exponential distribution (the waiting time between Poisson events is exponential), while the geometric is the discrete cousin of the exponential. The geometric is also a special case of the negative binomial distribution (waiting for the first success rather than the $r$-th).

Feature	Poisson	Geometric
Question answered	How many events in an interval?	How many trials until first success?
Random variable	Count $k=0,1,2,\dots$	Trial number $k=1,2,3,\dots$
Parameter	Rate $\lambda$	Success probability $p$
Pmf	$\dfrac{\lambda^k e^{-\lambda}}{k!}$	$(1-p)^{k-1}p$
Mean	$\lambda$	$1/p$
Variance	$\lambda$	$(1-p)/p^2$
Key property	Mean = variance	Memoryless
Related to	Binomial (limit), Exponential	Bernoulli, Negative binomial

Practice Questions

Recall

State the pmf, mean, and variance of both the Poisson and geometric distributions.

Answer: Poisson: $P(X=k)=\lambda^k e^{-\lambda}/k!$, mean $=$ variance $=\lambda$. Geometric: $P(X=k)=(1-p)^{k-1}p$, mean $1/p$, variance $(1-p)/p^2$.

Understanding

A bookshop sells on average $3$ copies of a rare title per week. Explain why Poisson is appropriate, and find the probability of selling zero copies in a week.

Guidance: Sales are rare, independent, at a constant rate over a fixed interval — Poisson fits. With $\lambda=3$, $P(X=0)=e^{-3}\approx 0.0498$, about 5%.

Application

A basketball player makes each free throw with probability $p=0.8$. What is the probability her first miss comes on the third attempt?

Answer: A "miss" is the event of interest, so its probability is $0.2$ and a "make" is $0.8$. $P(X=3)=(0.8)^2(0.2)=0.64\times 0.2=0.128$.

Analysis

A website gets $12$ visits per hour (Poisson). Compare the probability of exactly 12 visits in an hour with the probability of exactly 6 visits in a half hour. What do you notice about relative spread?

Guidance: For one hour, $\lambda=12$, $P(X=12)=12^{12}e^{-12}/12!\approx 0.114$. For a half hour, $\lambda=6$, $P(X=6)=6^6 e^{-6}/6!\approx 0.161$. The half-hour case peaks higher because a smaller mean has a relatively larger standard deviation ($\sqrt{6}/6\approx 0.41$ versus $\sqrt{12}/12\approx 0.29$) — smaller counts are proportionally more variable.

FAQ

Can the Poisson variable be larger than $\lambda$? Yes. $\lambda$ is only the average. Any non-negative integer is possible; values well above $\lambda$ just have small probability. With $\lambda = 3$ you can still observe 8 events in an interval.

Why do the mean and variance of the Poisson both equal $\lambda$? It falls out of the mathematics of the limit from the binomial. Practically, it gives you a quick sanity check: if real data have a variance much bigger than the mean, the events probably aren't independent (this is called overdispersion), and Poisson may be the wrong model.

Is there a version of the geometric that counts failures instead of trials? Yes. Some textbooks define $Y$ = number of failures before the first success, giving $P(Y=k)=(1-p)^k p$ for $k=0,1,2,\dots$ with mean $(1-p)/p$. It's the same idea shifted by one; always check which convention a problem uses.

When should I use Poisson instead of the binomial? Use Poisson when $n$ is large and $p$ is small (a common rule of thumb is $n\ge 20$ and $p\le 0.05$, or simply when you know the average rate but not $n$ and $p$ separately). It's easier and an excellent approximation.

What does "memoryless" really mean for the geometric distribution? It means past failures give you no information about how much longer you'll wait. After ten failed attempts, the expected additional number of trials is still $1/p$. This surprises people, but it follows directly because the trials are independent.

Quick Revision

• Poisson pmf: $P(X=k)=\lambda^k e^{-\lambda}/k!$; mean $=$ variance $=\lambda$.
• Use Poisson for counts of rare, independent events in a fixed interval; scale $\lambda$ with the interval length.
• Poisson is the large-$n$, small-$p$ limit of the binomial with $\lambda = np$.
• Geometric pmf: $P(X=k)=(1-p)^{k-1}p$; mean $1/p$, variance $(1-p)/p^2$.
• Use geometric for trials until the first success; it is memoryless.
• $P(X \le n)=1-(1-p)^n$ gives "at least one success within $n$ trials."
• History: Bortkiewicz's Prussian horse-kick deaths (200 corps-years, $\lambda\approx 0.61$) are the classic Poisson fit.

Related Topics

Prerequisites

• Probability Distributions overview
• Binomial distribution and Bernoulli trials
• The exponential function $e^x$

Related Topics

• Binomial and Bernoulli distributions
• Exponential distribution (continuous waiting times)
• Expected value and variance

Next Topics

• Negative binomial distribution
• Continuous distributions and the normal approximation
• Statistics and Probability branch