Poisson Distribution Calculator
Calculate Poisson probabilities for a given rate and count, with exact, cumulative and tail probabilities, and a full distribution table.
How to use this calculator
- 1Enter λ, the average number of events per interval, matching the interval length you care about.
- 2Enter k, the specific count you want the probability of.
- 3Read the exact probability, then the cumulative and tail probabilities beneath it.
How the calculation works
P(X = k) = (λ^k · e^(−λ)) ÷ k! mean = variance = λ SD = √λ- λ
- The average number of events per interval — the rate. Must be positive but need not be a whole number
- k
- The exact number of events whose probability you want. A non-negative integer
- e
- Euler's number, about 2.71828
- k!
- Factorial of k: k × (k−1) × … × 1, with 0! defined as 1
Mean and variance are both λ. No other common distribution ties them together like this, and it gives a quick diagnostic: if your data's variance greatly exceeds its mean, the process is overdispersed and Poisson is the wrong model.
P(X = 0) simplifies to e^(−λ), because λ⁰ = 1 and 0! = 1. That single term answers "what is the chance of a completely quiet interval?".
This calculator evaluates the formula in log space — ln P = k·ln λ − λ − ln(k!) — and exponentiates at the end. Computing λᵏ and k! directly overflows to Infinity well before λ = 200, returning NaN for probabilities that are perfectly ordinary numbers.
λ must match the interval you are asking about. A rate of 3 per hour becomes λ = 1.5 for a half-hour window; using the hourly figure for a half-hour question doubles the answer.
Worked example
A call centre averaging 3 calls an hour
- 1.λ = 3 calls per hour, and we want the probability of exactly 2.
- 2.P(X = 2) = 3² × e⁻³ ÷ 2! = 9 × 0.0497871 ÷ 2.
- 3.= 0.4480836 ÷ 2 = 0.2240418, so about 22.40%.
- 4.Cumulative: P(X ≤ 2) = P(0) + P(1) + P(2) = e⁻³(1 + 3 + 4.5) = 0.0497871 × 8.5 = 42.32%.
- 5.So there is a 57.68% chance of 3 or more calls in a given hour.
- 6.Standard deviation is √3 ≈ 1.732, so counts typically land between about 1 and 5.
Result: P(X = 2) = 22.40%
The chance of a completely quiet interval
- 1.A machine averages 0.5 breakdowns per week, so λ = 0.5.
- 2.P(X = 0) = 0.5⁰ × e^(−0.5) ÷ 0! = 1 × 0.6065307 ÷ 1.
- 3.So there is a 60.65% chance of a completely breakdown-free week.
- 4.This is the special case where the formula collapses to just e^(−λ).
- 5.The chance of at least one breakdown is the complement: 1 − 0.6065 = 39.35%.
- 6.Note that with λ below 1, zero is the single most likely outcome.
Result: P(X = 0) = 60.65%
When to use Poisson instead of binomial
Both distributions count things, and the choice between them turns on one question: is there a fixed number of trials? The binomial needs one — twenty coin flips, fifty inspected parts — and asks how many succeeded. The Poisson has no such number.
Ask how many customers walked into a shop this hour and there is no meaningful count of how many *could* have. The opportunity is continuous rather than a fixed set of trials. That is the Poisson's territory: events landing in an interval of time, length, area or volume at some average rate.
The two are closely related. A binomial with many trials and a small success probability converges on a Poisson with λ = np, which is why the Poisson is sometimes introduced as the "law of rare events". If n is above about 50 and p below about 0.1, the Poisson is an excellent and much easier approximation.
Mean equals variance, and why that is a useful test
The Poisson is unusual in having exactly one parameter. Fix λ and you have fixed everything — the mean is λ, the variance is λ, and the standard deviation is √λ. A normal distribution needs two parameters; a Poisson needs one.
That rigidity is diagnostically useful. Take your real data, compute its mean and its variance, and compare. If the variance is far larger than the mean, the data is overdispersed and something is violating the model's assumptions — usually events arriving in clusters rather than independently.
It also means busier processes are relatively more predictable. At λ = 4 the standard deviation is 2, half the mean. At λ = 400 it is 20, a twentieth of the mean. Absolute variability grows, but relative variability shrinks with √λ, which is why high-traffic systems are easier to forecast than low-traffic ones.
The assumptions, and how real data breaks them
The Poisson assumes events occur independently, at a constant average rate, and that two cannot occur at exactly the same instant. Real processes routinely break all three.
Independence fails when events cluster. One server failure causes a cascade; one customer arrives with three friends. Clustered data has variance well above its mean, and a Poisson model will badly understate the chance of a very busy interval — which matters, because those are the intervals that cause outages.
The constant-rate assumption fails whenever there is seasonality. A restaurant does not average the same arrivals at 3pm and 7pm, so a single λ across the whole day is a fiction that describes neither. The fix is to model each period separately with its own λ rather than to average.
The standard escape hatch when overdispersion is real is the negative binomial distribution, which adds a second parameter precisely so variance can exceed the mean.
What this assumes, and where it stops
Assumptions
- Events occur independently of one another.
- The average rate λ is constant across the interval being modelled.
- Two events cannot occur at exactly the same instant.
- λ is stated for the same interval length as the question being asked.
Limitations
- λ and k are capped at 1,000. Above that the distribution is very close to a normal with mean λ and standard deviation √λ, which is the better tool.
- Assumes a constant rate. Data with seasonality or time-of-day patterns needs a separate λ per period, not an overall average.
- Clustered events break the independence assumption and produce variance above the mean. Where that happens, a negative binomial distribution fits better.
- Gives probabilities from a rate you supply; it does not estimate λ from data or test whether your data is genuinely Poisson.
Common questions
When should I use the Poisson distribution?
When counting how many times something happens in an interval of time, space or volume, where events are independent and there is no fixed number of trials — calls per hour, defects per metre, arrivals per minute. If there is a fixed number of trials with a success probability for each, use the binomial distribution instead.
What is lambda in the Poisson distribution?
The average number of events per interval, and the distribution's only parameter. It is simultaneously the mean and the variance. It does not have to be a whole number — a rate of 2.7 defects per batch is perfectly valid — but it must refer to the same interval length as your question.
Can the Poisson distribution have a non-integer mean?
Yes. λ is a rate and can be any positive number, such as 0.5 breakdowns per week. What must be a whole number is k, the count of events, since you cannot observe half an event. The mean of the distribution is genuinely 0.5 even though no single observation can be.
Why do mean and variance both equal lambda?
It follows from the distribution's derivation as a limit of the binomial: variance np(1−p) tends to np as p tends to zero, and np is λ, which is also the mean. Practically it gives a quick model check — if your data's variance far exceeds its mean, the events are probably clustered and Poisson is the wrong choice.
Sources
Formula and content last reviewed on .
Results are estimates for information only, not professional advice.
Related calculators
Tools people commonly use alongside the poisson distribution calculator.