Poisson Distribution Calculator

P(X = k) = e^(−λ) · λ^k / k! with λ = eventRate × intervalLength, and Var(X) = λ

λ

The expected number of events in the interval — the rate multiplied by the length of the window, and the only parameter of the distribution. Everything on this page is a function of it, which is why a rate and an interval are asked for separately rather than as one number

k

The number of events you are asking about. It is a whole count from 0 upwards, and 0 is a real answer rather than an empty one: "none happened" is the most likely outcome whenever lambda is below about 0.7

e^(−λ)

The weight of nothing happening at all, before the count is worked in. It is the reason the whole distribution shrinks as lambda grows: the larger the expected count, the smaller the chance of seeing none

λ^k / k!

The part that grows with k and then falls: the k-th power of the mean divided by the factorial of the count. Its peak sits at k = λ, and that peak is where the distribution spends most of its mass

mean = variance = λ

The property that makes this distribution recognisable. The mean count and its variance are the same number, so the standard deviation is the square root of the mean — √4 = 2 for a mean of four events, √1000 ≈ 31.6 for a mean of a thousand

Use it when you are counting how many counts arrive per interval and the arrivals are independent of one another: how many calls in an hour, how many defects in a square metre of fabric, how many particles in a fixed counting window. Two conditions carry most of the weight. The events must be rare relative to the opportunity for them — a rate so high that the count is essentially capped by something else stops being Poisson — and they must not cluster, because a burst of arrivals from one cause breaks the independence that gives the distribution its shape. Reach for the binomial instead when the number of opportunities is fixed and countable, since a Poisson count has no upper bound while a binomial count cannot exceed its number of trials; reach for a normal approximation once lambda is large, where the two agree closely and the familiar standard deviation bands become readable. The mean equal to the variance is also the fastest diagnostic available: divide the variance by the mean of a set of counts and if the ratio is far from one, the process is over- or under-dispersed and this model is the wrong shape for it.

1. Four events per hour, asking about three

   1. The rate is 4 per hour over an interval of 1 hour, so lambda is 4
   2. P(X = 3) = e⁻⁴ × 4³ / 3! = 0.018316 × 64 / 6 = 0.1954, so 19.54%
   3. Adding the counts 0 through 3 gives 43.35% at most; the rest of the mass above 3 gives 76.19% at least
   4. The mean is lambda = 4 and so is the variance, which makes the standard deviation √4 = 2

   The two cumulative rows overlap here and that is not a mistake: 43.35% plus 76.19% comes to 119.54%, and the excess over 100% is exactly the 19.54% of three events, which sits in both rows. It is the same caution as with a binomial count, and it catches the common error of computing at least as one minus at most, which would say 56.65% here. Note also that three events is unremarkable when the average is four: it is half a standard deviation below the mean.

2. The centre of the distribution

   1. Asking about four events instead of three: P(X = 4) = e⁻⁴ × 4⁴ / 4! = 0.018316 × 256 / 24 = 0.1954
   2. The exact chance is the same 19.54% as for three events, because the peak of the distribution sits at k = 4 = lambda and the two counts either side of it are equally likely
   3. At most four is 62.88% and at least four is 56.65%
   4. The mean, variance and standard deviation are unchanged, since they depend only on lambda

   Two counts with the same probability is the clearest visible sign that the distribution is centred on lambda: with a mean of four events, seeing three and seeing four are equally likely, and every other count is less so. The accumulated rows are not symmetric even here — 62.88% against 56.65% — because the at-most row includes the whole of k = 4 while the at-least row includes it too, and below the peak the distribution is thinner than above it.

3. The same rate over three hours

   1. The rate is still 4 per hour, but the interval is now 3 hours, so lambda = 4 × 3 = 12
   2. P(X = 3) = e⁻¹² × 12³ / 3! = 0.0000061 × 1728 / 6 = 0.0018, so 0.18%
   3. At most three is 0.23% and at least three is 99.95%
   4. The mean is 12 and so is the variance, making the standard deviation √12 = 3.46

   This is the example that separates the two input fields: nothing changed except the length of the window, and the answer moved from 19.54% to 0.18%. Both fields are needed because the parameter is their product, and a page that took only the rate would give the same answer for an hour and for a day. The summary rows move with it — with twelve events expected, three is more than two standard deviations below the mean, which is why the at-least row is now nearly certain.

4. One event per hour, and nothing happens

   1. With lambda = 1, the chance of exactly zero events is e⁻¹ = 0.3679, so 36.79%
   2. Asking for at most zero asks the same question, so that row repeats the same number
   3. At least zero covers every possible count, so it is exactly 100% — not 99.99%
   4. The mean and variance are both 1, and the standard deviation is 1

   Lambda of one is the textbook starting point and it produces the most counter-intuitive number on the page: more than a third of the time, a process averaging one event per hour produces no events at all. The at-least row is worth reading as well — it is 100% exactly, because "zero or more" is the whole distribution, and a page that computed it by subtracting from a cumulative sum could easily return 99.99% instead. Zero events is also a legitimate answer to have asked for, which is why the count field starts at zero rather than at one.

The Poisson model rests on two assumptions that fail quietly, and both are worth checking before trusting the numbers. The events must be independent of one another, and they must arrive at a rate that does not drift within the interval you specify. Independence fails whenever arrivals share a cause — calls surge during an outage, defects cluster on one roll of fabric, accidents cluster in bad weather — and the symptom is that the real counts are more spread out than the page predicts, with more empty intervals and more very busy ones than the model allows. A drifting rate fails the other way: a call centre with a quiet night and a busy morning does not have one hourly rate, and averaging the two produces a number that describes neither. Beyond the assumptions, two limits are enforced rather than explained. The interval length and the rate may each be entered up to reasonable bounds, but their product is capped, because a mean count in the millions makes every probability on the panel round to zero and the page would rather decline than print a table of zeroes. The count itself must be a whole number. There is also no reference table of Poisson probabilities here: the table people want is one row per value of the mean, and the mean depends on the rate and interval you entered, so a printed table would answer a different question from the panel above it.

What is the difference between the Poisson and the binomial distribution?

The binomial counts successes in a fixed number of trials, so its count cannot exceed that number; the Poisson counts events that arrive on their own, so there is no upper bound and no trial count to enter. The binomial needs two inputs — how many trials and the chance of each — while this page needs a rate and an interval, which combine into the single expected count. The two agree in the narrow case where the number of trials is large and the chance of each is small, which is why rare events are described either way in practice. Use the binomial when you can name the opportunities, and the Poisson when events simply arrive.

Why does the mean equal the variance here?

Because that identity is what the model is built on rather than a coincidence of the examples: for a Poisson count the mean and the variance are the same number, lambda, so the standard deviation is always the square root of the mean. It gives a quick way to test whether the model fits: take a set of observed counts, divide their variance by their mean, and if the ratio is much above one the counts are clustered rather than independent, while a ratio well below one means they are more even than chance would make them. Neither case is fixed by entering different numbers — it means the process is not Poisson.

Why are at most k and at least k not complements?

Because both rows include the count of exactly k, so adding them counts that outcome twice. On the example above, at most three is 43.35% and at least three is 76.19%, which sums to 119.54% — the excess over 100% is exactly the 19.54% of three events. This is the same arithmetic as with a binomial count and the same trap: reading at least as one minus at most is wrong by the probability you asked about. If you want a genuine complement, use the at most k − 1 figure, or read the three rows together and check that exactly k fills the gap between them.

When is a count actually Poisson?

When the events are independent, the rate is steady over the interval, and the events are rare relative to the opportunities for them. Independence is the assumption that does the work and the one that fails most often, because clustering is easy to miss from summary numbers: a factory that reports the same monthly defect total every month may still be having bad days and quiet days, and that clustering will show up as counts more variable than this page predicts. The practical check is the one in the previous answer — compare the variance of a set of counts against their mean. If they are close, the model is a good description; if they are far apart, no choice of rate will fix it.

What happens when the rate is zero?

Nothing can occur in the interval, so the count of zero has probability 100% and every other count has probability 0%. The page reports that rather than failing: a rate of zero, or an interval of zero length, both give lambda of zero with no events expected, and asking about one or more events returns 0% while asking about none returns 100%. It is worth knowing this is handled explicitly, because the formula as written contains a factor of zero to the power of k and a negative infinity from the logarithm, and an implementation that evaluates it naively produces a meaningless result instead of the obvious answer.

Why is there no table of Poisson probabilities on this page?

Because every row of such a table belongs to a different value of the mean, and the mean here is the product of the two numbers you enter. A printed table would be right for one rate and one interval and wrong for every other combination, so it would answer a different question from the panel — and where the two disagreed, the panel would be the correct one. The panel is the table: the three probability rows are the exactly, at-most and at-least figures for the mean you specified, and changing the rate or the interval recomputes them. Reference tables earn their place on pages whose rows do not depend on any input, such as a grading scale or a set of cut-off bands.

• 4.3 Binomial Distribution — Introductory Statistics 2e (the fixed number of independent trials with a constant success probability, which is precisely the constraint a Poisson count does not have) — OpenStax, Rice University
• 1.3.6.6.1. Normal Distribution — e-Handbook of Statistical Methods (the density, the role of the mean and standard deviation, and the standard normal as the reference case a large-mean Poisson count approaches) — National Institute of Standards and Technology (NIST)
• 1.3.6.1. What is a Probability Distribution — e-Handbook of Statistical Methods (what it means for a count to have a distribution, and how probabilities are read off the outcomes it assigns them to) — National Institute of Standards and Technology (NIST)