Poisson Distribution Calculator

What this calculator does

Find the chance that something happens exactly k times, or at most, at least, fewer or more than k times. All you need is how often it happens on average. That's what the Poisson distribution describes. It counts events in a fixed stretch of time or space, when they happen on their own at a steady average rate. Think of calls reaching a help desk in an hour, typos on a page or cars passing a sensor in a minute. It needs just one number, λ (lambda), the average count for that stretch.

Enter λ and the count k you care about. The calculator gives the chance of exactly k events, and of at most, fewer than, at least and more than k. It also shows the mean (average), the variance and standard deviation (two measures of spread), the most likely count and a table of every likely count. The working puts your numbers into e^(−λ) λ^k ÷ k!, so you can follow the arithmetic.

Where it's used

Why this math matters outside the classroom:

• Call center staffing. A help desk averaging 12 calls an hour finds the chance of more than 20 calls in an hour to decide how many people to schedule.
• Website traffic. An engineer whose site averages 2.5 sign-ups a minute checks how unusual a minute with 9 sign-ups is before suspecting bots.
• Quality control. A printer averaging 0.3 defects per page uses 1 − e^(−0.3) ≈ 0.259 as the chance a page has at least one defect.
• Traffic counts. A planner expecting 4 cars a minute at a rural junction finds a 1.83% chance a given minute has no cars at all.
• Radioactive decay. A physics student models a detector averaging 5 clicks a second as Poisson and compares observed counts with the predicted table.

How to use it

1. Enter λ, the average number of events in the stretch you're asking about. If the rate is given for a different stretch, scale it first: 12 per hour means λ = 4 for 20 minutes.
2. Enter k, the number of events, as a whole number 0 or more.
3. Read P(X = k), the chance of exactly k, first. Then read the totals that match your question: 'at most' is P(X ≤ k), and 'at least' is P(X ≥ k).
4. Open the table to see how the chances spread across every likely count.

What the inputs mean

Average number of events in the interval (λ)

λ doesn't have to be a whole number. An average of 2.5 accidents per month is a perfectly good λ, even though each month has a whole number of accidents.

Number of events (k)

k is a count, so it must be a whole number. A very large k, far from λ, gives chances so small they're shown in scientific notation.

Formula and method

P(X = k) = e^(−λ) × λ^k ÷ k!
P(X ≤ k) = P(X = 0) + P(X = 1) + … + P(X = k)
P(X > k) = 1 − P(X ≤ k),  P(X ≥ k) = 1 − P(X ≤ k − 1)
mean = variance = λ,  standard deviation = √λ

Here's what the symbols mean. X is the number of events that actually happens, and P(…) is the chance of what's in the brackets. λ is the average count and k is the count you're asking about. e is a math constant, about 2.718. k! ("k factorial") means 1 × 2 × 3 × … × k, with 0! = 1.

The formula comes from slicing the stretch into a huge number of tiny pieces. Each piece has a tiny chance of one event, and the number of pieces keeps growing. So the Poisson distribution is the limit of the binomial distribution, with many trials and a small chance of success each time. Throughout, the average n × p (trials times chance) stays fixed at λ. The factor e^(−λ) is the chance of no events at all, and λ^k ÷ k! accounts for k of them.

The at-most totals add up the chances of each single count. For small k, the working shows the sum term by term. For larger k, it uses the regularized incomplete gamma function. That gives the same total without adding thousands of tiny numbers. The more-than and at-least chances are worked out directly, so they stay accurate even when they're very small.

Because the mean and the variance both equal λ, the spread grows with the rate. A count that averages 100 typically varies by about √100 = 10 either way.

Worked example

A bakery sells an average of 4 of its specialty cakes a day. Assume each sale happens independently. What's the chance it sells exactly 2 tomorrow, and the chance it sells more than 2?

1. Here λ = 4 and k = 2.
2. P(X = 2) = e^(−4) × 4² ÷ 2! = 0.0183156389 × 16 ÷ 2 = 0.146525.
3. P(X ≤ 2) = P(0) + P(1) + P(2) = 0.018316 + 0.073263 + 0.146525 = 0.238103.
4. P(X > 2) = 1 − 0.238103 = 0.761897.

Result: There's about a 14.65% chance of selling exactly 2 cakes, and a 76.19% chance of selling more than 2. The most likely sales are 3 and 4 cakes, which are equally likely. The standard deviation is √4 = 2.

Understanding the result

When λ is a whole number, the counts λ − 1 and λ are equally likely and tie for most likely. Otherwise, the most likely count is λ rounded down. For small λ the distribution is lopsided, with a long tail to the right (skewed right). As λ grows, it looks more and more like a bell curve.

Did the count you saw have a very small chance? That's a sign the rate may have changed, or the events may not be independent. For example, calls might arrive in bursts after an outage.

Assumptions and limitations

• Events happen one at a time, and each one doesn't affect the others.
• The average rate stays steady over the stretch, and λ is the average count for exactly the stretch you're asking about.
• Some counts cluster, where one event makes another more likely. Others have a hard upper limit. Other distributions describe those better, such as the binomial when there's a fixed number of trials.
• The table is shown only when the likely counts and k fit in about 100 rows.

Frequently asked questions

How this calculator is checked

Method: Poisson probability mass function e^(−λ)λ^k/k! evaluated in logarithms; cumulative probabilities from the regularized incomplete gamma function. Checked against hand values for small λ and a direct summation for λ = 100. The calculation is covered by automated tests with known reference values and edge cases. Last updated 2026-10-18. Found a problem? Tell us.

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