Poisson Distribution Calculator

What the Poisson distribution counts

The Poisson distribution models how many times something happens in a fixed window: calls per hour, typos per page, defects per batch, goals per match. It needs one number, the average rate, written as lambda (λ). If a help desk gets 4 calls per hour on average, lambda is 4, and the distribution tells you the chance of getting 0, 1, 2, or any other count in the next hour. NIST/SEMATECH defines it with the formula P(X = k) = e^(-λ) λ^k / k!, where e is the constant 2.71828 and k! is the factorial of k, in its Poisson distribution reference.

Three conditions make the model a good fit. Events happen at a constant average rate, one event does not change the chance of the next, and two events cannot happen at the same instant. Traffic that bunches up at rush hour breaks the constant-rate condition, so a single Poisson model over a whole day would mislead. Split the day into windows that each have a steady rate, and model each window with its own lambda.

One handy fact: the mean and the variance are both lambda. A rate of 4 per hour has a mean of 4 and a standard deviation of the square root of 4, which is 2. That is why counts with a small lambda look lumpy and skewed, while counts with a large lambda look close to a bell curve.

Entering values in the Poisson calculator

Enter the average rate lambda and the count k you care about, and the calculator fills in five probabilities at once. P(X = k) is the chance of landing on that exact count. P(X ≤ k) and P(X < k) are the chances of landing at or below it, and P(X ≥ k) and P(X > k) cover the other side. The at-or-below numbers matter in practice: a queue designed for 3 arrivals per minute wants P(X ≤ 3), not P(X = 3).

Use a rate measured over the same window as the question. If lambda comes from hourly data but the question is about 15 minutes, divide lambda by 4 first. If it comes from monthly data and the question is about a week, multiply by 7 divided by the average days per month. Getting the window wrong is the most common input mistake, and it scales the answers in a way that no single ratio describes.

The calculator rounds k down to a whole number, because counts are whole. Enter a rate above 0 and a count of 0 or more. Results are percentages rounded to 4 decimal places, so tiny tail probabilities stay visible.

Poisson examples worked out in full

Example 1: a small call center gets 4 calls per hour on average. With lambda = 4, the chance of a silent hour, P(X = 0), is e^(-4), which is 1.8316%. The chance of 2 or fewer calls, P(X ≤ 2), is e^(-4) (1 + 4 + 8), which is 23.8103%. The chance of a busy stretch, P(X ≥ 5), works out to 37.1163%, because 1 - P(X ≤ 4) = 1 - 0.6288 = 0.3712. More than a third of hours see 5 or more calls even though 4 is the average.

Example 2: a server logs 1.2 crashes per month on average. With lambda = 1.2, P(X = 0) is e^(-1.2), which is 30.1194%, so a clean month happens about 3 months in 10. P(X ≤ 1) is e^(-1.2) (1 + 1.2), which is 66.2626%, which is the chance you face at most one crash.

Check any example by hand with the formula. For lambda = 4 and k = 2: e^(-4) works out to 0.018316, times 4^2 = 16 gives 0.29305, divided by 2! = 2 gives 0.146525, or 14.6525%. The calculator returns the same number for P(X = 2).

Poisson versus binomial

The binomial distribution counts successes in a fixed number of trials, n. The Poisson counts events in a fixed window with no upper limit on the count. They connect: when n is large and the success chance p is small, a binomial with mean n × p behaves like a Poisson with lambda = n × p. A rare defect across 5,000 units with p = 0.0004 has lambda = 2, and Poisson answers will be within a rounding error of the binomial ones.

Use binomial when you can name the trials: 20 free throws, 50 coin flips, 100 surveyed customers. Use Poisson when events occur over a continuous stretch and the count could run arbitrarily high: arrivals, calls, breakdowns, typos. If you catch yourself asking "how many trials were there?", you probably want the binomial. If you catch yourself asking "how much time passed?", you probably want the Poisson.

For large lambda the Poisson also drifts toward the normal distribution, with mean lambda and standard deviation sqrt(lambda). Past lambda of about 20 or 30 the normal approximation is close enough for planning, but for small lambda the Poisson stays visibly skewed, with a long right tail, so the normal shortcut fails exactly where the interesting tail risk lives.

Common mistakes and assumptions

The first mistake is stacking the window wrong. A rate of 30 calls per 8 hours is not a lambda of 30 for the one-hour question; it is 3.75 per hour, assuming the rate is even across the shift. Recompute lambda for the window in the question before entering it.

The second mistake is using Poisson for data with clumping or variation in the rate. Emergency room arrivals spike on weekend nights, and manufacturing defects cluster when a machine drifts out of adjustment. When the variance of observed counts is much larger than the mean, the model is a poor fit and the tail probabilities will be too small. Statisticians switch to a negative binomial or a mixed model in those cases; Penn State's STAT 414 course notes cover the derivation and its limits.

The third mistake is reading P(X = k) as if it were cumulative. A chance of 9.16% for exactly 5 orders says nothing about 6, 7, or 8 orders. When a decision depends on capacity or budget, use the at-least or at-most probabilities, and add a margin rather than planning for the average alone.

Where the Poisson distribution shows up

Call centers and hospitals use it to staff for arrival counts. Quality engineers count defects per unit area of fabric or per solder joint batch. Web teams model requests per second and use the upper tail to size auto-scaling thresholds. Insurers use it for claim counts per policy period. Sports analysts model goals per match. Radioactive decay counters, which fire independently at a steady rate, follow it to a precision that makes it a physics lab standard.

The common thread is independent events at a steady average rate, counted in whole numbers. If your situation has that shape, enter the rate and the count above and you will get the same five probabilities that a textbook would give, computed locally in your browser.

Reading the shape of the distribution

The five numbers the calculator returns make more sense once you have seen the whole distribution at once. The table below lists every exact count probability for lambda = 4, and the first figure above draws the same list as bars. Together they show why a question like "how likely is exactly 4?" hides most of the story: the single most likely counts, 3 and 4, each carry under 20%, so no single outcome is ever a safe bet.

Two habits fall out of the picture. First, the mass spreads over several counts on either side of lambda, which is why planning with the average alone undersizes capacity about half the time. Second, the right tail thins out slowly: at lambda = 4 the chance of 10 or more events is still 0.84%, small per window but not negligible across thousands of windows. The second figure puts two rates side by side so you can see how a bigger lambda both shifts the peak right and stretches the spread.