Poisson vs Binomial: Which One Should Count Your Events?

Poisson vs binomial distribution compared on 10 points with worked numbers, where each one fails, and how the exponential covers the time between events.

What is the chance that a batch of 100 items, each defective with probability 2%, comes through with no defect at all? Two well-known models answer at once and do not quite agree. The binomial distribution says 13.26%. The Poisson distribution, with an average of 100 × 0.02 = 2 defects, says 13.53%. A quarter of a percentage point. Does it matter which one you use?

Here, hardly. But move the same question to a basketball court or a help line and one of the two models either gives a clearly wrong answer or cannot be used at all. This article puts the Poisson and the binomial side by side on ten points, runs both through the Poisson Distribution Calculator and the Binomial Distribution Calculator, and then adds the third piece people usually forget: the Exponential Distribution Calculator, which answers the question "how long until the next one?".

Poisson vs binomial distribution, side by side

The table is the center of the comparison. Every number in it comes from the two calculators; the rest of the article comments on its rows. The definitions follow the NIST/SEMATECH e-Handbook of Statistical Methods: the binomial distribution describes x successes in n trials when each trial has "exactly two mutually exclusive outcomes" and a fixed p, while the Poisson distribution models "the number of events occurring within a given time interval" from the average λ alone.

	Binomial	Poisson
What you need to know	number of trials n and the chance p of success in each	only the average count λ per period or area
Largest possible count	n, never more	no ceiling
Mean	n × p	λ
Variance	n × p × (1 - p), always below the mean	λ, equal to the mean
Batch of 100 at 2%: zero defects	13.26%	13.53% (λ = 2)
Batch of 100 at 2%: exactly 2	27.34%	27.07%
10 free throws at 75%: more than 10 made	0%, impossible	13.78% (λ = 7.5)
Help line, 4 calls an hour	cannot be used: no trials to count	exactly 4 calls: 19.54%
Excel function	BINOM.DIST(k, n, p, cumulative)	POISSON.DIST(k, mean, cumulative)
Time between events	counted in trials	exponential distribution with mean 1/λ

Two rows decide most real choices. The first one, what you need to know: the binomial needs a count of attempts, the Poisson only an average. And the row with the free throws, which shows what happens when a model has no ceiling and the data do.

When the binomial distribution wins

A fixed number of attempts with a large chance each

A player who makes 75% of free throws takes 10 shots. The chance of making at least 8 is 52.56% under the binomial distribution. Plug the same average into a Poisson, λ = 10 × 0.75 = 7.5, and it says 47.54%. Worse, the Poisson puts 13.78% on making more than 10 baskets out of 10 attempts. The binomial knows that the count stops at n; the Poisson has no idea there were only 10 shots.

The variance tells the same story in one line. For the binomial it is n × p × (1 - p) = 10 × 0.75 × 0.25 = 1.875, for the Poisson it would be 7.5, four times wider. When p is large, the Poisson spreads probability over counts that cannot happen.

Small numbers where the exact fraction matters

Six rolls of a die, at least one six. The binomial answer is exactly 31,031/46,656, about 66.51%. The Poisson with λ = 1 gives 1 - e^-1 = 63.21%, three points short. On a homework sheet or an exam key, only the first one is accepted, and the calculator returns it as a fraction you can copy.

Binomial Distribution Calculator: exactly 2 sixes in 10 rolls of a die, 29.07%, with the exact fraction and the formula filled in

Guessing on a quiz

Ten questions with four options each, pure guessing, at least 5 right. Binomial: 7.813%, or 40,961/524,288, about one attempt in 13. The Poisson with λ = 2.5 says 10.88%, which overstates the luck of a guesser by more than a third. Here the trials are known, p is not small, and the binomial is simply the right model.

When the Poisson distribution wins

There are no trials to count

A help line takes 4 calls an hour on average. How many "trials" is that? Every second in which a call could start? Every person who might call? There is no sensible n, and that is exactly the Poisson's home ground. Exactly 4 calls in the next hour: 19.54%. At least 6: 21.49%. The only input is the average.

Poisson Distribution Calculator: exactly 6 calls in an hour at an average of 4, 10.42%, with the distribution table

Rates that scale with the period

The same help line over an 8-hour shift has λ = 4 × 8 = 32. A shift with 40 or more calls happens 9.560% of the time, about one shift in ten, which is the kind of number you staff for. The Poisson calculator takes the average per hour and the number of hours and multiplies them for you, so the period of the average never gets mixed up with the period of the question.

Huge n, tiny p, and too many digits

One part in a thousand is faulty and a run has 1,000 parts. The binomial and the Poisson give 36.77% and 36.79% for a run with no faulty part. With numbers like these the binomial is correct but heavy, and the Poisson gets the same answer from one input. The NIST e-Handbook gives the rule of thumb for swapping one for the other: n of at least 20 and p of at most 0.05, with an excellent approximation when n is 100 or more and n × p stays at or below 10. Its own check, n = 200 and p = 0.025, gives 0.1399995 against 0.1403739 for exactly 3.

How close is the shortcut? Seven cases checked

The table below runs the same question through both calculators. The first three rows sit inside the NIST rule and the two models agree. The last four break it, with few trials and a p well above 0.05, and the Poisson drifts by three to five percentage points, in both directions.

Situation	Binomial (exact)	Poisson with λ = np	Verdict
n = 1,000, p = 0.001, zero events	36.77%	36.79%	interchangeable
n = 200, p = 0.025, exactly 3 (NIST)	0.1399995	0.1403739	interchangeable
n = 100, p = 0.02, exactly 2	27.34%	27.07%	close enough
n = 6, p = 1/6, at least one	66.51%	63.21%	Poisson 3.3 points low
n = 10, p = 0.3, exactly 3	26.68%	22.40%	Poisson 4.3 points low
n = 10, p = 1/4, at least 5	7.813%	10.88%	Poisson 3.1 points high
n = 10, p = 0.75, at least 8	52.56%	47.54%	Poisson wrong, and allows 11+

The pattern is easy to remember. Poisson is the binomial with the ceiling taken away and the variance stretched to the mean. When n is large and p small, neither change matters. When p is large, both do.

The third model: time between events

The Poisson counts events in a period. Turn the question around and ask how long until the next event, and the answer is the exponential distribution. NIST puts it in one sentence in its reliability chapter: the exponential model "works well for inter arrival times (while the Poisson distribution describes the total number of events in a given period)". Same process, two views of it.

The two calculators should therefore agree, and they do. At 4 calls an hour the average gap is 15 minutes. The chance of 30 minutes without a call is, in the Poisson view, zero events at λ = 2: 13.53%. In the exponential view it is a wait longer than 30 minutes at a mean of 15: e^-2 = 13.53%. At least one call within 10 minutes: 48.66% both ways.

Exponential Distribution Calculator: the next event within 10 minutes at a mean of 15 minutes, 48.66%, with the waiting-time table

Where the exponential earns its place is the planning question. Half of all gaps end before 10.4 minutes, but to be 90% sure the next call has come you have to allow 34.54 minutes, and 44.94 for 95%. The average of 15 minutes hides both of those numbers.

Where all three models fail. They assume independent events at a steady rate. Calls that arrive in bursts after an outage, defects that come in whole bad batches, or a help line busier at 9 a.m. than at 2 p.m. all break that assumption. A quick check on your own counts: if the variance is far above the mean, split the data into calmer periods before using any of these tools.

The verdict: which one should count your events?

There is no single winner, but there is a short decision rule, and it starts with one question: can you count the attempts? If yes, and each attempt has the same chance, the binomial is exact and should be the default. If all you have is a rate per hour, per page or per mile, the Poisson is the only one of the two that works. If the question is about time rather than counts, it is the exponential.

Your situation	Use	Why
You can count the attempts and each has the same chance	Binomial	exact, capped at n, works for any p
Only an average per hour, day, page or mile	Poisson	needs no trial count
Many trials, tiny p (n at least 20, p at most 0.05)	either	NIST rule of thumb: Poisson is a good approximation
p above about 0.1 with few trials	Binomial	Poisson drifts several points and ignores the ceiling
How long until the next event, or how long to wait for 90% certainty	Exponential	the time side of the same Poisson process
Variance far above the mean in your data	neither as is	both models understate clusters

Tools discussed in this article

Poisson Distribution Calculator - exactly, at most, at least or between a and b events from an average rate λ, with rate scaling over several periods and the full distribution table.
Binomial Distribution Calculator - k successes in n trials with p as a fraction, a decimal or a percentage, the exact fraction, the formula filled in and the distribution table.
Exponential Distribution Calculator - the chance the next event comes within, after or between given times, and the wait that covers 90% or 95% of cases, from a mean or a rate.