Poisson vs binomial

Are you counting successes in a fixed number of opportunities, or arrivals during an exposure window? Both answers produce counts, but their assumptions and possible values differ.

Question	Binomial	Poisson
What is fixed?	Number of trials n	Expected count λ for the exposure
Mechanism	Independent binary trials with a common p	Constant-rate Poisson process counts in a window
Possible counts	0 through n	Every nonnegative integer
Mean	np	λ
Variance	np(1 − p)	λ

The Poisson distribution also arises outside time processes, but integer-valued observations alone do not justify it.

Hold the mean fixed

Binomial: n=20, p=0.1250. Both means equal 2.5; binomial variance is 2.1875, versus Poisson variance 2.5.

The same event in both models: count ≤ 3
Binomial (filled, left)	Poisson (outline, right)	Absolute difference
0.765332	0.757576	0.007756

Number of opportunities 20.00 Fixed expected count 2.50 Compare counts at most 3.00

Increase n while holding the expected count fixed: p decreases automatically. The binomial has a finite maximum n; Poisson has no upper limit. The displayed count window omits a small Poisson tail; the table uses the full cumulative probability.

The filled bar on the left is the binomial probability; the outline on the right is Poisson. Both models have the same mean. Increasing opportunities lowers p so that the expected count stays fixed. The probability table compares the same event, count at most k, under both models.

A worked comparison

Imagine 20 independent opportunities, each with success probability 0.125. The expected count is 2.5. For at most 3 successes, the exact binomial probability is about 0.7653; the Poisson approximation gives about 0.7576. Their absolute difference is about 0.0078, or 0.78 percentage points.

Now use 500 opportunities and p = 0.005. The mean remains 2.5, and the difference for this event drops to about 0.000269. Set the controls to verify both examples.

Make a prediction

With n = 5 and mean = 5, what does the binomial model say about exactly five successes?

Explore the answer

Its p is 1, so all five trials succeed with probability 1 and its variance is zero. A Poisson variable with mean 5 still varies and can exceed 5. Equal means alone cannot justify substituting one distribution for the other.

The approximation is a particular limit

The binomial approaches Poisson when n grows, p shrinks, and their product stays at λ:

n\to\infty,\quad p\to0,\quad np=\lambda

The binomial variance np(1 − p) then approaches λ too. Increasing n while keeping p fixed is a different limit: both the mean and variance grow. A large trial count by itself is not enough to establish a good Poisson approximation.

The table reports absolute error for one cumulative event. A small absolute error may still be a large relative error for a very rare outcome. When the exact binomial calculation is available, there is no need to discard it merely because an approximation exists.

Check the mechanism

Sampling without replacement from a finite collection changes success probabilities and induces dependence: consider hypergeometric. Distinct trial probabilities lead away from a simple binomial model. A time-varying or clustered arrival process may also miss the simple constant-rate Poisson model.

Explore binomial trials, the Poisson event timeline, or the distribution chooser.

References

NIST: binomial distribution, for the PMF and moments.
NIST: Poisson distribution, for the count probabilities and moments.

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