Some counts have no fixed number of trials: emails in an hour, typos on a page, meteors in a night. When events happen independently and at a steady average rate, the count follows the Poisson distribution. Its formula comes from the series $e^{\lambda} = \sum \frac{\lambda^{k}}{k!}$ from known series.
🔑 Key idea
$X$ is Poisson with mean $\lambda > 0$ when $$P(X = k) = e^{-\lambda}\frac{\lambda^{k}}{k!}, \quad k = 0, 1, 2, \ldots$$ Its mean and its variance are both $\lambda$. If events occur at rate $r$ per unit of time, the count in a window of length $t$ has $\lambda = rt$.
💡 Why it works
The probabilities add to 1: $\sum_k e^{-\lambda}\frac{\lambda^{k}}{k!} = e^{-\lambda}e^{\lambda} = 1$. For the mean, the $k = 0$ term vanishes and $\frac{k}{k!} = \frac{1}{(k - 1)!}$: $$E[X] = \sum_{k \ge 1} e^{-\lambda}\frac{\lambda^{k}}{(k - 1)!} = \lambda\sum_{j \ge 0} e^{-\lambda}\frac{\lambda^{j}}{j!} = \lambda.$$ The same trick gives $E[X(X - 1)] = \lambda^{2}$, so $\operatorname{Var}(X) = \lambda^{2} + \lambda - \lambda^{2} = \lambda$. (The next lesson shows where the formula comes from.)
When is a count Poisson? Three conditions are enough in practice: events in separate stretches of time (or space) happen independently, the average rate stays the same, and two events never happen at exactly the same instant. Calls to a large help desk, decays of a radioactive sample and flaws along a cable fit well. Customers who arrive in groups, or a rate that changes through the day, do not, unless you shorten the window until the rate is roughly constant.
Random arrivals at rate 3 per unit of time. Drag the window of length 1 and count; many windows build a histogram that matches the Poisson pmf with mean 3.
Worked examples
✏️ Worked example: Typos
A page has on average 2 typos, Poisson. Find the probability of no typos, and of at most one.
✏️ Worked example: From one probability to the mean
For a Poisson $X$, $P(X = 0) = \frac{1}{4}$. Find $\lambda$.
$P(X = 0) = e^{-\lambda} = \frac{1}{4}$.
$\lambda = \ln 4 \approx 1.386$.
Answer: $\ln 4$
Neighboring probabilities have a simple ratio: $\frac{P(X = k + 1)}{P(X = k)} = \frac{\lambda}{k + 1}$. So the probabilities rise while $k + 1 < \lambda$ and fall after, and the most likely count is the whole-number part of $\lambda$. The ratio also recovers $\lambda$: if $P(X = 3) = P(X = 2)$, then $\frac{\lambda}{3} = 1$ and $\lambda = 3$.
⚠️ Common mistake
Keeping the same $\lambda$ when the window changes. A rate of 3 per hour means $\lambda = 6$ for two hours and $\lambda = 1.5$ for half an hour. Check: $\lambda$ is the expected count in the window you are asking about, so it scales with the window's length.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.
Grinstead and Snell, Introduction to Probability (2nd revised edition, American Mathematical Society; free from Dartmouth College), chapter 5 (Distributions and Densities), section 5.1, the Poisson distribution
MIT OpenCourseWare 18.440, Probability and Random Variables (Spring 2014), lecture notes on the Poisson distribution