Picture a quiet bakery. Some hours nobody walks in; other hours three or four customers turn up close together. You can't say exactly when the next one arrives, but you know the shop averages, say, about two customers an hour. The Poisson distribution is simply the math for "how many rare, scattered things happen in one fixed stretch of time or space?" — calls in a minute, raindrops on a tile, typos on a page.

There is really only one knob: the average count, written $\lambda$ ("lambda") and called the rate. If a call centre averages $\lambda=3$ calls a minute, the chance of getting exactly $k$ calls in the next minute is $P(X=k)=\dfrac{e^{-\lambda}\lambda^{k}}{k!}$. Try one number with $\lambda=2$: the chance of a totally quiet hour is $P(0)=e^{-2}\approx 0.135$, about a 1-in-7 chance. The chance of exactly two is $e^{-2}\cdot\dfrac{2^{2}}{2!}=0.135\cdot 2\approx 0.271$.

For events that occur independently at a constant average rate, the count in an interval is exactly $\text{Poisson}(\lambda)$. Its signature property: the mean and the variance are both equal to $\lambda$, so the spread widens as the rate grows. Poisson is also the limit of the binomial — chop time into $n$ tiny slices, each with success probability $p=\lambda/n$, and as $n\to\infty$, $\text{Bin}(n,\lambda/n)\to\text{Poisson}(\lambda)$. In the sim above, the λ slider sets the rate (watch the $E[X]=\text{Var}[X]=\lambda$ cards move together), and the k slider picks which outcome you are asking about, so $P(X=k)$ and $P(X\le k)$ update live.

1. Set $\lambda$ small (around 0.5): the bars pile up near $k=0$ — rare events mean mostly zeros. 2. Slide $\lambda$ up toward 15: the distribution becomes broad and bell-shaped, showing how Poisson approaches a Normal for large $\lambda$. 3. Open the "vs Binomial" tab and watch the $\text{Bin}(n,\lambda/n)$ bars close in on the Poisson dots as $n$ climbs from 10 to 200.

Divide interval into $n$ sub-intervals, each with $p=\lambda/n$ probability. As $n\to\infty$:

$$P(X=k)=\binom{n}{k}\left(\frac{\lambda}{n}\right)^k\left(1-\frac{\lambda}{n}\right)^{n-k}\to\frac{e^{-\lambda}\lambda^k}{k!}$$

using $\lim_{n\to\infty}(1-\lambda/n)^n=e^{-\lambda}$ and $\binom{n}{k}/n^k\to1/k!$.

$E[X]=\sum_{k=0}^\infty k\frac{e^{-\lambda}\lambda^k}{k!}=\lambda e^{-\lambda}\sum_{k=1}^\infty\frac{\lambda^{k-1}}{(k-1)!}=\lambda e^{-\lambda}e^\lambda=\lambda$

$\text{Var}(X)=E[X(X-1)]+E[X]-(E[X])^2=\lambda^2+\lambda-\lambda^2=\lambda$

If events arrive at rate $\lambda$ per unit time, independently, then in any interval of length $t$: $N(t)\sim\text{Poisson}(\lambda t)$. The inter-arrival times follow $\text{Exponential}(\lambda)$.