TITLE: What is the intuition behind the Poisson distribution's function?
QUESTION [21 upvotes]: I'm trying to intuitively understand the Poisson distribution's probability mass function. When $X \sim \mathrm{Pois}(\lambda)$, then $P(X=k)=\frac{\lambda^k e^{-k}}{k!}$, but I don't see the reasoning behind this formula. In other discrete distributions, namely the binomial, geometric, negative binomial, and hypergeometric distributions, I have an intuitive, combinatorics-based understanding of why each distribution's pmf is defined the way it is. 
That is, if $Y \sim\mathrm{Bin}(n,p)$ then $P(Y=k)=\binom{n}{k}p^k(1-p)^{n-k}$, and this equation is clear - there are $\binom{n}{k}$ ways to choose the $k$ successful trials, and we need the trials to succeed $k$ times and fail $n-k$ times. 
What is the corresponding intuition for the Poisson distribution?

REPLY [6 votes]: Let $p_k(t)$ be the probability of $k$ events in time $t$. We first find $p_0(t)$. Let $h$ be small. By independence $p_0(t+h)=p_0(t)p_0(h)$. The probability of an event in time $h$, where $h$ is very small, is roughly $\lambda h$. More accurately, $\lim_{h\to 0^+}\frac{1-p_0(h)}{h}=\lambda$.  So $p_0(h)\approx 1-\lambda h$. Substitute. We get
$$\frac{p_0(t+h)-p_0(t)}{h}\approx -\lambda p_0(t).$$
Let $h\to 0$. We conclude that $p_0'(t)=-\lambda p_0(t)$. This is a familiar differential equation. Since $p_0(0)=1$, it has solution $p_0(t)=e^{-\lambda t}$.   
Now do a similar argument for general $k$. If $h$ is a small time interval, then the probability of $2$ or more events in time interval $h$ is negligible in comparison with the probability of $1$ event. Thus
$$p_k(t+h)\approx p_{k}(t)(1-\lambda h)+p_{k-1}(t)\lambda h.$$
Simplifying, and letting $h\to 0$, we find that
$$p_k'(t)=-\lambda p_k(t)+\lambda p_{k-1}(t).$$
This DE can be solved, using the induction hypothesis $p_{k-1}(t)=e^{-\lambda t}\frac{(\lambda t)^{k-1}}{(k-1)!}$. Or else we can verify by substitution that the standard expressions do satisfy the DE.