Do this experiment. Draw 10000 random number in $[0,1]$ according to the uniform distribution. Order them in the increasing order. The difference between two neighbouring numbers follows a Poisson distribution.
Why?
I checked it numerically, but do not know how to prove it.
Consider $10000$ random variables in $[0, 1]$ given by $Y_{(i)}$, where $i = 0, 1, \dots, 10000$ and $Y_{(s)} \leq Y_{(t)}$ for $s < t$. Then we wish to find the distribution of $X = Y_{(j)} - Y_{(j-1)}$ for all $j$.
For any $k$, recall that
$$\begin{align} f_{Y_{(k-1)}, Y_{(k)}}(s,t) &= \dfrac{10000!}{(k-1)!\left[k-(k-1)-1\right]!(10000-k)!}\left(\dfrac{s-0}{1-0}\right)^{(k-1)-1}\\ &\left[\dfrac{t-0}{1-0} - \dfrac{s-0}{1-0}\right]^{k-1-(k-1)}\left(1-\dfrac{t-0}{1-0}\right)^{10000-k}(1)(1) \\ &= \dfrac{10000!}{(k-1)!(10000-k)!}s^{(k-1)-1}\left(1-t\right)^{10000-k} \end{align}$$ where $0 \leq s \leq t \leq 1$. Now using the CDF method of transformations,
$$\begin{align}\mathbb{P}\left(X \leq x\right) &= \mathbb{P}\left(Y_{(j)} \leq Y_{(j-1)} + x \right) \\ &= \int\limits_{0}^{1-x}\int\limits_{x}^{1}\dfrac{10000!}{(j-1)!(10000-j)!}s^{(j-1)-1}\left(1-t\right)^{10000-j}\mathrm{d}s\mathrm{ d}t \\ &= \dfrac{10000!}{(j-1)!(10000-j)!}\left(\dfrac{x^{10001-j}-1}{10001-j}\right)\left(\dfrac{x-1}{j-1}\right) \end{align}$$ which I got after doing some $u$-substitution and algebra.
My algebra might be incorrect. However, I have no reason to believe that the differences follow a Poisson distribution since the differences are going to be from a continuous distribution since you're drawing the $Y_{(i)}$ from the interval $[0, 1]$ (and the Poisson distribution is a discrete distribution).