# Distribution of gaps between points

If you choose $n-1$ points uniformly and independently at random from the unit interval, what is the distribution of the lengths of the $n$ intervals without points in them?

To make it a little more concrete. If $n=4$ and I choose points at $0.9$, $0.2$ and $0.5$ then the four intervals are $0.2$, $0.3$, $0.4$ and $0.1$.

They each have the same distribution, with mean $1/n$. You can imagine choosing $n$ points on a circle of circumference $1$, and then choosing one of them to split the circle, leaving $n-1$ points and $n$ intervals.
So let's look at the distribution of the first interval. The probability all the $n-1$ points are above $x$ is $(1-x)^{n-1}$ so the density of the length of the first (and so each) interval is $(n-1)(1-x)^{n-2}$. This is a Beta distribution with parameters $\alpha=1$ and $\beta=n-1$.
Although the intervals are identically distributed, they are not independently distributed, since their sum is $1$.
• @Lembik Try this: Break the interval in two places and you are likely to have one big piece and two small pieces: the probability of being able to make a triangle is only 25%. So small pieces are more likely than big pieces. The effect is in fact not extreme: the mean length here is about $0.33333$ while the median length ("as likely as not above or below") is about $0.29289$ Aug 4, 2013 at 18:11