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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 $0.1$.

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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 $1$.

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  • $\begingroup$ Thank you. It seems odd to me that the pdf is monotone and that maximum is at zero. Is there an intuitive explanation for why the intervals are more likely to be very short than not? $\endgroup$ – user66307 Aug 4 '13 at 17:57
  • $\begingroup$ @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 $0.29289$ $\endgroup$ – Henry Aug 4 '13 at 18:11

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