3 uniform points on a circle 
Suppose 3 (distinct) points are uniformly and independently distributed on a circle of unit length (smaller than a unit circle!). This is really circle and not disc. Call one of these points $B$. Let $E$ be the random variable denoting the distance (along the circle) of the point $B$ to its anti-clockwise neighbour. Find the pdf of $E$.

There's an argument that goes:

*

*Call these 3 points $A,B,C$ s.t. they are iid $\sim \ Unif(0,1)$ (or $[0,1)$ or whatever).


*The pdf of $E$ is $f_E(e)=2(1-e) 1_{(0,1)}(e)$ (the pdf of a minimum of 2 iid Unif(0,1) or of the absolute value difference of 2 iid Unif(0,1)). This can be seen as like 2 ways to choose the anti-clockwise neighbour and then '$1-e$ ways' to choose the position of the clockwise neighbour.
My concerns: Since we're not dealing with discrete, I find this '$1-e$ ways' to be very heuristic. I don't even see where exactly we use independence or uniformity. I'm hoping for something more rigorous like a precise formula relating $E$ to to $A,B,C$. (However, I think this makes perfect rigorous sense in the discrete case like 10 points on a circle separated $\frac1{10}$ each and then $A,B,C$ are discrete uniform.)
Question: How do we make this argument precise? I mean, what is $E$ exactly in relation to $A,B,C$?
 A: I am going to generalise your problem to allow any number of points.  The problem is just as easy to solve with this generalisation.
Since you are measuring from a pre-specified point, and since all points are uniform, without loss of the structure of the problem we can set the reference point at $0$.  This can easily be achieved by rotating the circle so that the reference point is at the top after the points are distributed; see additional information in the section below.
Now suppose you have $n \in \mathbb{N}$ other points, which I will label as $X_1,...,X_n \sim \text{IID U}[0,1]$, where the measure of distance is taken in an anti-clockwise direction.  (In your exposition of the problem you have called the other points $A$ and $C$, but I will generalise from this.)  The anti-clockwise distance from the reference point to the first neighbour is:
$$E \equiv \min(X_1,...,X_n).$$
For all $0 \leqslant e \leqslant 1$ the cumulative distribution function for this distance is:
$$\begin{align}
F_E(e)
\equiv \mathbb{P}(E \leqslant e)
&= 1 - \mathbb{P}(E > e) \\[12pt]
&= \prod_{i=1}^n \mathbb{P}(X_i > e) \\[6pt]
&= 1 - \prod_{i=1}^n (1-e) \\[12pt]
&= 1 - (1-e)^n, \\[6pt]
\end{align}$$
so the density function for this distance is:
$$\begin{align}
f_E(e) = \frac{dF_E}{de}(e) 
&= n (1-e)^{n-1} \cdot \mathbb{I}(0 \leqslant e \leqslant 1) \\[12pt]
&= \text{Beta}(e | 1, n). \\[6pt]
\end{align}$$
This shows that we have $E \sim \text{Beta}(1, n)$ in the general problem.  Your problem has three points, so you have $n=2$ points that are not the reference point, which gives the density you stated in your question.

More information on the transformation: Suppose we let $Y_0, Y_1, ..., Y_n \sim \text{IID U}[0,1]$ be points uniformly distributed on the circle prior to re-framing (taken as anti-clockwise distances around the circle).  Taking $Y_0$ as the reference point (which you have labelled as $B$) we can define the anti-clockwise distances from the reference point as:
$$X_i \equiv (Y_i - Y_0) \text{ mod } 1.$$
By definition we have $X_0 = 0$, which means that using the values $X_0,X_1,...,X_n$ re-frames the analysis to set the reference point at zero.  It is simple to show that $X_1,...,X_n \sim \text{IID}[0,1]$, so re-framing to put the reference point at zero does not change the underlying distributional framework of the problem.
