Probability - Length of an arc which contains a fixed point On the circumference $x^2+y^2=1$ one randomly chooses (uniformly and independently) $3$ points. These points split the circumference, forming $3$ arcs.
What's the expected value of the length of the arc which contains the point $(1,0)$?
Is there any quick way to solve this? I kind of already used some symmetries but it seems too complex to me. (It's supposed to be an easy exercise, so maybe there's some clever thing to consider.)
 A: You can model this on the interval $[0,2\pi)$ and set the point $(1,0)$ equal to $0$ on this interval. Then we draw 3 uniform variables and are interested in the sum of the length below the minimal draw and the length above the maximal draw, ie the value $X_{min} + 2\pi - X_{max}$. For the min and max of $n$ uniform variables we have
\begin{align}
E(X_{min}) &= 2\pi\frac{1}{n+1} \\
E(X_{max}) &= 2\pi\frac{n}{n+1}.
\end{align}
Therefore our total expected arc length equals
$$
2\pi\frac{1}{4} + 2\pi - 2\pi\frac{3}{4} = 2\pi(1/4+1-3/4) = \pi.
$$
A: Hint: Add the expected distance to the first point clockwise to the expected distance to the first point counterclockwise. Now we have reduced it to the expected value of the largest and smallest of three uniformly random numbers on the interval $[0,2\pi]$. Specifically, to the length of the first and the last of the four segments of the same interval that you get from the three points.
A: Hint:
If you change the question into:
"Randomly choose (uniformly and independently) $4$ points so that there are $4$ arcs. What is the sum of the expected lengths of the two arcs that are bordered by (e.g.) the first chosen point?" 
then there is no essential difference with the original question. 
To understand that be aware of the "coincidental" choice for $(0,1)$. Why not just some randomly chosen point here? That would not change things, would it?
This approach will lead easily to $\pi$ as answer on base of linearity of expectation and symmetry.
