TITLE: Probability - Length of an arc which contains a fixed point
QUESTION [11 upvotes]: 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.)

REPLY [13 votes]: 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.