# 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.)

• Am I missing something? It seems that $2\pi / 3$ is the obvious solution, given that by symmetry in their construction all arcs have the same expected length – Mariuslp Aug 1 '17 at 8:52
• I think it should be half of circumference ie $\pi$. The length of the arc containing the point can vary from 0 to $2\pi$ linearly – samjoe Aug 1 '17 at 8:53
• @Mariuslp The average length is $2\pi/3$, yes, so if you choose between the three segments uniformly, that would be the answer. But you are more likely to be in a longer segment than a shorter one, so the expected length is longer than that. – Arthur Aug 1 '17 at 8:57
• @Arthur: You're right, my bad. – Mariuslp Aug 1 '17 at 9:10
• Spelling: length, not lenght. – James K Aug 1 '17 at 11:16

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.

• This answer was downvoted. But without any explanation. Sad. – drhab Aug 1 '17 at 11:16
• Why do we choose the "first chosen point" (the one by which our two arcs are bordered) uniformly? What would happen if the distributions of the other three points were non-uniform? Would we still have the added point chosen uniformly (as I would think) or would its distribution be something else? – Alessio Di Lorenzo Aug 2 '17 at 22:19
• @AlessioDiLorenzo There is indeed no reason to choose the first chosen point uniformly. We can use any distribution for it. That is the essence of the answer. The author of the question came up with $(0,1)$, but how did he get that? By means of a choice based on some distribution? All irrelevant. To gave things a symmetric look in my answer I proposed an underlying uniform distribution, but again: any will do. On your second and third question not much can be said. If the uniformity dissappears for the three points then the question changes intrinsically. – drhab Aug 3 '17 at 6:32

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 $$2\pi\frac{1}{4} + 2\pi - 2\pi\frac{3}{4} = 2\pi(1/4+1-3/4) = \pi.$$

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.