# Probability of a point lying in longest arc when the circle is divided into $3$ random arcs.

There is a unit circle centered at origin on which $3$ points are chosen randomly thus we have $3$ arcs. What is the probability of $(1,0)$ lying on the longest arc?

I tried the intuition that the point may lie in any of the three arcs thus the answer may be $1/3$. Am I correct?

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• Zero! Did you pose the question correctly. The unit circle goes through $(1,0)$. So only if that point is randomly selected will the arc (2 in this case) go through it. Since there are infinite number of points on a circle, the chance of that goes to zero. – Dan Sp. Jul 28 '18 at 18:59
• @DanSp. I didn't got what you said. Can you elaborate? – Ashish Ranjan Jul 28 '18 at 19:01
• The point $(1,0)$ is ON the unit circle centered at the origin so the only way it can be included on an arc of the circle is if the point $(1,0)$ exactly is selected. – Dan Sp. Jul 28 '18 at 19:04
• @DanSp. the unit circle itself is split up in $3$ arcs. One of them contains $(0,1)$. What is the probability that this arc is the longest of the $3$ arcs. That is the question. – Vera Jul 28 '18 at 19:07

Let the circle have circumference $1$. The probability $p$ we are looking for then is the expected length of the longest arc. We may assume that one of the three points is at $0$, and that the other two points $x$ and $y$ are independently uniformly distributed in $[0,1]$, in other words: that $(x,y)$ is uniformly distributed in $[0,1]^2$. This leads to $$p=2\int_0^1\int_0^x\max\{y, x-y,1-x\}\>dy\>dx\ .$$ This is an elementary, but nasty integral to compute. Therefore I let Mathematica compute it numerically. The result was $0.611111$. Since the result would have to be a rational number with small divisor this lead to the conjecture that $$p={11\over18}\ .$$ I then drew a figure and computed the integral "by hand". The conjectured value was thereby confirmed.

Partial answer: 3 points $(a,b,c)$ chosen uniformly at random from (0,1). Assume $a<b<c$. Let $f=b−a,\ g=c−b,\ h=1−c+a$.Get the distribution of $L=max(f,g,h)$. That is the distribution you want.

Slight simplification: Choose 2 points $(a,b)$ uniformly from (0,1). Assume $a<b$. Let $f=a,\ g=b-a,\ h=1-b$. As before, get the distribution of $L=max(f,g,h)$. That is the distribution you want.

I suspect this will not be too easy.

This answer provides in a probabilistic foundation of the answer of Christian.

Let $A$ denote the longest arc and let $M$ denote its length. Then:$$\mathbb P((0,1)\in A)=\mathbb E1_A((0,1))=\mathbb E[\mathbb E[1_A((0,1))\mid M]]=\mathbb E\left(\frac{M}{2\pi}\right)=\frac{\mathbb EM}{2\pi}$$

So the expectation of the length of the longest arc is essential.

Finding RHS comes to finding the expectation of the longest of the $3$ intervals that arise if two points are selected independently and according to uniform distribution on $[0,1]$.

For that see the answer of Christian.

• Need link to answer of Christian! – herb steinberg Jul 29 '18 at 15:27