0
$\begingroup$

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?

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ – Shaun Jul 28 '18 at 18:57
  • 1
    $\begingroup$ 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. $\endgroup$ – Dan Sp. Jul 28 '18 at 18:59
  • $\begingroup$ @DanSp. I didn't got what you said. Can you elaborate? $\endgroup$ – Ashish Ranjan Jul 28 '18 at 19:01
  • $\begingroup$ 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. $\endgroup$ – Dan Sp. Jul 28 '18 at 19:04
  • $\begingroup$ @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. $\endgroup$ – Vera Jul 28 '18 at 19:07
2
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
1
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
1
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Need link to answer of Christian! $\endgroup$ – herb steinberg Jul 29 '18 at 15:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.