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If three points are randomly chosen on the boundary of a circle, what is the probability that there exists a diameter of the circle such that all three points lie on the same side of it?

I have a solution, but I'm very curious to see how others do it (and whether we get the same answer)!

Thanks.

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  • $\begingroup$ Are the points chosen uniformly in area? If it is not specified one would usually think so, but you could also choose $r, \theta$ uniformly. This would make points near the center more probable. I'm not sure it matters for this problem. $\endgroup$ – Ross Millikan Mar 27 '13 at 0:55
  • $\begingroup$ Sorry--the three points are on the BOUNDARY of the circle! Edited. $\endgroup$ – Matt E Mar 27 '13 at 0:59
  • $\begingroup$ I think the radius doesn't matter. The argument gives doesn't depend on that. $\endgroup$ – Ross Millikan Mar 27 '13 at 1:40
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Choose the points on the unit circle in the complex plane, with arguments $\theta_1, \theta_2, \theta_3$. If it happens that the differences $\theta_2 - \theta_1$ and $\theta_3 - \theta_1$ are both between $0$ and $\pi$ modulo $2\pi$, then the points are on the same half circle. From independence of the variables, the probability of this happening is $\frac {1} {4}$. Now we need to consider symmetrical cases - the above is the case that the first point is "to the right" of the next two points, but either point could be "on the right" of the other two points. The three cases are symmetrical and disjoint, so the total probability is $\frac 3 4$.

The general case of $n$ points can be solved the same way and it results in $\frac n {2^{n-1}}$.

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    $\begingroup$ I would add that not only are the three $\theta$s independent, but the two differences are independent. That can be shown by considering the conditional distribution of the pair of differences given the value of $\theta_1$, and seeing that that does not depend on $\theta_1$. $\endgroup$ – Michael Hardy Mar 27 '13 at 1:33
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    $\begingroup$ You could extend this argument to the probability of $n$ random points lying in a circular segment with angle $\phi \le \pi$, to give a probability of $$n\left(\dfrac{\phi}{2\pi}\right)^{n-1}.$$ I suspect the position is harder for $\pi \lt \phi \lt 2\pi \frac{n-1}{n}$. $\endgroup$ – Henry May 22 '15 at 8:20
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    $\begingroup$ The question Centre in N-sided polygon on circle recently asked about the case with arbitrary $n$. I only found your post here after answering there. $\endgroup$ – MvG Oct 13 '16 at 8:59
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Here's my answer:

Suppose points $A$ and $B$ are placed first. The measure $\theta$ of the minor arc that has $A$ and $B$ as endpoints is uniformly distributed over the interval $(0,\pi)$. Given a value for $\theta$, though, we can know that the probability that point $C$ will be placed so that all three points are on the same side of a diameter is equal to $1-\frac{\theta}{2\pi}$. Averaging over the interval $(0,\pi)$ gives $$ \frac{1}{\pi} \int_0^\pi (1-\frac{\theta}{2\pi})d\theta=\frac{1}{\pi}\left[\theta-\frac{1}{4\pi}\theta^2\right]_0^\pi=\frac{3}{4} $$

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  • $\begingroup$ Hi Matt, of the many answers to the various versions of this question posted around SE, yours matches my thinking most closely. I'm still confused as to why we have to average the integral over the interval (0, $\pi$). Any explanation would be much appreciated. $\endgroup$ – advait May 9 '13 at 21:14
  • $\begingroup$ @advait we integrate over interval (0, π) because the value of theta( angle between A and B) can range from 0 to π $\endgroup$ – Atul Vaibhav Jun 9 '17 at 5:37

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