Three points are chosen at random on a circle. What is the probability that they are on the same semi circle?

If I have two portions $x$ and $y$, then $x+y= \pi r$...if the projected angles are $c_1$ and $c_2$. then it will imply that $c_1+c_2=\pi$...I have assumed uniform distribtuion so that $f(c_1)=\frac{\pi}{2}$...to calculate $P(c_1+c_2= \pi)$ I have integrated $c_2$ from $0$ to $\pi-c_1$ and $c_1$ from $0$ to $\pi$..but not arriving at the answer of $\frac 3 4$


The first two points must be on the same semicircle. They are separated by an angle between $0^{\circ}$ and $180^{\circ}$ with uniform probability.

If the first two points are the same, then the third point must lie on the same semicircle as the first two (probability $1$). If the first two points approach defining a diameter, then the probability that the third lies on the same semicircle approaches $1/2$. The probability decreases linearly from $1$ to $1/2$ as the separation of the points goes from $0^{\circ}$ to $180^{\circ}.$

So the probability averaged over all angles is $3/4$.

  • $\begingroup$ Strictly, the probability jumps from just-over-½ at just-under-180° back to 1 at exactly 180°. (This doesn't affect the average.) $\endgroup$ – Gnubie Feb 2 '16 at 15:14
  • $\begingroup$ How can you be so sure that the probability variation is LINEAR? $\endgroup$ – arya_stark Mar 17 '18 at 7:00
  • $\begingroup$ @schrodinger_16 Pick some separations for the first two points, and calculate the probability explicitly. $\endgroup$ – John Mar 17 '18 at 16:38

Given a semi circle, the probability that a uniform random point will lie on that semi circle is $\frac{1}{2}$. If three points lie on a semi circle then they definitely lie on a semicircle starting at exactly one of the points. There are 3 points, thus three semi circles and the events that the all the points lie on a semi circle starting at each of the points are mutually exclusive. Therefore the probability is $3\times\frac{1}{2^2}$. This is easily extended to $n$ points as $\frac{n}{2^{n-1}}$.

P.S: I read the solution on this blog by Saurabh Joshi

  • $\begingroup$ I didn't quite get the generalisation, could you explain in more detail? $\endgroup$ – arya_stark Mar 17 '18 at 7:05
  • $\begingroup$ @schrodinger_16. the event is equivalent to that there exists a point A, and all the rest points stay in the clockwise semicircle. If we pin down point A, there are n-1 points left, and the probability of them staying the the clockwise semicircle is $\frac{1}{2^{n-1}}$. Since we have n points, so the overall probability is $\frac{n}{2^{n-1}}$ $\endgroup$ – Albert Chen Dec 1 '18 at 19:59

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