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 $\frac 3 4$

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3 Answers3


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 $3/4$.

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  • Strictly, the probability jumps from just-over-½ at just-under-180° back to 1 at exactly 180°. (This doesn't affect the average.) – Gnubie Feb 02 '16 at 15:14
  • How can you be so sure that the probability variation is LINEAR? – connected-subgroup Mar 17 '18 at 07:00
  • @schrodinger_16 Pick some separations for the first two points, and calculate the probability explicitly. – John Mar 17 '18 at 16:38
  • @arya_stark, the probability increases linearly because probability of any point on the circular line is uniform. The fact that the line is curved doesn't effect the given distribution. – jds Jan 19 '20 at 22:57

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

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  • I didn't quite get the generalisation, could you explain in more detail? – connected-subgroup Mar 17 '18 at 07:05
  • @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 $\frac{n}{2^{n-1}}$ – Albert Chen Dec 01 '18 at 19:59

Without loss of generality, we can assume the circumference of the circle to be equal to $1$. Cut the circle at the first point $A$ and spread it out as a line. Let the other two points $B$ and $C$ be located at distances of $x$ and $y$ from $A$. The midpoint of this line is $M$ and $0 \leq x,y \leq 1$.


Points $B$ and $C$ can be both located on same side of $M$ or on either side of $M$. If both points lie on the same side of $M$, then all $3$ points lie on the same semi-circle.
$$ P(\text{same side}) = 2 * \frac{1}{4} = \frac{1}{2} $$ We are multiplying by 2 as the points can lie on both sides of $M$.

They can also be located on the either side of $M$ and in this case, the conditions for all 3 points to lie on the same semi-circle are $$ x + 1 - y < 0.5 \Rightarrow y > x + 0.5 \text{ if } y > x\\ y + 1 - x < 0.5 \Rightarrow x > y + 0.5 \text{ if } x > y $$ Representing these two regions graphically, we get the following area:


The area of the shaded region is $\frac{1}{4}$. Combining all results, we get the probability that $A$, $B$ and $C$ all lie on the same semi-circle is $$\frac{1}{2} + \frac{1}{4} = \frac{3}{4}$$.

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