Three points on a circle

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.

Solution 1:

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 $\frac n {2^{n-1}}$.

Solution 2:

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 $$ \frac{1}{\pi} \int_0^\pi (1-\frac{\theta}{2\pi})d\theta=\frac{1}{\pi}\left[\theta-\frac{1}{4\pi}\theta^2\right]_0^\pi=\frac{3}{4} $$