Probability that n points on a circle are in one semicircle

A variation on @joriki's answer (and edited with help from @joriki):

Suppose that point $i$ has angle $0$ (angle is arbitrary in this problem) -- essentially this is the event that point $i$ is the "first" or "leading" point in the semicircle. Then we want the event that all of the points are in the same semicircle -- i.e., that the remaining points end up all in the upper halfplane.

That's a coin-flip for each remaining point, so you end up with $1/2^{n-1}$. There's $n$ points, and the event that any point $i$ is the "leading" point is disjoint from the event that any other point $j$ is, so the final probability is $n/2^{n-1}$ (i.e. we can just add them up).

A sanity check for this answer is to notice that if you have either one or two points, then the probability must be 1, which is true in both cases.

See

https://mathoverflow.net/questions/33112/estimate-probability-0-is-in-the-convex-hull-of-n-random-points

for the general problem (when the points have any distribution that is invariant w.r.t. rotation about the origin) and

https://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re/2016#2016

for a nice application.

As a curiosity, this answer can be expressed as a product of sines:

Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$

Here's another way to do this:

Divide the circle into $2k$ equal sectors. There are $2k$ contiguous stretches of $k$ sectors each that form a semicircle, and $2k$ slightly shorter contiguous stretches of $k-1$ sectors that almost form a semicircle. The number of the semicircles containing all the points minus the number of slightly shorter stretches containing all the points is $1$ if the points are contained in at least one of the semicircles and $0$ otherwise; that is, it's the indicator variable for the points all being contained in at least one of the semicircles. The probability of an event is the expected value of its indicator variable, which in this case is

$$2k\left(\frac k{2k}\right)^n-2k\left(\frac{k-1}{2k}\right)^n=\frac k{2^{n-1}}\left(1-\left(1-\frac1k\right)^n\right)\;.$$

The limit $k\to\infty$ yields the desired probability:

$$ \lim_{k\to\infty}\frac k{2^{n-1}}\left(1-\left(1-\frac1k\right)^n\right)=\lim_{k\to\infty}\frac k{2^{n-1}}\cdot\frac nk=\frac n{2^{n-1}}\;. $$