I am aware of many forms of the solutions posted on SE. I am only interested in an approach using order statistics. The correct probability is $\frac{n}{2^{n-1}}$.
So n points are chosen randomly on a circle and I am interested in the probability that there exists a semi circle that covers all the points. We can think of it as unit circumference circle, and the sample of $n$ points can be modeled as $n$ iid standard uniform random variables, $U_1, ..., U_n$. Also let the two order statistics of this sample to be $U_\textrm{max}$ and $U_\textrm{min}$.
Here is my logic. In order for $n$ points to fall in a semicircle, $U_\textrm{max} - U_\textrm{min} \le 0.5$.
According to the First course in probability by Ross (sec. 6.6), $$\mathbb{P} \left(U_\textrm{max} - U_\textrm{min} \le x \right) = n (1 - x) x^{n-1} + x^n.$$
But using $x = 0.5$, I get probability of $\frac{n+1}{2^{n}}$, which is not the correct answer. In fact using the same distribution for order statistics of $U_1, ..., U_{n-1}$ leads to a correct answer according to this link, which says that I need to condition on $U_\textrm{max}$. But I don't get this idea. Why do we need to condition on the maximum value?
If we let one of the n points be 0.5, then the semi circle would never cross the zero point, because any semi circle crossing the zero point would not contains 0.5 . Furthermore, any semi circle not crossing the zero point would surely contains 0.5 .
The n points in a semi circle.
= The other n-1 points in a semi circle which contains 0.5
= The other n-1 points in a semi circle not cross zero.
= The other n-1 points satisfy Umax−Umin≤0.5