TITLE: Probability that n points on a circle are in one semicircle
QUESTION [44 upvotes]: Choose n points randomly from a circle, how to calculate the probability that all the points are in one semicircle? Any hint is appreciated.

REPLY [2 votes]: Another simpler approach,
1) Randomly pick $1$ out of $n$ points and call it $A$ : $\binom n1$ ways
2) Starting from $A$, mark another point $B$ on circumference, such that $length(AB) = \frac12(Cirumference)$ [so that $AB$ and $BA$ are two semi-circles]
3) Now out of remaining $(n-1)$ points, each point can lie on either $AB$ or $BA$ with probability $\frac12$
4) For ALL the remaining $(n-1)$ points to lie on EITHER $AB$ OR $BA$ (i.e., all $(n-1)$ lie on same semi-circle), the joint probability is $\frac12*\frac12 ...(n-1) times$ $=$ $(\frac12)^{(n-1)}$
Since #1 above (randomly picking $A$) is an independent event, $\therefore$ $(\frac12)^{(n-1)}$ (expression in #4) will add $\binom n1$ times
$\implies$ Required probability is $\binom n1(\frac12)^{(n-1)}$ $=$ $n(\frac12)^{(n-1)}$