TITLE: A "Geometric" probability problem
QUESTION [3 upvotes]: n points are chosen randomly from a circumference of a circle. Find the probability that all points are on the same half of the circle. My intuition was that the probability is $1/2^{n-1}$ since if the first point is placed somewhere on the circumference, for each point there is probability of $1/2$ to be placed on the half to the right of it and $1/2$ to the left..but i feel i might be counting some probabilitys twice. Any ideas?

REPLY [1 votes]: Hint:
If a half-circle exists containing all $n$ points then also a half-circle exists that starts from one of the points, goes in counter-clockwise direction and contains all points.
If the points are numbered with $1,\dots,n$ and $E_i$ denotes the event that the half-circle as described above and starting at point $i$ contains all points then $E_1,\dots E_n$ are mutually disjoint events.
By symmetry we have $P(E_1)=\cdots=P(E_n)$.
Actually it remains now to find $P(E_1)$.
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