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.

Question

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.

Gustavo Zimmerman · Accepted Answer

Step 1If 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,…,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    ,  …      E    n   are mutually disjoint events.Step 2By symmetry we have   P  (      E    1    )  =  ⋯  =  P  (      E    n    ).Actually it remains now to find   P  (      E    1    ).

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