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?

  • $\begingroup$ What you have done is assume a division of the circle into a right half and a left half before you distribute the points, and calculate the probability that all the points end up in the same half. Your answer is therefore too small, because you haven't taken into account that you're free to choose a division of the circle into halves after all the points have been placed. Specifically, for two points, the probability should be $1$. $\endgroup$ – Arthur Mar 2 '18 at 11:58
  • $\begingroup$ You may distribute the points on a line segment of length $2\pi$. Then calculate the probability that distance between the largest and smallest value is less than $\pi$. $\endgroup$ – Michael Hoppe Mar 2 '18 at 12:43
  • $\begingroup$ @MichaelHoppe That is not good enough. If the points concentrate at utmost left and utmost right then the value you mention is more than $\pi$ but still it is possible that a half-circle exists that contains them all. $\endgroup$ – drhab Mar 2 '18 at 12:55
  • $\begingroup$ @drhab Yes, it’s not that easy. Then just find a tangent such that all projections of the points on the tangent lie on the same side of the point of contact (just kidding). $\endgroup$ – Michael Hoppe Mar 2 '18 at 13:21


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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