🚀 go-pugleaf

In article <1992Nov22.020105.27398@hubcap.clemson.edu>, mjfreem@hubcap.clemson.edu (Matthew J. Freeman) writes:
> isaak@aurora.com (Mark Isaak) writes:

>> Four points are randomly selected from the surface of a sphere.

Without further information, I'll assume the distribution over the
surface of the sphere is uniform - that is, the expected number of
points falling inside an area is directly proportional to the area,
independent of where it is on the sphere.

>> What is the probability that all four lie in the same hemisphere?

> Think of placing 4 points on an otherwise featureless sphere.  The
> first three points will define a hemisphere no matter where they are
> placed, and the probability that the fourth is in the same hemisphere
> is 1/2.

No, they will all fall in the same hemisphere[%], but they do not
*define* a hemisphere.  Consider three points clustered near one
another: provided the fourth point doesn't fall in a small area
antipodal to the triangle formed by the cluster, all four will still be
in the same hemisphere.

Given three points, the fourth point will fall so as to put all four
points in the same hemisphere with probability equal to the fraction of
the total spherical surface area which lies outside the (spherical)
triangle formed by the three points.  This is because any point which
is not antipodal to a point inside that triangle will permit a
hemisphere that includes all four points.

This lies between 1/2 and 1, depending on how closely the first three
points cluster to one another.  So answering the question amounts to
computing the expected fraction of the sphere's area covered by the
triangle formed by the first three points.  My knowledge of spherical
geometry is insufficient for this.

[%] Throughout this posting, I am ignoring boundaries.  Cases where the
    first three points all fall on the same great circle, or the fourth
    point falls antipodal to a boundary, I dismiss as a set of measure
    zero, not affecting the probability.

					der Mouse

			    mouse@larry.mcrcim.mcgill.edu