🚀 go-pugleaf

isaak@aurora.com (Mark Isaak) writes:

>Four points are randomly selected from the surface of a sphere.
>What is the probability that all four lie in the same hemisphere?
>How about for 5 points?  6?  more?

One simple-minded solution (which doesn't worry about boundaries) is as follows.
The first two points picked, together with the center of the sphere, define a
plane that divides the sphere into two hemispheres.  The two points are on the
boundary dividing the hemispheres.  With probability 1, the third point lies
strictly in the interior of one of the hemispherical surfaces, and this becomes
our chosen hemisphere (the first two points, being on the boundary, are included
by courtesy in the chosen hemisphere.)  Thus, assuming that the points are
picked with a uniform distribution over the surface of the sphere, the
probability that the fourth point lies in the chosen hemisphere is 1/2, the
probability that the fifth point also lies in the chosen hemisphere is 1/4, etc.