Four points are chosen at random on the surface of a sphere. Determine the probability that the center of the sphere lies inside the tetrahedron described by these four points.

I recall this post.  The problem asked the probability that all 4 points chosen randomly on a sphere lied on the same hemisphere.  The answer turned out to be 7/8.  In other words, there was a 1/8 chance that not all points would be on the same "half" of the sphere.

I can visualize drawing all lines between these points and seeing the tetrahedron would not contain the center of the sphere.

If, on the other hand, the points do not lie on one hemisphere, the points will form a tetrahedron that will contain the center of the sphere.  I can't prove that if the points are not on one hemisphere, the tetrahedron formed must contain the center, but conversely, I can't visualize an exception.

Therefore, the probability is 1/8.

Posted by Dustin on 2012-01-22 20:10:53