lew@ihuxr.UUCP (09/23/83)
Here's a "proof" that the probability of 4 points on a sphere falling in a hemisphere is 7/8 (in agreement with the formula John Pedersen gave.) Three points form a spherical triangle. If you extend the sides all the way around the sphere, you will divide the sphere into eight triangles. A fourth point will fail to be in a hemisphere common to the other points only if it falls in the "antipodal" triangle, which is congruent to the orginal triangle. A little thought (maybe more) will show that the average triangle chosen with random vertices covers 1/8 the sphere. To see this consider that the three points and their antipodes divide the sphere into eight triangles, as noted above. Each one is identified with one of the eight (pode, antipode) combo's of the three points and each of these comb- inations generates the same distribution. I don't know if this line of thought can be generalized to the whole formula or not, but it can probably be extended to the case of N = n+1 pretty easily. Lew Mammel, Jr. ihuxr!lew