Newsgroups:rec.puzzles,sci.mathSubject:Re: Random Points on a SPHEREFrom:kubo@zariski.harvard.edu (Tal Kubo)Date:3 May 93 03:49:58 EDTOrganization:Dept. of Math, Harvard Univ.

In article <C6FGv2.6Dn@cantua.canterbury.ac.nz> wft@math.canterbury.ac.nz (Bill Taylor) writes: > >Choose points A,B,C,D independently and (uniformly) randomly on a unit sphere, >what is the probability that short-arc AB intersects short-arc CD ? > >I had given a slick proof that the answer is 1/8, by... > >(i) considering the placement of A & B conditional on knowing their great > circle; >(ii) taking these to be uniform (& independent) on the great circle; >(iii) showing the probability that short-arc AB contains a given point is 1/4; >(iv) looking at the intersection of circles AB & CD, and deducing 1/8. >[...] Actually the answer of 1/8 doesn't really require uniform distribution, only a distribution invariant with respect to 180-degree rotation through the center of the sphere. The idea is that among the 16 points (+/- A, +/- B, +/- C, +/- D), exactly two correspond to "good" configurations. This is not hard to see in the probability 1 "generic" case were A,B,C,D are distinct and no three of them lie on a great circle. To make this argument completely rigorous one needs only the finite additivity of the measure, that the non-generic cases have probability 0, that the set of "good" cases *has* a measure, and that the measure be invariant with respect to 180 degree rotation in any coordinate. Tal kubo@math.harvard.edu