```Newsgroups:     rec.puzzles,sci.math
Subject:        Re: Random Points on a SPHERE
From:           kubo@zariski.harvard.edu (Tal Kubo)
Date:           3 May 93 03:49:58 EDT
Organization:   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
```