🚀 go-pugleaf

matt@severian.chi.il.us (Matt Crawford) writes...
>That's two questions in a row dealing with randomly selected points
>in or on a sphere and neither one stated the distribution.  Assuming
>uniform distribution, I have a half-answer to this one:

I get 21/20 for the other one, but the solution is not enlightening.

>>Four points are randomly selected from the surface of a sphere.
>>What is the probability that all four lie in the same hemisphere?
> 
>If A is the solid angle subtended by 3 points on the sphere, the
>probability that a 4th point chosen uniformly on the sphere is, uh,
>cohemispheric with the 3 is 1-A/4pi if A!=2pi; 1 if A=2pi.

Neat. So we need the expectation value of A. I think it is pi/2.
Let B be chosen along the equator and along the prime
meridian. Let C be chosen along the equator, and let D be chosen in
the northern hemisphere. The great circles BD and CD divide the hemisphere
into 4 regions which can be shown to have equal expected areas. But the
expected areas must add to 2pi.

So the final probability is: 7/8

Dave Ring
dwr2560@zeus.tamu.edu