Given 4 points ( not all on the same plane ), what is the probability that a hemisphere exists that passes through all four of them ?



What is the probability that 4 points determine a hemisphere ?
See [J. G. Wendell, "A problem in geometric probability", Math. Scand. 11 (1962) 109-111](http://www.mscand.dk/article/view/10655/8676). The probability that $N$ random points lie in some hemisphere of the unit sphere in $n$-space is


$$p\_{n,N} = 2^{-N+1} \sum\_{k=0}^{n-1} {N-1 \choose k}$$


and in particular you want


$$p\_{3,4} = 2^{-3} \sum\_{k=0}^2 {3 \choose k} = {7 \over 8}$$.


**A second solution:** A solution from *The Annals of Mathematics*, 2 (1886) 133-143 (available from [jstor](http://www.jstor.org/stable/1967203)), specific to the (3,4) case, is as follows. First take three points at random, A, B, C; they are all in the same hemisphere and form a spherical triangle. Find the antipodal points to those three, A', B', C'. Now either the fourth point is in the same hemisphere as the first three or it is in the triangle A'B'C'. The average area of this triangle is one-eighth the surface of the sphere.


This gets the right answer, but I'm not sure how I feel about it; why is the average area one-eighth the surface of the sphere? One can guess this from the fact that three great circles divide a sphere into eight spherical triangles, but that's hardly a proof. Generally this solution seems to assume more facility with spherical geometry than is common nowadays.


