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. 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), 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.

We chose n points on $S^{n-2}$ and want to show that the probability for them to be in one half-sphere is $1-2^{1-n}$. A simple way to solve this question is to notice that up to a linear transformation there exsits a unique collection of generic $n$ lines in $R^{n-1}$ through 0. This reduces whe problem to a combinatorial one. End of solution.

Here are details. Namely, Instead of chosing points on the whole shpere, it is sufficient to chose these points among $2n$ points of intersection of the sphere with generic lines $L1,...,L_n$. We just need to chose one point on one line. We call these $2n$ points $P_1, -P_1,...,P_n, -P_n$

Lemma. For generic $L_i$ there will be only two choices of n points $\pm P_i$, such that the obtained simplex is not contained in the demi-sphere.

Proof for n=4. It is sufficient to check this statement for the verticies of the regular cube. Indeed, for generic 4 lines in $R^3$ there is a linear transformation that takes these lines to the axes of the cube.

"Proof" for any n. For n generic lines in $R^{n-1}$ it is alway possible to send them to the lines generated by vectors $1,0,...,0$,... $0,0,...,1$ and $1,1,...,1$. It is sufficient to check the lemma for 2n points repersenting intersections of these lines with $S^{n-2}$.

From this lemma we get the answer. Number of all choices of $n$ points is $2^n$, two choices are bad, so we get $(2^n-2)2^{-n}$.