Four random points on a sphere, all in one hemisphere?
Four points are chosen independently and uniformly at random on the surface of a sphere.
What is the probability that all four lie in some single hemisphere? Equivalently, what is the probability that the tetrahedron they form does not contain the centre of the sphere?
First warm up with three points: is it always possible to find a hemisphere containing three given points?
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Three points always fit in a hemisphere (the great circle through any two of them can be rotated slightly). For four, use the same trick as for random points on a circle: choose four random lines through the centre first, then flip a coin for each line to decide which of its two ends to use. Count how many of the 16 end-choices give a tetrahedron containing the centre.
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