Nice effort Dan, I do like a nice maths-debate...(ahhh humor at last). I do have a couple of concerns about what you wrote...
your part 1) contradicts 2), as in part 1) all darts will lie on the same great circle
When three darts land on a great circle then, as an entire sphere can be broken into 2 hemispheres with no volume remaining then the three darts must still be in one of the hemispheres. In this case the darts can be allocated to EITHER hemisphere. i.e. 1 hemisphere consists of the infinitesimally thin great circle and the hemisphere above and the other everything below the great circle. As the great circle has a width not dissimilar to 0 then both great circles will still be of the same volume.
The probability is not dependent on the first 3 darts, i.e. the question is NOT: after throwing 3 darts what is the prob the fourth will land on the same hemisphere, which the solution to will obviously be 50%.
But, yes the tighter the spherical triangle of the first three darts (i.e. smaller SA) the higher then chance the 4th will land in the same hemisphere.
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The proof in annals of mathematics states:
Build a spherical triangle where the vertices are any three darts (i.e. kind of like a triangular based prism but the base is the surface of the sphere and the top of the prism is located at the center of the circle). Then reflect this spherical triangle through the center of the sphere (i.e. identical shape directly opposite). If the remaining dart is on the surface of this reflected spherical triangle then the 4 darts do not share the same hemisphere. I do not know how to prove this..... bummer as it is the crux of the proof.
If you randomly throw three darts at a sphere then on average the surface area of the spherical triangle made using the three darts as vertices will be 1/8th of the sphere volume. This result can be proved quite easily using calculus.
