pr.probability - What is the probability that 4 points determine a hemisphere ?

The probability is 7/8ths.

Consider throwing 3 darts at a sphere on average the darts will land with one on each of the ends of the Cartesian coordinates, i.e. (0,0,1), (0,1,0), (1,0,0). Or on average the SA of the spherical triangle made using the three darts as vertices will be 1/8th of the sphere volume.

This is easy to verify. The position the dart lands can be described using 2 coordinates and the equation for a sphere in Cartesian. Randomly choose an x value between -1 and 1, randomly choose a y value between -1 and 1 and the equation for a sphere will give you the z value.

On average x will be 0, y will be 0 and z will be 1.

Do this using the others and you can see why the triangle will on average have a SA of 1/8th of sphere volume.

Now consider placing a great circle around any 2 points. There are 3 ways this can be done.
On average These 3 great circles will make up the x, y and z planes. Or better to describe as, on average these three great circles will be on orthogonal planes.

So there are 8 octant to choose from, Sa of 7 of these octants can be included in the same hemisphere as the 3 points by choosing different great circles. So only if the 4th dart lands in the 8th octant do we not have a great circle that can be used to split the sphere into 2 hemispheres encompassing all 4 darts.

The 8th octant will be the dipodal spherical triangle of the "average position of the darts landing. that is draw lines through the darts, through the centre of the sphere and make another spherical triangle using the intersection of the before mentioned lines on the opposite side of the sphere.

Think about it, that is the only octant that can not be encompassed using any of the three great circles.

and again

throw 3 darts, 3 great circle can be drawn, on average the planes these three great circles lie on will be orthogonal. The intersection of these three planes creates 8 octants. There can be a spherical triangle drawn around the three darts. If and only if the fourth dart lands on the antipodal spherical triangle of the first three darts will it not be in the same hemisphere.