Can Manifold Markets solve this math problem?

Ṁ100Ṁ912

resolved Feb 16

Resolved as

88%

ALL

4 points are chosen uniformly at random on the surface of a sphere. What is the probability that there is a hemisphere containing all 4 points? This question uses Manifold Markets' PROB resolution, and resolves to the probability asked for in the above problem. Feb 16, 12:01am: The correct answer (as many commenters pointed out, and as the market seems to have inferred) is 7/8 = 87.5%. I rounded this to 88% when resolving. The question is equivalent to the probability that the tetrahedron formed by the 4 points does not contain the sphere's center, which appeared as question A6 on a past Putnam Exam. The very elegant solution is presented by 3Blue1Brown in this video: https://www.youtube.com/watch?v=OkmNXy7er84

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I don't understand how the PROB resolution works but it looks like it actually is not in my favor to buy NO to correct it down to 87%?

https://www.mathpages.com/home/kmath327/kmath327.htm

The first 3 points always fit in a hemisphere, and they define a triangle on the sphere. If you draw the great circle for each edge, they intersect again on the opposite side to form the opposite of the original triangle. If the 4th point falls outside this opposite triangle, it works (take the hemisphere defined by one of the great circles you drew). If it falls inside the opposite triangle, then there is no hemisphere that includes it and the 3 original points. So it boils down to what is the average area of a triangle on a sphere?

That's what I thought at first too, but then I imagined 4 points all equidistant to each other (i.e. forming an internal tetrahedron) - you can't do it. Starting by putting the points opposite to each other actually makes it easier in the long run!

With two points, you can choose the hemisphere to contain both, since the farthest apart they could be is directly opposite. You could always include a third point by rotating the hemisphere that contains the first two. In fact, since you can put either edge of the hemisphere on the third point, you can additionally cover either half of the sphere, meaning any possible fourth point. I think this means the answer is 100%! Cool math problem.