The original question is here:
"A–6 Four points are chosen at random on the surface of a sphere. What is the probability that the center of the sphere lies inside the tetrahedron whose vertices are at the four points? (It is understood that each point is independently chosen relative to a uniform distribution on the sphere.)"
An YT video by 3Blue1Brown first shows the case in 2d case,when you choose 3 point from the circumference of a circle,What is the probability that the center of the circle lies inside the triagle whose vertices are at the three points?
The answer of this question is $\frac{1}{4}$ ,and the answer of A6 is $\frac{1}{8}$.It's easy to notice that when we choose 2 pionts on a line ,the probability that the center of the line lies inside the line whose vertices are at the two points is $\frac{1}{2}$.So I have this question:
- When the question is in a 4D or more Ds sphere,What is the probability that the center of the sphere lies inside the tetrahedron whose vertices are at the five points?And how to prove that?
My guess is like:$\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32}...$
This question is too hard for me .
