Randomly choose $n+1$ points on a $S^{n-1}$, probability of $n$-simplex containing center

calculusgeometric-probabilityprobability theory

Randomly choose $n+1$ points on a $S^{n-1}$(surface of ball in $n$-dim space). What's the probability that the $n$-simplex formed by these $n+1$ points contain the center of the sphere?

I conjecture that the result is $\frac{1}{2^n}$.

For $n=2$, it's easy:

enter image description here

$$\int_0^{2\pi} \frac{1}{2\pi} \frac{\pi-|\pi-\theta|}{2\pi}d\theta = \frac{1}{4}$$

For $n=3$, I compute and the result is $1/8$ :

By the same method, but need to use spherical triangle area formula and spherical triangle cosine rule

The detailed computation is following:

enter image description here

Very complicated function integration, I can hardly imagine that such a numerical integration has analytical solution $1/8$. And this complicated method is hard to be generalized to higher dimension. There should be some simple explanation and can be generalized to higher dimension case.

Best Answer

In your picture, you've first chosen two points on the circle, then drawn the diagonals, then found the probability that the third point is on the correct cord. This will get difficult to generalize.

Instead, first choose the two diagonals, then choose the third point, then find the probability that the first two points are on the "correct ends" of the two diagonals. The probability is immediately seen to be $\frac14$.

This method generalizes without much issue to higher dimensions: pick $n$ diagonals and the $(n+1)$th point at random, then place the first $n$ points on the ends of their respective diagonals. This gives you the answer $\frac1{2^n}$ that you conjectured.

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