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

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:

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

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:

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.

• Any reason you have that conjecture? – Arctic Char Apr 8 at 19:49
• This is the problem considered in this 3blue1brown video. It is only solved for $n=2$ and $3$, but the method generalizes easily. – Arthur Apr 8 at 19:53
• @ArcticChar I compute by brute force the case of $n=2$ the result is $1/4$ and the case $n=3$ the result is $1/8$ – maplemaple Apr 8 at 19:53
• @Arthur Yes, I compute above the case of $n=2, 3$. In $n=3$, by numerical integral is $0.125 = 1/8$. But I can hardly imagine this method can get analytical solution. And this method can hardly to generalize to higher dimension case. – maplemaple Apr 8 at 19:58
• Check out Wendel's Theorem for a generalization. – Mike Earnest Apr 9 at 0:20

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.