Tetrahedron Inception

Solution

Let the four random points on the unit sphere be \[ P_1,P_2,P_3,P_4. \] We want the probability that the origin lies inside the tetrahedron they determine. The key geometric idea is this: The origin is inside the tetrahedron spanned by \(P_1,P_2,P_3,P_4\) exactly when the four points are not all contained in any open hemisphere. Why is that true? If all four points lie in some open hemisphere, then there is a plane through the origin with all four points on the same side. That means the convex hull of the points is entirely on one side of that plane, so it cannot contain the origin. Conversely, if they are not all contained in any open hemisphere, then the origin must lie in their convex hull. Since four generic points on the sphere determine a tetrahedron, this is exactly the event we want. So the problem becomes: \[ \Pr(\text{origin is inside the tetrahedron}) = \Pr(\text{the 4 points are not all in one hemisphere}). \] Now we use a standard result: For \(n\) random points on the sphere \(S^{d-1}\), the probability that they all lie in some hemisphere is \[ 2^{-(n-1)} \sum_{k=0}^{d-1} \binom{n-1}{k}. \] Here we are in \(\mathbb R^3\), so \(d=3\), and we have \(n=4\). Thus \[ \Pr(\text{all 4 points lie in some hemisphere}) = 2^{-3}\left[\binom{3}{0}+\binom{3}{1}+\binom{3}{2}\right]. \] Compute this: \[ \binom{3}{0}=1,\qquad \binom{3}{1}=3,\qquad \binom{3}{2}=3, \] so \[ \Pr(\text{all in some hemisphere}) = \frac{1+3+3}{8} = \frac{7}{8}. \] Therefore, \[ \Pr(\text{origin lies inside the tetrahedron}) = 1-\frac{7}{8} = \frac{1}{8}. \] So the probability is \( \frac{1}{8} \).