Four (non-coincident) points on the unit sphere determine a tetrahedron. What is the expected value of the volume of such a tetrahedron--the volume of the sphere itself being $\frac{4 \pi}{3} \approx 4.18879$?
This is a variant--stripped of the quantum-information-theoretic details there--of a problem I just posed in the last paragraph of QuantumSteeringEllipsoidVolume .
In TetrahedronVolume, the volume of a tetrahedron is expressed as \begin{equation} \frac{1}{6} |\mbox{det}(\bf{a-d},\bf{b-d},\bf{c-d})|, \end{equation} where the vertices of the tetrahedron are ${\bf{a}}=\{a_1,a_2,a_3\}, {\bf{b}}=\{b_1,b_2,b_3\},{\bf{c}}=\{c_1,c_2.c_3\},{\bf{d}}=\{d_1,d_2,d_3\}$.
Can/should each vertex be considered as independent of the other three?