Four points around the centre

Idea. A uniform point on the sphere is a random diameter plus a fair choice of one of its ends. Fix the four diameters: among the 16 equally likely choices of ends, exactly 2 give a tetrahedron around the centre.

Step 1: two-stage choice. Pick four independent random lines through the centre, with unit direction vectors $v_1,\dots ,v_4$, then independently a sign ${\epsilon }_i=\pm 1$ for each. The points ${\epsilon }_i{v}_i$ are four independent uniform points on the sphere.

Step 2: when is the centre inside? The centre is inside the tetrahedron with vertices $P_1,\dots ,P_4$ when $0=\sum {\lambda }_i{P}_i$ with all ${\lambda }_i>0$. Four vectors in space are linearly dependent: with probability 1 there is a single relation $\sum c_i{v}_i=0$, up to scaling, and no $c_i$ is zero.

Step 3: count the signs. With $P_i={\epsilon }_i{v}_i$, the relation reads $\sum (c_i{\epsilon }_i)P_i=0$. The centre is inside exactly when all the numbers $c_i{\epsilon }_i$ have the same sign, which happens for exactly 2 of the 16 sign choices: ${\epsilon }_i=\mathrm{sign}(c_i)$ for all $i$, or the opposite for all $i$.

Step 4. So the probability is $\frac{2}{16}=\frac{1}{8}$, whatever the lines are.

In the plane. The same argument with three points on a circle gives $\frac{2}{8}=\frac{1}{4}$ for the triangle to contain the centre, which matches $1-\frac{3}{4}$ from the semicircle problem.