Tesseract Pentatope Picking

First, some definitions are given:

Simplex is the generalization of line-segments ($1$d) or triangles ($2$d) to $n$-dimensions. Tetrahedron is the well-known $3$-simplex. Pentatope then is the $4$-simplex.

Similarly, hypercube is the generalization of squares ($2$d) or cubes ($3$d) to $n$-dimensions. Tesseract is the $4$-dimensional hypercube.

Then the question builds up as follows:
In line line picking, the expected length of picked line-segment is shown to be $\frac{1}{3}$.
Up one dimension, namely in square triangle picking, the expected area of picked triangle is analytically given as $\frac{11}{144}$.
Up one more, namely in cube tetrahedron picking it gets a bit complicated, but there is again an analytic answer for the expected volume of picked tetrahedron, $\frac{3977}{216000}-\frac{{\pi }^2}{2160}$.

What about going one more up, namely tesseract pentatope picking? Since an analytic answer is hard to find, devise a numerical solution that gives the expected content of picked pentatope rounded to $6$ decimal places.

There might be some inaccuracy in my calculation for the time being...