Geometric Probability Problem

This is equivalent to the probability that a point chosen randomly from a 2x2x2 cube will be inside the unit sphere, but outside its inscribed octahedron.

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The answer is then

((Volume of unit sphere)-(Volume of inscribed octahedron))/(volume of 2x2x2 cube)

Plugging things in, we get

((4π/3) - (4/3))/(8) = (π-1)/6 ~ .36

A twenty-five million point Monte Carlo:

$l=\text{Select}[\text{RandomReal}[\{-1,1\},\{25000000,3\}],\text{\$\#\$1}.\text{\$\#\$1}<1\&]$

$r=\text{Table}[\text{Boole}[\text{Total}[\left| p\right| ]>1.],\{p,l\}]$

$\frac{\text{Total}[r]}{25000000} \Rightarrow \frac{8922091}{25000000} \approx 0.356884$

The question that I had because of the wording was whether the denominator was 25000000 or 13086730 (the length of the list $r$ ).