Hyper-intriguing Ornament VI - A gentle ending

Since this one is very similar to Part V, I will submit a very similar solution, Comrade ;)

At this point, it is all just a matter of "crunching the numbers". Let's summarize what we found in problem I, at the beginning of the saga. If $B$ is the volume of the ball and $C$ is the volume of the cube, then $C=8^4=4096,\ B\approx 3084,\ B\text\ C=8V_{caps}\approx 379,\ B\cap C\approx2705,\ C\text\ B=1391,\ B \cup C= 4475$ The probability we seek is $\frac{B\text\ C}{B \cup C}\approx \boxed{8}\text\%$

The Hyper-intriguing Ornaments have been fun (for a crowd of three,essentially), but every good thing must come to an end. Thank you, Comrade!

$\text{cap}_4(r,h)=\frac{1}{12} \pi \left(-4 \sqrt{-h^7 (h-2 r)}-2 r \left(r \sqrt{-h (h-2 r)} (h+3 r)-6 \sqrt{-h^5 (h-2 r)}\right)+6 r^4 \tan ^{-1}\left(\frac{h-r}{\sqrt{-h (h-2 r)}}\right)+3 \pi r^4\right)$

Because this is 4 dimensional, there are 8 caps, $8\ \text{cap}_4(5,1) = \frac{2}{3} \pi \left(1875 \pi -2232-3750 \tan ^{-1}\left(\frac{4}{3}\right)\right) \approx 379.356021363$ .

$\frac{2 \pi \left(1875 \pi -2232-3750 \tan ^{-1}\left(\frac{4}{3}\right)\right)}{3 \left(4096+\frac{2}{3} \pi \left(1875 \pi -2232-3750 \tan ^{-1}\left(\frac{4}{3}\right)\right)\right)} \approx 0.0847656$