Thanks for editing your answer though. The thug got you in the end, but you saved me in the process.
int 0..1 2 pi c int 0..(1-c) (2 pi r)^3 dr dc / pi^3
int 0..1 2 pi c int 0..(1-c) (2 r)^3 dr dc
int 0..1 2 pi c 2 (1-c)^4 dc
-4 pi int 0..1 (1-g) g^4 dg
4 pi (1/6 - 1/5)
4 pi / 30
2 pi/ 15
(The article had already covered the r^3 in another part of the calculation.)
I'm trying to figure out an intuitive explanation as to why the work with the inner Jacobian is needed or an argument as to why it isn't.
Anyone want to simulate this accurately enough to distinguish between 40% and 41.9% probability? 5000 samples should be more than enough.
Doesn’t it have half its circle centers have 0 < c < ½, while that covers only a quarter of the area of the unit circle?
You've done
∫₀¹ ∫₀¹⁻ᶜ (2π r)³ dr dc
However,
- the `dr` integral is assuming that the radii are uniformly likely in [0, r]
- the `dc` integral is assuming that the centers are uniformly likely in [0, c]
You need to wait these integrals by the conditional probability distributions.
Much more elegant: consider every circle that fits inside the unit circle, and we will work backward to find combinations of points. We only need consider centers on the x axis by symmetry, so these are parameterized by circle center at (0,c) and radius r with 0<c<1 and 0<r<1-c. Each circle contributes (2 pi r)^3 volume of triples of points, and this double integral easily works out to 2 pi^3/5 which is the answer (after dividing by the volume of point triples in the unit circle, pi^3)