EDIT: ok this was nagging at me for a while as something being off, I think this is actually wrong (in some way that must cancel out to accidentally get the right answer) because I need to multiply by 2 pi c to consider all rotations of centers around (0,0) at a given radius, but then my integral no longer works. Ah well, that's what I get for trying to method act and solve quickly, I guess the hooligan stabs me. I think at least this approach done properly could save some dimensions out of the Jacobian we need to calculate. Original post below:

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)

▲ Someone 5 hours ago | parent | next [-]

One can discuss what “choosing three points independently and uniformly at random from the interior of a unit circle” means, but whatever you pick, I don’t think that method is doing it.

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?

▲ wedog6 11 hours ago | parent | prev | next [-]

I think it's fairly straightforward to adapt your method. Given circle center c you just need to multiply by 2 pi c to get all the circles.

    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

Genuinely not sure if this is wrong or if TFA is.

▲

wedog6 10 hours ago | parent [-]

This result is out from the article by a factor of pi/3. This is the multiplicative difference between his inner integral with all the sins 24pi^2 and the GP's observation that 3 points on the chosen circle have density (2 pi r)^3 = 8pi^3 r^3.

(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.

	▲	quibono 9 hours ago \| parent [-]
		There's actually a second post on exactly that [0] https://blog.szczepan.org/blog/monte-carlo/

▲ clutter55561 11 hours ago | parent | prev | next [-]

Damn! I read your answer before bed and actually had trouble sleeping trying to understand it!

Thanks for editing your answer though. The thug got you in the end, but you saved me in the process.

▲ ccvannorman 15 hours ago | parent | prev [-]

took me a few reads but this is indeed correct (lol)