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.
reply 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
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.
reply(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.
There's actually a second post on exactly that [0]
replyDamn! I read your answer before bed and actually had trouble sleeping trying to understand it!
replyThanks for editing your answer though. The thug got you in the end, but you saved me in the process.
took me a few reads but this is indeed correct (lol)
reply