On 19/10/2014 04:47, Bob Hearn wrote:
Fun problem from the new book Inside Interesting Integrals: given a circle, pick three points inside it at random (uniformly). These points define another circle. What is the probability that this circle lies entirely inside the original circle? The theoretical answer is 2pi/15 = 0.418879... But Monte-Carlo simulations strongly suggest that the answer is in fact exactly 2/5. So what gives?
Original derivation from A Treatise on the Integral Calculus, Edwards, 1922:
https://dl.dropboxusercontent.com/u/70818776/edwards-circles.png
My MC source code (Nahin, author of Inside Interesting Integrals, also did MC simulations, which give the same result, and pointed out the discrepancy):
https://dl.dropboxusercontent.com/u/70818776/interesting-circles.cpp
Initial pure-prejudice reaction: A mathematical error is more likely than a numerical error of such magnitude. Two independent MC runs are unlikely to have the same mathematical error in them. So my guess is that Edwards made a mistake. Also, I kinda wouldn't expect a factor of pi in the probability. Initial slightly-more-than-prejudice reaction on looking at Edwards's alleged proof: That looks really fishy to me; it wouldn't surprise me if what he's actually computed has (e.g.) the probabilities implicitly weighted in proportion to the radius of the "inner" circle or something like that. So my money is on the simulations' answer being correct. (Either because the exact probability is 2/5, or because it's something else close to 2/5 and not so close to 2pi/15.) (But I haven't looked at the MC code, nor have I thought carefully enough about Edwards's proof to be sure it can't be made rigorous without changing the answer.) -- g