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)
ccvannorman 3 hours ago [-]
took me a few reads but this is indeed correct (lol)
I would calculate that the probability of a mathematician doing anything practical like operating a gun is even lower than the probability that I could solve the riddle (even with pen, paper, wikipedia and a liter of coffee on a good day), and choose to sprint off.
layman51 4 hours ago [-]
When I first read the title, I thought it was gonna be about a book similar to one I heard about called “Street Fighting Mathematics” and it would be about like heuristics, estimation, etc. but this one seems to be about a specific problem.
tzs 2 hours ago [-]
I've got an idea for a simpler approach, but I've forgotten too much math to be able to actually try it.
The idea is to consider the set A of all circles that intersect the unit circle.
If you pick 3 random points inside the unit circle the probability that circle c ∈ A is the circle determined by those points should be proportional the length of the intersection of c's circumference with the unit circle.
The constant of proportionality should be such that the integral over all the circles is 1.
Then consider the set of all circles that are contained entirely in the unit circle. Integrate their circumferences times the aforementioned constant over all of these contained circles.
The ratio of these two integrals should I think be the desired probability.
mehulashah 3 hours ago [-]
So, I’m left wondering why he did it the hard way.
fancyswimtime 3 hours ago [-]
I'd prefer a world like this; higher levels of whimsy accompanied with greater danger
derelicta 5 hours ago [-]
What's even scarier than such encounter, is that I personally know some people who would survive it. Unfortunately, I'm not one of them.
Rendered at 07:22:09 GMT+0000 (Coordinated Universal Time) with Vercel.
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)
Really enjoyed this keep writing!
The idea is to consider the set A of all circles that intersect the unit circle.
If you pick 3 random points inside the unit circle the probability that circle c ∈ A is the circle determined by those points should be proportional the length of the intersection of c's circumference with the unit circle.
The constant of proportionality should be such that the integral over all the circles is 1.
Then consider the set of all circles that are contained entirely in the unit circle. Integrate their circumferences times the aforementioned constant over all of these contained circles.
The ratio of these two integrals should I think be the desired probability.