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)

The intro strongly reminded me of https://existentialcomics.com/comic/604

Really enjoyed this keep writing!

We were told a (kind of) similar story in high school: https://medium.com/intuition/explain-this-or-i-will-shoot-yo...

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.

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.

Ah, 24, reminds me of ole days the lattice of those math alleys had a monstrous moonshine leeching into reality stranger than we’d care to code…

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.

It’s funny because it’s true.

So, I’m left wondering why he did it the hard way.

I'd prefer a world like this; higher levels of whimsy accompanied with greater danger

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.