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:<p>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)
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.<p><pre><code> 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
</code></pre>
Genuinely not sure if this is wrong or if TFA is.
Damn! I read your answer before bed and actually had trouble sleeping trying to understand it!<p>Thanks 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)
The intro strongly reminded me of <a href="https://existentialcomics.com/comic/604" rel="nofollow">https://existentialcomics.com/comic/604</a><p>Really enjoyed this keep writing!
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.
We were told a (kind of) similar story in high school: <a href="https://medium.com/intuition/explain-this-or-i-will-shoot-you-mathematics-saved-a-physicists-life-568c3d9f5bc5" rel="nofollow">https://medium.com/intuition/explain-this-or-i-will-shoot-yo...</a>
I've got an idea for a simpler approach, but I've forgotten too much math to be able to actually try it.<p>The idea is to consider the set A of all circles that intersect the unit circle.<p>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.<p>The constant of proportionality should be such that the integral over all the circles is 1.<p>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.<p>The ratio of these two integrals should I think be the desired probability.
I like this reasoning. Define a probability distribution on all circles of (x,y,r>0) based on how likely a given circle is. Then we can just sum the good circles and all the circles.<p>And the probability distribution is simple: a given (x,y,r) is as likely as its circumference in the unit circle.<p>Reasoning: Let C:(x,y,r) a given circle. We want to know how likely is it that the circle on 3 random points are close to it, closer than a given value d. (A d wide ball or cube around C in (x,y,r) space. Different shapes lead to diffferent constants but same for every circle.) The set of good 3 points is more or less the same as the set of 3 points from the point set C(d): make C's circumference d thick, and pick the 3 points from this set. Now not any 3 points will suffice, but we can hope that the error goes to 0 as d goes to 0 and there is no systematic error.<p>Then we just have to integrate.<p>ChatGPT got me the result 2/3, so it's incorrect. I guess the circumference must not be the right distribution.
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.
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…
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
It’s funny because it’s true.
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.