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Sep 14 at 2:21 comment added Dan @Tyo There is already a link to the cross-post at the end of the question :)
Sep 14 at 2:14 comment added Tyo Cross posted here. math.stackexchange.com/questions/5089940/…
Aug 18 at 2:44 answer added Theguyabovemeislying timeline score: 0
Aug 14 at 4:35 history edited Dan CC BY-SA 4.0
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Aug 12 at 6:34 answer added Kilian timeline score: 2
Aug 11 at 1:57 history edited Dan CC BY-SA 4.0
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Aug 11 at 0:16 comment added Dan @Lucenaposition The vertices of a pentagram are random points on a circle. Let $a,b,c$ be the areas of three consecutive triangular "petals". Which is more likely to be larger: $a^2$ or $bc$? Simulations suggest that $bc$ is larger approximately $50.005\%$ of the time. (I don't think it's exactly $1/2$.)
Aug 10 at 23:14 comment added Lucenaposition @Dan You could also post some where they were not exactly equal but very close to equal. These could also be interesting.
Aug 10 at 23:12 comment added Dan @Lucenaposition Yes, I knew by integration. First, I did simulations that suggested they were equally likely. Then I worked out the integrals. Then I posted the question.
Aug 10 at 23:10 comment added Lucenaposition @Dan you knew by integration they were equally likely, right?
Aug 10 at 23:10 comment added Dan @Lucenaposition I knew they were equally likely before I posted the question.
Aug 10 at 23:08 comment added Lucenaposition @Dan did you know they were equally likely when you posted the question, or did you only find out later?
Aug 10 at 17:12 answer added Dan timeline score: 9
Aug 10 at 16:32 comment added Benjamin Wang @Dan maybe this question should also be on Math StackExchange.
Aug 10 at 10:11 comment added Lucenaposition @Dan Post the integration answer as an answer.
Aug 10 at 6:28 comment added Pranay Ah I see. I found some different integral expressions for the probabilities and Mathematica gave me their exact values in terms of polylog function, which is also what you see. The numerical value you quote agrees with what I got.
Aug 10 at 6:20 comment added Dan @Pranay rot13 V nyfb hfrq vagrtengvba gb fubj gung gur cebonovyvgvrf ner gur fnzr, ohg V fhfcrpg gurer vf n zber ryrtnag zrgubq. Fb vg'f n chmmyr sbe zr, gbb. Ol gur jnl, obgu cebonovyvgvrf ner $\frac{2}{\pi^2}\int_0^\pi\operatorname{arccot}\left(2\sqrt2\sin x\right)dx=\frac{2}{\pi^2}\left(\frac38\left(\pi^2-\ln^22\right)-\operatorname{Li}_2\left(-\sqrt{2}\right)-3\operatorname{Li}_2\left(\frac{1}{\sqrt{2}}\right)\right)=0.3871287106\dots$
Aug 10 at 5:28 comment added Pranay @Lucenaposition. My answers also agree with yours up to the first three decimals.
Aug 10 at 5:22 comment added Pranay @Lucenaposition. I see. I found integral expressions for both the probabilities (it’s much simpler for BC than AB) and then computed them numerically. They are equal to whatever decimal places I chose. So I believe they are the same. I have some ideas on showing the integrals are the same but it feels far too complicated and not what Dan intended.
Aug 10 at 5:20 comment added Lucenaposition It was, and I hoped nothing went wrong.
Aug 10 at 5:18 comment added Pranay @Lucenaposition. Was this result obtaining by picking points on the circles (pseudo)randomly?
Aug 8 at 17:35 vote accept Dan
Aug 8 at 21:33
Aug 8 at 15:52 answer added Nuclear Hoagie timeline score: -3
Aug 8 at 9:25 history became hot network question
Aug 8 at 1:33 history edited Lucenaposition CC BY-SA 4.0
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Aug 8 at 1:05 comment added Lucenaposition [Not an answer]: If my code works, I think the probability of intersection is the same (or very similar) and around 0.387.
Aug 7 at 23:04 history edited bobble CC BY-SA 4.0
removed unnecessary MathJax https://puzzling.meta.stackexchange.com/questions/7421/a-brief-guide-to-mathjax-for-pse-users
Aug 7 at 22:55 history asked Dan CC BY-SA 4.0