The vast majority of human evolution happened in non-humans

Don't take this the wrong way, but it seems like you did not actually learn to cope

Cope, yes. Thrive, no. Surviving forty seven years alone at least counts as coping.

Let me put it this way, you do not seem to have learned to cope very well. Actually focusing on and learning to cope was a big improvement for me.

Can you elaborate on your hypothesis? Would them being "still there" imply the possibility of treatment to enable their effectiveness?

How does WSL1 do it then?

Anyway, the section you are quoting makes no claim as the the permitted granularity.

Explained in the very first sentence, "This is the quarterly links ‘n’ updates post, a selection of things I’ve been reading and doing for the past few months."

I missed that sentence too. I guess the large heading starting with "(1)" drew my eyes first and felt like the natural place to start reading, while the short sentence or two above it in smaller text subconsciously felt skippable (and was skipped). Even when I went back to read the first sentence, I had to kind of force myself to read the stuff above the first large heading. How odd!

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)

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.

    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

Genuinely not sure if this is wrong or if TFA is.

This result is out from the article by a factor of pi/3. This is the multiplicative difference between his inner integral with all the sins 24pi^2 and the GP's observation that 3 points on the chosen circle have density (2 pi r)^3 = 8pi^3 r^3.

(The article had already covered the r^3 in another part of the calculation.)

I'm trying to figure out an intuitive explanation as to why the work with the inner Jacobian is needed or an argument as to why it isn't.

Anyone want to simulate this accurately enough to distinguish between 40% and 41.9% probability? 5000 samples should be more than enough.

There's actually a second post on exactly that [0]

https://blog.szczepan.org/blog/monte-carlo/

You pointed out one error in your own answer, here's another. I haven't figured out the adjustments, but presumably they cancel.

You've done

∫₀¹ ∫₀¹⁻ᶜ (2π r)³ dr dc

However,

- the `dr` integral is assuming that the radii are uniformly likely in [0, r]

- the `dc` integral is assuming that the centers are uniformly likely in [0, c]

You need to wait these integrals by the conditional probability distributions.

Sorry, this was incorrect. In this integral we're just calculating the geometrical question, what is is the 5d volume of the collection of 3 points whose circumcenter is on {(x, 0) : x in [0, 1]}. So the "overall" probability of the problem has nothing to do with it.

One can discuss what “choosing three points independently and uniformly at random from the interior of a unit circle” means, but whatever you pick, I don’t think that method is doing it.

Doesn’t it have half its circle centers have 0 < c < ½, while that covers only a quarter of the area of the unit circle?

Damn! I read your answer before bed and actually had trouble sleeping trying to understand it!

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)

MSR is a somewhat independent org; you should be making predictions based on other MSR projects

They may be trying to suggest that, that claim does not follow from the quoted statement.

Alternative POV: the better they profile, you the better they can slip the scams past your defenses