/sci/ - Science & Math

>>15714580
>random point in a circle
Is that even possible? In terms of a distribution that isn't biased by the coordinate system you use?

>>15714613
uniformly distributed

Anyway, I did some solving:
P=0 when n=1

A spiral would have the most lines with no intersection.

>>15714619
But how do you even pick a random point that way? Polar skews it one way and cartesian skews another.

Lol. Also P=0 when n=2

>homework thread

>>15714634
You pick a random number between 0 and 1 for R and then you pick a random number between 0 and 2pi for theta. Alternatively, you pick a number between 0 and 1 for y and then pick a random number between -(1-y^2) and (1-y^2) for x.

>>15714580
[math]
≈2^{-(n-2)^{3/2}}
[/math]

>>15714649
>≈

>>15714643
But if you pick using polar coordinates like that, then 50% of your points will be in 25% of the circle.

>>15714696
If "n" is a random number between zero and one, then "sqrt(n)" is the radius and n*2*pi is the angle. So the coordinates of a uniformly random point inside a unitcircle are then (sqrt(n)*cos(n*2*pi), sqrt(n)*sin(n*2*pi)).

Here's a video for more insight
https://www.youtube.com/watch?v=4y_nmpv-9lI

>>15714663
100% sure the exponent is between 5/4 and 3/2. Might be too hard to prove, though.

>>15714770
The guy in the video just says how polar coordinates work lol. I'm saying that if you draw 2 small circles of equal area in the main circle, 1 around the center and 1 out by the circumference, then picking random polar coordinates in the way described should oversample the inner circle and undersample the outer circle. How could it not?

>>15714803
If the distance between the point and the center of the circle is a random number between zero and one (n), and the angle is also a random number between zero and 2pi radians, then yes you're right. But you can correct it by making the distance a square root of 'n' instead of just 'n'.

>>15714819
That makes sense, thank you.

>>15714634
let starting area be a circle
1. cut the area vertically and horizontally so that all resulting shapes are of same area
2. pick a random shape and say it will contain your random point
3. back to step 1.

[math]\sum_{i=1}^n = 100/i[/math]

>>15714580

can someone actually post their working out for their solution? am dumb

For n=3. Is it fair to argue:
The probability of picking 4 points such that the lines can't intersect in any order is 35/12/π^2 = X.
Otherwise, there are 4! ways to order the points, 8 of which cause an intersection.
So (1-X)/3 = 1/3 - 35/36/π^2 ~ 0.2348.. ?

>>15714580
its 50 50 either it diss or it dissnt, and this cause you missed to define a lot of properties of your riddle.

>>15714927
idk bro but riddes with no bounderys doesn exist for me.

>>15714898
>>15714920
The problem is symmetric for n=3, implying 50% chance.

>>15714634
if you are not convinced about the arguments for the scaled samples
just sample from a unit square and discard the sample if it is outside the circle and repeat until you get a sample inside it

>>15714580
instructions unclear, gone full Diogenes and drew a peener and a smiley on it.

>>15714580
it's an abstract le happy merchant meme

>>15714936
How so? At n=3 you've picked 4 points at random. You'll either have a triangle with a point inside (probability X) or a convex quadrilateral (probability 1-X). X is a well known solution. If it's the triangle with a point inside, it doesn't matter which order you picked the points in, they'll never form an intersecting chain. Throw out X. If it's the convex quadrilateral, the chain intersects 1/3 of time depending on the order you pick the points.

>>15714949
Stop thinking about 4 points. The problem for n = 3 is 2 random lines, AB and CD, because you can ignore the line BC (the chance to intersect that line is 0).

>>15714970
That's a fair description but AB and CD are still 4 points picked in a disk at random. The probability that AB will intersect with CD if ABCD is concave is 0, if convex, 1/3, and the probability that ABCD is convex is (1-X).

>>15714970
I'm wondering if the same logic used in n=3 could be somehow utilized to solve n=4

>>15714997
fair point. should be around 1/3 minus some small number, then. The probability that ABCD is convex is much higher.

I propose we first work on a special case of this problem
Suppose you can only pick points on the outside edge of the circle
Seems doable, each point you add leaves you with 2 smaller parts of the circumference to choose from
I conjecture the probability to be p(n) = (n-3)/(n-2)

>>15715042
X = 1 - 35/12/π^2 ~ 0.7044..

>>15715047
*I mean X = 35/12/π^2 . 1-X ~ 0.7044..

>>15715046
n was supposed to be the number of points with my conjecture btw

>>15715053
I totally typed shit wrong into wolfram alpha and somehow didn't realize
This is what I was going for with the conjecture:
https://www.wolframalpha.com/input?i=product+%282%5Ex-1%29%2F2%5Ex%2C+x%3D1..n

The solution brings up the "q-Pochhammer function" which is related to enumerating partitions, which might be useful for what we're doing :https://en.wikipedia.org/wiki/Q-Pochhammer_symbol

>>15714580
Every choice of n points is equally likely.
The question can kind of be rephrased as:
Given n points in a circle, how many ways can you assign an ordering to the points so the lines between sequential points don't intersect?
If the answer is o(n!) then the probability tends to 0.
There will always be at least 3 points on the boundary of the convex hull in general (all points colinear has measure 0) so you can bound it from below by 3/n!.
The argument here is pick a point on the boundary of the convex hull then rotate things so that it has the largest x coordinate.
Then choose the ordering according to lexicographic order.

You can do even better. Choose the first point on the boundary of the convex hull then consider the convex hull of the remaining points. There will be at least 2 points on the boundary of this remaining convex hull which are visible to the first point.
Choose one of them as the next point and repeat.
The better lower bound is 3*2^(n-2)/n!

>>15715024
problem is not that easy. X should be
[math]X = 35/(48*\pi^2)[/math]

>>15715052
>>15715092
https://math.stackexchange.com/questions/1243160/expected-area-of-triangle-formed-by-three-random-points-inside-unit-circle

>>15715091
If you want, you can try to find the probability for exactly k points being on the boundary of the initial convex hull given n points to get a lower bound of:
ΣP(k|n)*k*2^(n-2)/n!

>>15715096
Here's a short history of the 4th point
https://www.jstor.org/stable/2689482

i ran a simulation for n=3 and got [eqn]p \approx 0.76734 [/eqn] after a million runs

>>15715114
*hundred thousand runs

>>15715114
How did you solve for intersection

>>15715117
stole the code from https://stackoverflow.com/questions/3838329/how-can-i-check-if-two-segments-intersect

>>15715114
That's within spitting distance of 2/3 + 35/36/π^2.
Your p is that they don't intersect, yes?

>>15715125
yup. also for n=4 it drops to [eqn]p \approx 0.48516 [/eqn] (100k runs)

>>15715127
Get these for the next few n and we can try to do some asymptotic witchcraft

>>15715141
100k run. not >>15715127

[math]{3: 0.76325, 4: 0.48664, 5: 0.26764, 6: 0.1292, 7: 0.05651, 8: 0.02115, 9: 0.00794}[/math]

>>15715141
I'd expect [math]p[/math] to approach 0 asymptotically but here it is anyway (each one 100k trials):
[eqn]
n = 3: p = 0.764030000000000 \\
n = 4: p = 0.490520000000000 \\
n = 5: p = 0.267830000000000 \\
n = 6: p = 0.127820000000000 \\
n = 7: p = 0.0560100000000000 \\
n = 8: p = 0.0219500000000000 \\
n = 9: p = 0.00780000000000000
[/eqn]
>>15715172
beat me to it

>>15715178
>>15715172
looks like [math](2+\epsilon)^{x-3}[/math] for higher n

>>15715207
*[math](2+\epsilon)^{-(x-3)}[/math]

>>15715178
>>15715212
just to see what would happen I used random disjoint segments instead of the snake shape but kept the collision-checking logic the same (it doesn't check neighbouring segments in the segment array) and got this:
[eqn]
n = 3 , p = 0.765340000000000 \\
n = 4 , p = 0.471100000000000 \\
n = 5 , p = 0.247820000000000 \\
n = 6 , p = 0.110100000000000 \\
n = 7 , p = 0.0446400000000000 \\
n = 8 , p = 0.0160700000000000 \\
n = 9 , p = 0.00505000000000000 \\
[/eqn]
so i'm thinking maybe we could start off by solving this (presumably easier) problem and then add terms to get to the actual solution. but i'm a mathlet so idk

>>15715172
>>15715178
Here's the recipe boys:
https://www.youtube.com/watch?v=axL3qJO9q-8
Don Zagier is a beast.
We can also let p=1 for n=2.
If you suppose p(n) = A^n * n^a * (c0 + c1/n + ... )
p(n+1)/p(n) = A * (1 + a/n + ... )
Do his "n^8" trick (we only have enough for n^7) on the ratios to get A and a.
Once you have these you can plug them in and get c0 from p(n)/(A^n * n^a).

>>15715308
Once we get a formula we can test it at n=20 or something.

>>15714580
For n=1,2 it's 1. For n = 3, we basically want to know how likely it is for two line segments in a circle to intersect. This can be Googled, but we can easily see it's greater than 0.5; we can establish a bijection between intersecting lines and a strict subset of nonintersecting lines by reflecting the fourth endpoint about the first line with an appropriate scale factor.

>>15714580
easy

kino thread

>>15715244
good idea. If line intersection is independet we get

[math]p(n)=(p(n=3))^{n-2}*p(n-1)[/math]
with
[math]p(n=3) = 0.765 [/math]

by just multiplying independent intersections

>>15715660
>>15714580
lower bound for the problem:
[math]p(n)≈(1-\dfrac{35}{48\pi})^{\dfrac{n}{2}(n-3)+1}[/math]

>>15714580
It's 50%.

>>15715718
It is not.

>>15715864
It either intersects or it doesn't.

>>15714613
of course it's possible lmfao

>>15714613
ignore the retards, probability is just geometry as the rest of math.
uniform means constant density, or height.
Imagine cylinder of height one, base being the circle.
if the circle has area A, then the volume of the cylinder is Ah = A.
So normalize the height to 1/A such that the total volume is one.
now to find "probabilities" you integrate.
Shouldn't surprise you that the chance to pick a point, any point, is zero. the chance of subset A will be ratio of the area of A to that of the whole circle and so on.

>>15714770
But then the angle and radius are correlated.

>>15714997
>>15715127
>>15715172
>>15715178
For n=4.
There are three convex shapes you can pick with 5 points. 3 sides with 2 points inside, 4 with 1, 5 with 0. Call that (3,2) (4,1) (5,0). The probabilities of picking those shapes in a disk are a result of R. E. Miles (1971) https://doi.org/10.2307/1426176

P(3,2) = 15/(16 π^2)
P(4,1) = 65/(12 π^2)
P(5,0) = 1 - (P(4,1) + P(3,2))

Permute the point order to find how often the lines intersect in each shape.

X(3,2) : 0 = 96/120, 1 = 24/120
X(4,1) : 0 = 64/120, 1 = 42/120, 2 = 12/120, 3 = 2/120
X(5,0) : 0 = 40/120, 1 = 50/120, 2 = 20/120, 3 = 10/120

Then plug the combinatorics into Miles.

p(0) = 1/3 + 73/(48 π^2) ~ 0.4874..., which is also in spitting distance of all the simulations.

Other probabilities for 1 to 3 intersections.

p(1) = 5/12 - 325/(576 π^2) ~ 0.3594...
p(2) = 1/6 - 149/(288 π^2) ~ 0.1142...
p(3) = 1/12 - 253/(576 π^2) ~ 0.0388...

>>15716293
can confirm. i ran a better simulation for n=4 with one billion trials, got [math]p \approx 0.487436934[/math]

>>15716293
>>15716316
For n>4.
Marckert (2017) has a solution for the P of picking shapes (3, n-2) (4, n-3) ... (n, 1) (n+1, 0) in a disk.
https://doi.org/10.1214/16-BJPS315

Unfortunately, if there's more than 5 points, there's more than one X distribution for some of the shapes. So you'd need to refine Marckert before you could get any more closed forms.

Best you could do is intervals.

>>15714696
What no it's not? What are you talking about? It's two dimensional so you draw two random numbers.

>>15715308
that's a really cool method. you're welcome to implement it but >>15715691 already got a lower bound
you can use this high precision table (one billion trials)
n = 3, p = 7.652142e-01 (234785814 intersects)
n = 4, p = 4.874369e-01 (512563066 intersects)
n = 5, p = 2.672459e-01 (732754052 intersects)
n = 6, p = 1.291986e-01 (870801367 intersects)
n = 7, p = 5.601638e-02 (943983622 intersects)
n = 8, p = 2.203642e-02 (977963582 intersects)
n = 9, p = 7.953838e-03 (992046162 intersects)
n = 10, p = 2.647217e-03 (997352783 intersects)

>you should be able to solve this
Yes I should but I haven't been training for these olympiad style questions since I gave up most of my youth in my college years doing this stuff

>>15714634
P = 0 for n = 1,2,3 since the straight line after n = 2 is a set of measure zero

>>15716430
Not him but it looks like the exponent could be tweaked a bit more. Unless I'm reading it wrong, the ratio of that simulation to the bound is ~ 1.0, 1.0, 1.0, 1.1, 1.3, 1.8, 2.9, 5.6, 13, 36

>>15716496
You would have to show the exponent grows less than [math]n^2[/math] for n->inf.
Might be possible but not my area.

>>15716829
Simulation already showed it. The first difference looks like a missing log.

>>15717008
You can always choose the best fitting line for the data. That doesn't tell you anything about the problem for any n that hasn't been simulated.
Extrapolating to unknown values is a gamble that could or could not pay of.

The advantage is that we know p(n->inf)=0 so it might pay of (however, it is impossible to be certain)

>>15717087
True but it's not a gamble that there's a missing log.

>>15717100
show or prove it. otherwise it doesn't mean anything. There could be a missing gamma-, beta-, bessel-, gaussian-, log, sqrt function.

>>15717109
I solved n=3, n=4, and showed that a solution to n>4 isn't in the canon. There's a log missing in 1.0, 1.0, 1.0, 1.1, 1.3, 1.8, 2.9, 5.6, 13, 36. I don't care enough about asymptotes to spend time on it. I'm happy to point it out.

>>15717146
*at least one log missing (presumably a stack of them)

>>15717146
again whats the point if you can't show or prove it?
>>15715660 assumes independent intersections, which is obviously not true but is the simplest case for estimating a lower bound

>>15717162
I solved everything that I care about. I don't care what >>15715660 assumes. If that's you, I'm telling you there's a log missing. If you're picking a random post, ask him, not me.

>>15717177
>I'm telling you there's a log missing
There is nothing missing. It's a lower bound. Adding logs without proof is interpolation. If you have proof or an approach, I will listen. Otherwise, I don't care.

>>15717222
Don't then.

>>15714580
>I was bored so I
Why do people so often begin what they want to say with this?

>>15717308
You surround yourself with insecure people.

>>15717308
give 5 examples

>>15717311
Not him but any 5 poems by Baudelaire are sufficient.

>>15717308
Maybe because it's true?

>>15717361
It's almost never true. That phrase really only makes sense if you're doing something impulsive and quick. Something like "I was bored, so I threw a pencil at my coworker just to see how he'd react". I could also have prefaced this message with "I was bored so I replied to you", but it's not true in any meaningful sense, even though I technically probably would have been bored had I chosen to do literally nothing else instead.

>>15717385
>I'm bored
The root of math, the history of math, the opera of math, the commedia dell'arte of math, and the future of math.

>>15716409
Allowing r to range over [0, 1], the ring from r = 0.5 to r = 1 has greater area than the disc r < 0.5, yet your way of picking points makes the odds of selecting a point in the ring the same as in the disc.

>>15717322
Not to be a fag but Baudelaire's "Ennui" is not really boredom in the common sense. There's a soul crushing dread rather than idleness.

>>15717479
Everything he wrote was out of boredom. It's not the title in this case but the words he juxtaposed. (For him, boredom was a soul crushing dread that squeezed out soul, perhaps. If math is your soul, boredom squeezes out more math.)

Is it possible to put 6 points in a disk that don't have a path that intersects itself twice? I can't see how that would work but I'm very low IQ.

>>15717728
Sorry I mean random points. Nothing where 3 are collinear.

>>15717735
>colinear
This is an American board, you're welcome

>>15714580
Don't know. Too complex for an analytical solution, at least when taken at face value... but since the result doesn't depend on any particular arrangement of points, shouldn't there be an equivalent geometric problem that doesn't involve random coordinates but only the underlying aspects that affect the result?

>>15714580
can't you use pic related and calculate the area of the space that the next point is allowed to fall onto given the previous points?

>>15718663
>calculate the area of the space that the next point is allowed to fall onto given the previous points
That space is whatever is not in the "shadow" cast by the polyline with respect to the last point. Good luck calculating that...

>>15718663
exactly what >>15715660 did while assuming all conditional probabilities are 1. If you can find an expression for the conditionals, you've solved the problem. Good luck

>>15718676
>all conditional probabilities are 1
how can they be one retard. You mean assuming independet events

>>15714580
this is the type of question where the answer is either trivial but you need to be exposed to it to actually get it or it's fiendishly hard and still an open question

>>15715660
>>15718691
Anyway, it's possible to arbitrarily lower the bound with enough samples.
With the model (assuming a_i is independent of n):
[eqn]p(n)= (\prod_{i = 1}^{n-2} a_{i})p(n-1)
[/eqn]
with a_1 = p(n=3) and a_(i>1) representing the conditional probabilities and a_(i+1) >= a(i) which is trivial to show you can always construct a better lower bound for p(n) if you fit:
[math]a_i = p(n=i+2)/(p(n=i+1)\prod_{k = 1}^{i-1} a_{k})[/math] and setting all a_(l>i_max) = a_i_max

>>15718663
You will have fun time calculating that

>>15715691
Here's a simple lower bound that's better for n>8. There are 2^(k-2)k/k! ways to connect the vertices of a convex k-gon without crossing your path. Any shape with a point inside has more paths that don't cross, so convexity is lower bound on that. n is k-1 here, due to an artefact of OP's question, so

2^(n-1)(n+1)/(n+1)!

It's still a very bad lower bound because your chance of picking a convex polygon crashes to 0. But it crashes much less quickly than the n^2 exponent.

>>15719841
You could technically squeeze a log out of it by putting 1 point inside, which is still an easy solution for paths that don't cross.

2^(n)(n)/(n+1)!

But it doesn't really matter in this case because the crash is just as bad either way.

>>15714580
take a discrete set of points over the sphere, sample from a discrete uniform over thos points, take the limit as number of points goes to infinity. might get you somehere

>>15716293
>Permute the point order to find how often the lines intersect in each shape.
I don't understand this part. Care to elaborate?

>>15720272
Sure. If there's 5 points and you want to connect them, you have 5! ways to do it.

>>15720284
Ok but how do you get X(n,m)?

>>15720288
By connecting points.

>>15720291
Did you draw every permutation to find out?

>>15720295
Pretty much.

>>15720295
>>15720297
k=5 is tame.
k>5 is wild.
Is that what you're asking about?

>>15714580
Maths is fake and gay
The answer is OP=faggot when OP!=dead

>>15720301
well i want to get an interval for k>5

>>15720310
So do I lol. It annoys me.

2/15 + [ 587/(432 π^2) - 1004003/(1036800 π^4), 1219/(864 π^2) - 287287/(2073600 π]^4) ] ~ [0.2610..., 0.2748...]

The billion that anon simulated for for k=6 goes right into the heart of that interval.

>>15720319
Very nice. But I guess there's no way to make progress from here, right?
I wonder if there's at least a good numerical method to converge on p faster than just rolling the dice.

>>15720379
There's always a way even if I'm too dumb to see it :) First thing would be to explain how 2 points inside convex k-2 works.

For example, you can pick a 6 point (4,2) shape where the 5 point subsets are {(3,2), (3,2), (4,1), (4,1), (4,1), (4,1)} and the 3 point subsets are {(3,1), (3,1), (3,1), (3,1), (3,1), (3,1), (3,1), (3,1), (4,0), (4,0), (4,0), (4,0), (4,0), (4,0), (4,0)} but they have two different X distributions.

I've only thought about this for a couple days so I'm not an expert but I don't think anyone's an expert. It's a cool crossover problem.

>>15720428
*3 4 you get it

>>15719841
>>15720097
good shit. Now we can construct an even better bound

using >>15718854 and >>15716293 and >>15715125 the current lower bound should be:

[eqn]
p(n) >= \begin{cases}
(2/3 + 35/(36\pi^2))^{(n-2)(4-n)}(1/3+73/(48\pi^2))^{(n-2)(n-3)/2} &\quad n \leq 14 \\
2^{n}*n/(n+1)! &\quad n > 14
\end{cases}
[/eqn]

for the upper bound its easy to show that:

[math]p(n) <= (2/3 + 35/(36\pi^2))^{(n-2)}[/math]

>>15720758
The n^2 in the exponent seems unnatural.
From >>15715091 >>15719841 >>15720097
I would expect both the lower bound and exact formula to be roughly e^O(n) /n!.
A positive n^2 in the exponent would quickly overtake the n!
At most it is O(nlog(n)) with the nlog(n) coefficient <=1.

The exponential term appears to be between 2^n and 4^n from >>15716430
but it might be different as the lower order terms settle at larger n.

Assuming there is a nlog(n) contribution, it looks like roughly e^(0.35*nlog(n)).

>>15720892
I wouldn't say it's unnatural, but it's definitely wrong for higher n.

If you compare >>15720758 and >>15719841 then adding 1 sample to the model improved the bound for 6 values compared to 2^(n-1)(n+1)/(n+1)!


>I would expect both the lower bound and exact formula to be roughly e^O(n) /n!

I agree with that (it might grow slower).
My current idea is to find a function for the conditional probabilities in the case n->inf and roughly estimate a lower bound from there. Might be impossible though.

>>15714580
the probability is random

>>15716293
>>15716393
How tf did you even find those?
I wish I wasn't a poorfag.
>Best you could do is intervals.
So basically input the best/worst X for each (k, N-k) and just use the Marckert probabilities?
Drawing these things and counting paths is getting tiresome.
If there was a nice way to systematically catalog the X's then maybe we could at least figure out what the best/worst ones always are.
Best bet is to find them all for 6 (and maybe 7) points then see what oeis spits out.

I attempted to beat that bound while assuming conditional independence (the condition being the length of lines) and couldn't beat >>15720758.

+I ran a simulation for n=20 (50 million) and >>15720097 was still 3-4 orders of magnitude off the simulation.

Although the upper bound could be improved, I'm calling it quits here. There's probably a better bound out there, but I'm too retarded.

>>15723705
*(500 million)

>>15723705
I misremembered. I did it for n=15. The result was:
[math]p(n=15) = 4*10^{-6}[/math]
so >>15720097 is off by 10-11 orders of magnitude.

>>15722500
Here's Marckert.
>input the best/worst X for each
Yeah. Change the notation from (k-j, j) to (k, k-j) so (5,0) (4,0) (3,2) is (5,5) (5,4) (5,3). For (k, k) the input is always 2^(k-2)k/k! For (k, k-1) it's 2^(k-1)(k-1)/k!

>systematically catalog
Consider that 1 shape of k points is also k shapes of k-1 points... k choose j of k-j points. You can measure the convexity a shape up to its full set of subshapes. For example, the minimal and maximal X distributions for 6 points.

min 0 in X(6,j)

X(6; 3, 333335, 333333333344444) : 0 = 360/720, 1 = 260/720, 2 = 80/720, 3 = 20/720
X(6; 3, 333444, 333333333444444) : 0 = 360/720, 1 = 228/720, 2 = 72/720, 3 = 36/720, 4 = 24/720

X(6; 4, 344445, 333333444444444) : 0 = 248/720, 1 = 240/720, 2 = 124/720, 3 = 68/720, 4 = 32/720, 5 = 4/720, 6 = 4/720

max 0 in X(6,j)

X(6; 3, 333333, 333333333333444) : 0 = 456/720, 1 = 240/720, 2 = 24/720
{0, 268}, {1, 288}, {2, 112}, {3, 48}, {4, 4}}

X(6; 4, 334444, 333333334444444) : 0 = 268/720, 1 = 288/720, 2 = 112/720, 3 = 48/720, 4 = 4/720
X(6; 4, 334444, 333333334444444) : 0 = 268/720, 1 = 284/720, 2 = 124/720, 3 = 36/720, 4 = 8/720

Unfortunately, there's no name for the 2 different max X(6,4) shapes. You need a new measure. Something beyond convexity.

>>15722500
Miles is too big to post.
>figure out what the best/worst
I suspect you can always find a minimal (k, k-j) in the subset of shapes with a (k-1, 0) subshape. It makes sense and seems to hold for 7 points. But I'm not sure that my data's complete here.

empirical min 0 in X(7,j)

X(7; 3, 3333336) : 0 = 1272/5040, 1 = 1680/5040, 2 = 1152/5040, 3 = 648/5040, 4 = 192/5040, 5 = 72/5040, 6 = 24/5040

X(7; 4, 3444446) : 0 = 920/5040, 1 = 1356/5040, 2 = 1116/5040, 3 = 806/5040, 4 = 420/5040, 5 = 226/5040, 6 = 136/5040, 7 = 40/5040, 8 = 14/5040, 9 = 4/5040, 10 = 2/5040

X(7; 5, 4445556) : 0 = 616/5040, 1 = 1180/5040, 2 = 1128/5040, 3 = 958/5040, 4 = 548/5040, 5 = 318/5040, 6 = 188/5040, 7 = 64/5040, 8 = 30/5040, 9 = 8/5040, 10 = 2/5040
X(7; 5, 4455556) : 0 = 616/5040, 1 = 1086/5040, 2 = 1042/5040, 3 = 918/5040, 4 = 602/5040, 5 = 342/5040, 6 = 252/5040, 7 = 98/5040, 8 = 50/5040, 9 = 20/5040, 10 = 14/5040

If it is, you could add n=6 into >>15720758 along with n=5. Not as solutions but as a tighter lower bounds.

p(n=5) > 2/15 + 587/(432 π^2) - 1004003/(1036800 π^4) ~ 0.2610...
p(n=6) > 2/45 + 1481/(1728 π^2) - 2285489/(2073600 π^4) ~ 0.1199...

>oeis
It won't. It's new combinatorics. Even the (relatively) simple triangle of X(k,0) distributions isn't there. Neither in full nor up to symmetry (it's always divisible by 2k).

1
2, 1
4, 5, 2, 1
8, 17, 15, 12, 4, 3, 1
16, 49, 69, 79, 53, 40, 29, 14, 8, 2, 1
32, 129, 252, 382, 387, 346, 316, 246, 190, 110, 61, 37, 23, 5, 3, 1
64, 321, 804, 1527, 2073, 2287, 2408, 2364, 2261, 1838, 1336, 991, 800, 462, 281, 162, 96, 55, 19, 8, 2, 1

The first 3 columns, outer 2 diagonals, and row lengths are there. The rest isn't.

>>15723810
*(k-1, k-1) subshape

>>15723810
The lower bound for p(n=7) is p_7. As a result, the absolute difference is 0.001 when compared to >>15716430.

If I only consider p(n=5), the lower bound on p(n=6) is:

p(n=6) > 0.117568104119

so not too far away from:

>p(n=6) > 2/45 + 1481/(1728 π^2) - 2285489/(2073600 π^4)

picrel. calculated with >>15718854

>>15723874
*d^2 was missing resulting in a higher abs. diff.

>>15723874
>>15723884
How long until it crosses 2^n(n)/(n+1)!

>>15714580
the probability is 0, and my proof is that i made it the fuck up, now fuck off

>>15723890
that shit is starting to hurt.

If I haven't made a mistake, 2^n(n)/(n+1)! is better for n>19

brap