Probability Thread

[Username: Username:]

Probability of the first layer solved with two oriented edges?

 

[Cuberstache]

Probability of the first layer solved with two oriented edges?

What do you mean "with two oriented edges"? Two pieces outside the first layer oriented?

 

[Username: Username:]

Yes.

What do you mean "with two oriented edges"? Two pieces outside the first layer oriented?

 

[Cuberstache]

Yes.

Exactly two or at least two?

 

[Username: Username:]

Exactly two or at least two?

Two exactly

 

[Cuberstache]

Two exactly

Ok, I'll let someone else answer now since I don't really know how to calculate that, hahaha.

 

[Username: Username:]

Lol no one answered my question for days.

 

[Kit Clement]

Lol no one answered my question for days.

It would help if you gave some motivation for why anyone would want to know that probability. I could calculate it but I'm 0% interested in the result.

 

[Username: Username:]

It would help if you gave some motivation for why anyone would want to know that probability. I could calculate it but I'm 0% interested in the result.

I'm just curious.

 

[Devagio]

I'm just curious.

A back of the envelope calculation gives me 1.1x10^-9; I expect it to be off by less than 5%
Basically 1 in 24x22x20x18x24x21x18x15 scrambles will have a solved white face.
Approximating that all scrambles with one solved face is disjoint means about 6 in that many scrambles have a solved face. Then simply (8 choose 2)/128 EO cases have exactly 2 oriented edges.

 

[Username: Username:]

A back of the envelope calculation gives me 1.1x10^-9; I expect it to be off by less than 5%

Thanks! even if it is not accurate.

 

[Devagio]

Thanks! even if it is not accurate.

Not precise*
It is accurate.

 

[N's-cvt]

I was re-watching J-perm's all 77 F2L cases video and he claimed that there are 167 total cases but he wouldn't explain why so over the years I have tried to get that number and I can't get it so leave a comment below so I can see you math, here is mine so far:
F2L consists of the cross edges being done with 4 edges and 4 corners that need to be solved. Because placement is an addition problem and orientation is a multiplication problem and solved positions are a subtraction problem I think the math would look like this (8+7+6+5+8+7+6+5) * 2 * 3 - 4 = 308

 

[kubesolver]

Assuming that the two cases that differ only by the U layer are the same case I counted 167 cases excluding the solved state.

3 * 2 * (4 + 4 + 4 + 4*4) - 1

3 corner orientations *
2 edge orientations * (
4 relative edge.corner positions when edge and corner are in U layer:
+ 4 cases of edge in U layer, corner in D layer:
+ 4 cases edge in E layer, corner in U layer
+ 4 * 4 cases corner in D layer and edge in E layer )
- 1 solved position

 

[SenorJuan]

In practice, a competent cuber would know far more than 167 cases. He/she would know how to solve every case from the y2 orientation, so only y or y' rotations would be needed to solve the case, and never y2, and in some cases a seperate algorithm for those y , y' cases would be known. Then add in good use of empty slots, and a considerable number on new ways to solve the same basic case become possible. On the other hand, not many people solve those cases known as 'Jeff2L' , where neither edge or corner are on the U layer. I did work out algs for a lot of these cases many years ago, mainly for adjacent slots, but only a few of them made it into my speedsolving repertoire. I have no idea how many 'algs' I actually use in the F2L stage, it could be over 500, far too many to try and total up.
There's also all the LL edge control variants. Even if you just use basic sledgehammer insert as an alternative to your normal insert, you're adding a lot of cases to your repertoire.

 

[AMKuber]

The probability of getting cross solved is 1/190,080

This is because there are 12 edges on the cube and each has 2 possible orientations. this means each edge has a 1/24 chance of being solved, but because the first edge is already there the second edge has a 1/22 chance of being solved as one edge place is taken up already the third edge has a 1/20 chance of being solved and the fourth one has a 1/18 chance.


So the math would be
1/24 x 1/22 x 1/20 x 1/18 = 190,080



To put this in perspective a LL skip on 3x3 is 1/15,552 and think about how many LL skips you have gotten, not many I assume. and you would need to get 12 (really 12.22222....) LL skips before you should have gotten a cross solved skip.

 

[vidcapper]

Probability of x number of randomly paired edges on a 4x4?

I've rarely seen more than one...

 

[xyzzy]

Probability of x number of randomly paired edges on a 4x4?

I've rarely seen more than one...

tl;dr See the table at the bottom of this post.

First-order approximation: There are 12 places you can get a free edge pair, and the probability of each one actually having a free edge pair is 1/23 (whatever the first piece is, there are 23 possibilities for its partner and only one of those 23 will give you a free edge pair). The probabilities are not independent, but since we're approximating, let's pretend they are, so we can use a binomial distribution.

# pairs	probability
0	0.586595
1	0.319961
2	0.079990
3	0.012120
4	0.001240
5	0.000090
6	0.000005
≥7	<0.0000002

How good/bad is this approximation? Let's do a simulation.

import random

def test(num_samples, N=12):
    a = list(range(N))*2
    counts = [0]*(N+1)
    for i in range(num_samples):
        random.shuffle(a)
        counts[sum(a[j] == a[j+N] for j in range(N))] += 1
    return counts

>>> test(1000000)
[593992, 308682, 80931, 14315, 1845, 213, 22, 0, 0, 0, 0, 0, 0]

Unsurprisingly, the approximation breaks down near the tails (that's where the independence assumption happens to be least applicable), but at least for small number of pairs, it's not too far off. (We can further approximate the binomial distribution with a Poisson distribution with mean 12/23, which very counterintuitively actually improves (?!) the approximation a lot, as if most of the approximation errors cancel each other out. I don't understand this. There is probably some fancy group theory lurking behind this.)

It is in fact possible to compute the exact distribution via the inclusion-exclusion principle as well.

Let \( n \) be the number of edge pair locations in total. (12 for reduction, 9 for Yau after centres and 3/4 cross, 8 for Yau after centres and cross.) The number of ways of pairing up \( 2n \) items, labelled with two copies of \( 0 \) through \( n-1 \), such that no pair has items of the same label, is:
\( \displaystyle a_n:=\sum_{k=0}^n(-1)^k\binom nk(2n-1-2k)!! \)

(Note: !! means double factorial. The number of ways of pairing up \( 2n \) items, ignoring the order of the pairs and the order within the pairs, is \( \frac{(2n)!}{n!2^n}=(2n-1)!! \).)

If we want to count the number of ways to have \( m \) free pairs out of \( N \) total, we first choose which pairs are free (\( \binom Nm \)), then treat the remaining \( N-m \) pairs as an instance of the above problem (\( a_{N-m} \)). To wit:
\( \displaystyle (\text{number of ways to have }m\text{ free pairs out of }N)=\binom Nm a_{N-m} \)

Now we can tabulate the values with \( N=12 \) and varying values of \( m \):

# pairs	# cases	probability
0​	187778717632​	0.59379647​
1​	97779555840​	0.30919987​
2​	25548595776​	0.08079012​
3​	4469273600​	0.01413280​
4​	589382640​	0.00186375​
5​	62574336​	0.00019787​
6​	5580960​	0.00001765​
7​	430848​	0.00000136​
8​	29700​	0.00000009​
9​	1760​	0.00000001​
10​	132​	0.00000000​
11​	0​	0.00000000​
12​	1​	0.00000000​

As a sanity check, these values are in good agreement with the distribution obtained via sampling.

edit: Fixed some incoherent nonsense. The calculations are still correct, of course.

 

[vidcapper]

Thank you very much for this - very comprehensive!

They do seem to correspond with my observational experience. Offhand I don't recall getting more than 2, but I daresay others here have stories of getting rather more than that.

 

[ProStar]

What are the chances of having F2L-1 solved?