bizarre G-perms

[David Pritts]

EDIT:

<I decided to make this post easier/faster to read, and to correct the error I had made.>

The revised version is as follows:

On previous occasions, I have been surprised by how many G-perms I got. However, this is probably because I was in error (by a factor of 2) regarding how often they occur.

So, today I did 50 solves and counted the G-perms, getting 18 of them.

I thought this was still kind of cool.


David

 

[hait2]

get a larger sample
i'd be happy you got 2 PLL skips in 50 solves

as for math:
probability of Gperm according to you = 1/9 = p
probability of 18 Gperms in 50 solves = 50c18 p^18 q^32 = 0.0002775%
if tim's right, and p is 2/9, then we get 1% here

probability of PLL skip according to internet = 1/72 = p
probability of 2 PLL skips in 50 solves = 50c2 p^2 q^48 = 12%

haven't done stats in a while but that should probably be right

also
adjacent 4edge cycle is Zperm
uf/ur ufl/ubr is Vperm
ul/ur ufl/ufr is Fperm

4 corners? i suppose Eperm. could also be Hperm i guess, depends what you mean
3edge cycles are Uperms
number 7 is probably Rperm, need to be more specific there

final edit: no it has nothing to do with the way you scramble the cube. there's nothing subconscious going on here, it's not the cubing statistic gods aligned against you to give you all the Gperms in the world. just a small sample

 

[Pedro]

didn't you have any N perms?
UL - UR, UBL - UFR
UL - UR, UFL - UBR

 

[Tim Reynolds]

Hey David,
I'm pretty sure the probability of each G Perm is 1/18, not 1/36, so the total probability would actually be 2/9. Since G isn't rotationally symmetric at all, there's 4 ways you can get each G out of 72 total, so that has probability 4/72=1/18.

 

[David Pritts]

Pedro: no N perms (thank God). The algorithms I have for those 2 really blow.

Hait2: Yes, I was of course happy to get 2 PLL skips out of 50 solves And I guess your calculation is off because, as Tim pointed out (thanks Tim), each G-perm has a 1/18 chance of occurring, thus I was wrong.

I will continue a larger sample... and hopefully I'll get lots of those nice T-perms :-|

 

[hait2]

so revised # would be about 1% for the Gperms then. not that unlikely at all. i never get PLL skips though, i'm going to record my next solves and see my distribution of PLLs too for fun

 

[Lt-UnReaL]

Out of 50, around 11 G's would be normal...18 is a bit high...

 

[Dene]

final edit: no it has nothing to do with the way you scramble the cube. there's nothing subconscious going on here, it's not the cubing statistic gods aligned against you to give you all the Gperms in the world. just a small sample

I wouldn't be too sure about that lol, I find when I do OH, I get way more U perms than should be expected

 

[Pedro]

I'm doing some counts too...

did 50 so far...

no H (!), lots of A, G, J, R, some U, N, V, Y...

I'll do maybe more 50 or 100 tomorrow...

 

[hait2]

final edit: no it has nothing to do with the way you scramble the cube. there's nothing subconscious going on here, it's not the cubing statistic gods aligned against you to give you all the Gperms in the world. just a small sample

I wouldn't be too sure about that lol, I find when I do OH, I get way more U perms than should be expected

selective memory
when you see a Uperm you think "hey omg another Uperm!"
but when you dont, you don't think twice and just go for another solve

 

[Dene]

final edit: no it has nothing to do with the way you scramble the cube. there's nothing subconscious going on here, it's not the cubing statistic gods aligned against you to give you all the Gperms in the world. just a small sample

I wouldn't be too sure about that lol, I find when I do OH, I get way more U perms than should be expected

selective memory
when you see a Uperm you think "hey omg another Uperm!"
but when you dont, you don't think twice and just go for another solve

I don't think so, I find that every average of 12 I do, I get at least 3 U's and 2 J's, it's really good!

 

[cmhardw]

G-perms--18

Since each G-term has a 1/36 chance of occurring on any given solve, the probability of any solve ending in a G-perm should be 1/9. With this in mind, I'd expect between 5-6 G-terms out of 50 solves.

What are the chances of this happening?

What's happened is that you have miscalculated the probability of ending in a G perm. There are 4 unique G perm PLL cases, each one has a probability of 1/18 to occur for a combined chance of 2/9.

The chance that within 50 solves you get exactly 18 G perm cases is
(50 C 18) * (2/9)^18 * (7/9)^32 = 1.01% chance.

Where (50 C 18) means 50!/[18!*(50-18)!]

The probability to get the G perm 18 or more times is:
summation(i=18 to i=50, (50 C i) * (2/9)^i * (7/9)^(50-i)) = 1.86% chance.

You getting 18 G perms out of 50 solves seems perfectly reasonable to me.

Chris

 

[David Pritts]

I see.

I guess with a 2% chance of 18+ G-perms, I shouldn't be too surprised.

Though I really should edit the original post so that people see i've already realized my initial mistake -.-

David

 

[Kenneth]

Hmm, I wonder....

If you take the pieces of the U layer out of the cube and then put them back in, oriented but in random order, purposly placing the three last pieces so unsolvable paritys does not occure.

Then distribution will be as you calculate it.

But we do not do that!

The creation of a "case" is a iterative process. What we do is Fxn where n is the number of iteriations done and Fx is a selectable function = the turns we choose to do.

If we set max n to for example 20 turns, execute all possible combinations Fx in every iteriation = scrambling, would we stil have the same ditrubution of U, N, T... PLL's for all scrambles that has got a solved F2L and LL oriented?

I'm not really clear about what I'm talking about (many of you guys knows maths much better than me, I'm a self learned) but I'm sure if yo set max n to 12 you will surley have no Y-PLL but a number of G-PLL's, if you set it to 15 you will have a few Y-PLL's and a great lot of G-PLL's.

Maybe I'm totaly wrong about this, but it is a thought =)

 

[hait2]

@kenneth
it all depends on how the probabilities of getting a specific permutation are calculated. for instance, apparently getting any G is 2/9, but this is given what? there's obviously assumptions made here that i dont know. i mean clearly, if you permute corners while orienting them every time, you're not going to get a G. likewise if you insert your last f2l pair such that you orient the cross, you're not going to have 50+ OLL cases, only 7.

actually i'm interested in how they came up with those numbers anyway. most likely they just took every reachable permutation of LL pieces, assumed each is equally likely, and see what PLLs came up, taking into account symmetries, self-inverses and so on. i'm not sure if assuming each is equally likely is a fair one, as that makes more hidden assumptions.. oh well =)

 

[Kenneth]

I'm actually thinking there is chaos lurking somewhere in the darks of twisting the cube.

Earlier I worked a lot with fractals, not so much to understand the maths about it but to create pretty images (many fractal images at the Wikipedia are my creations).

Take this system : the for corners of a square as points, start in the middle x,y = origo of the square. then do this:

1 Randomly choose any of the for corner points
2 Move half the way there from current position
3 Repeat from 1 (iteriate)

Using this system the whole surface is reachable but some points are wery hard to reach and others are visited often (if resolution is not eternal, if so, then you will only get near the earlier points but newer hit exacly the same one)

For cubes one could do this : do a random turn : if it's a PLL? (six faces possible) then wich? : repeat

What would the ditrubution of A - Z PLL's be then? I assume the ones that are easy to solve (uses few turns) occoure more often than others but I may be wrong

 

[cmhardw]

The assumptions I made for my calculation are these:

1) The centers are fixed in space and define both the solved position and solved orientation of each piece.

2) I assume that at no point during the solving of the first two layers or the orientation of the last layer pieces after the completion of the first two layers did the solver attempt to affect the permutations of the last layer pieces.

2) There are 288 distinct possible positions of the last layer (4!*4*3) once the first two layers have been solved and the pieces of the last layer oriented correctly with respect to the first two layers orientation. I also assume each of those 288 positions has an equal probability to occur because of assumption 2.

If you want to know the odds based on if you use COLL and edge control then I could calculate those as well, as yes they would indeed change the probability of every PLL case.

This is actually pretty interesting, unless someone beats me to it, and of course after I get off work that is ;-) , I'll find the odds of each PLL case assuming perfect edge conrol (never having an OLL case with all 4 last layer edges incorrectly flipped), and the use of COLL whenever this is possible.

Chris

 

[Lucas Garron]

For cubes one could do this : do a random turn : if it's a PLL? (six faces possible) then wich? : repeat

What would the ditrubution of A - Z PLL's be then? I assume the ones that are easy to solve (uses few turns) occoure more often than others but I may be wrong

The PLL distribution calculations only assume that the first two layers are solved and the PLL state wasn't influenced. This qualifies even more for such conditions than a regular cube.
This will approach the standard distribution very quickly. I'd guess that it gets within 1% (for each PLL) soon after 20 moves.

 

[cmhardw]

Edge control / COLL influenced PLL probabilities:

Simplifying assumptions:
1) The centers of the cube are fixed in space

2) The first two layers are solved such that there is no purposeful influence on the last layer pieces, except for the avoidance of the cases where all 4 last layer edges are flipped incorrectly after solving the first two layers (edge control).

3) If after solving the first two layers using assumption 2 all four edges of the last layer are correctly oriented, then a COLL alg is used to solve the position and orientations of the corners. No purposeful influence is applied to the permutations of the edges (ZBLL is not allowed).

4) There are 14 distinct PLL cases including the solved case where all last layer pieces are permuted correctly in relation to each other and not necessarily in relation to the first two layers.

-----

U, G: 4/21 (probability for each individually)
A, R, J: 2/21 (probability for each individually)
V, Y, F, T, Z: 1/21 (probability for each individually)
E, N, H, solved: 1/42 (probability for each individually)

Chris