Probability Thread

[ruffleduck]

WV only works when you have a cross solved on top.

You can force EO solved.

I am talking about WV (and let's assume that I will use WV whenever I can)

So more specifically, you use WV whenever EO is solved. In that case, you will have 1/72 chance of LL skip whenever EO is solved. Depending on your solutions for other LS cases, your probability for non-EO solved cases may vary.

 

[Jaym-er]

Can someone explain why certain PLLs (eg N, Z, H) are rarer than others? Just curious.

An easy way is to imagine the number of ways you can draw the top layer - each way you draw it is a different possible position and can contribute to its probability of occurring.

(For below examples consider white base)

With an F perm, you can draw it with the "bar" on 4 different sides, and the bar can be 4 different colours (red,blue,green or orange). This is 16 different positions.

With the h perm, you can draw it only with 4 different corner colours facing the front. All other top down drawings will give the same image. This is only 4 different positions.

This method is basically just an easier way of visualising. I can't really give a simple math version other than saying that the pieces just can't permute (ie. be drawn) in any new way.

 

[Jaym-er]

What are the chances of getting the same EG-1 case?
I got this scramble, F R' U F2 R' F R U2 F', I saw it was an EG-1 case, I learnt it and continued solving. Then some 15 solves later, I get this scramble, U' F U' F' R F' U2 F' U' and it's the exact same case that I just learnt some minutes ago. What are the chances of this happening, I don't know if it's rare of quite common. If it helps, I only have like 1470 solves.

It sounds like you're not exactly sure what the scenario you want a probability for is.

I'll dissect this into two interesting bits:

Chance of any solve having a premade EG-1:

Approx 6*4*4!*3^3/(7!*3^7) = 4/2835 (there are intersections where they can be 2 solved EG-1 faces on a scramble so the actual figure is slightly lower than this) This is approx 1 in 709 solves that you will have a premade EG-1 case.

Chance of not having any premade EG-1 scrambles in 1470 solves:

(2831/2835)^1470 = 12.5%

Expected premade EG-1 faces in 1470 solves:

4/2835*1470 = 2.07

Again those figures were approximate.

Now let's say you do get a premade EG-1 scramble, what are the chances of getting the same case within 15 solves?

Chance of not getting a specific EG-1 scramble in any solve (this is different for the solved and diag oriented cases):

(7!3^7-6*4*4)/(7!3^7) = 11022384/11022480 (not gonna bother as % here, trust me it's low)

Chance of getting a specific EG-1 at least once within 15 solves:

1- (11022384/11022480)^15 = 0.013%

So approx 1 in 7600 times that you get a premade EG-1 scramble you will get that same case within your next 15 solves.

 

[Bruce MacKenzie]

Can someone explain why certain PLLs (eg N, Z, H) are rarer than others? Just curious.

There are 4! * 4! / 2 = 288 PLL cube arrangements. These are of 22 types when reduced by the orientation of the cube to which they are applied. Apply a pattern then rotate the cube 90˚ and apply it again and you get two different patterns which are deemed the same. However, many of the PLL patterns have symmetry. Rotating the cube gives the same exact pattern not a new similar one. A pattern may have two fold, four fold or no symmetry. Patterns with no symmetry will be twice as likely as patterns with two fold symmetry and four times as likely as patterns with four fold symmetry. Here are the 21 PLL patterns and their probability:

​	Name​	Symmetries​	Probability​
​	​	​	​
1​	Aa​	1​	0.0556 (1:18)​
​	​	​	​
2​	Ab​	1​	0.0556 (1:18)​
​	​	​	​
3​	E​	2​	0.0278 (1:36)​
​	​	​	​
4​	F​	1​	0.0556 (1:18)​
​	​	​	​
5​	Ga​	1​	0.0556 (1:18)​
​	​	​	​
6​	Gb​	1​	0.0556 (1:18)​
​	​	​	​
7​	Gc​	1​	0.0556 (1:18)​
​	​	​	​
8​	Gd​	1​	0.0556 (1:18)​
​	​	​	​
9​	H​	4​	0.0139 (1:72)​
​	​	​	​
10​	Ja​	1​	0.0556 (1:18)​
​	​	​	​
11​	Jb​	1​	0.0556 (1:18)​
​	​	​	​
12​	Na​	4​	0.0139 (1:72)​
​	​	​	​
13​	Nb​	4​	0.0139 (1:72)​
​	​	​	​
14​	Ra​	1​	0.0556 (1:18)​
​	​	​	​
15​	Rb​	1​	0.0556 (1:18)​
​	​	​	​
16​	T​	1​	0.0556 (1:18)​
​	​	​	​
17​	Ua​	1​	0.0556 (1:18)​
​	​	​	​
18​	Ub​	1​	0.0556 (1:18)​
​	​	​	​
19​	V​	1​	0.0556 (1:18)​
​	​	​	​
20​	Y​	1​	0.0556 (1:18)​
​	​	​	​
21​	Z​	2​	0.0278 (1:36)​

 

[Aluminum]

Chances for an Xcross to already be finished?

1 in 43 million

 

[Imsoosm]

1 in 43 million

No, there are many different x-crosses that are possible. For each x-cross on white face, there are four different pairs that can be put in. And that's not to account for all the other pieces. You can also have 24 orientations for each x-cross, and you have 6 faces, so to start off you would get 576 different x-crosses NOT including the other pieces.

 

[hyn]

No, there are many different x-crosses that are possible. For each x-cross on white face, there are four different pairs that can be put in. And that's not to account for all the other pieces. You can also have 24 orientations for each x-cross, and you have 6 faces, so to start off you would get 576 different x-crosses NOT including the other pieces.

24 orientations?

 

[Anto]

24 orientations?

6 possible faces at the bottom times 4 possible rotations
But do we count cube rotations actually ?

 

[Thom S.]

6 possible faces at the bottom times 4 possible rotations
But do we count cube rotations actually ?

You need to.
Let's say we take Cross + FR as an xCross on the YO orientation. Take that times(divided by) 24, then you have all possible ways an XCross can occur. And that is how we need to calculate chance in this case.

 

[Bruce MacKenzie]

1 in 43 million

Chances for an Xcross to already be finished?

For any particular manifestation of the XCross, one has one corner cubie and five edge cubies. One may calculate the total number of arrangements these six cubies may be in:

24 * ( 24 * 22 * 20 * 18 * 16 ) = 24 * 3,041,280 = 72,990,720

So the probability of getting a randomly generated cube with that pattern is 1 in 72,990,720

There are six faces and the pattern may be applied to a face four ways, giving a total of 24 XCross patterns. Thus the probability of randomly having an XCross cube is 1 in 3,041,280.

 

[Aluminum]

chances of pll skip with no auf? I just got one

 

[SpeedCubing RDJ]

For any particular manifestation of the XCross, one has one corner cubie and five edge cubies. One may calculate the total number of arrangements these six cubies may be in:

24 * ( 24 * 22 * 20 * 18 * 16 ) = 24 * 3,041,280 = 72,990,720

So the probability of getting a randomly generated cube with that pattern is 1 in 72,990,720

There are six faces and the pattern may be applied to a face four ways, giving a total of 24 XCross patterns. Thus the probability of randomly having an XCross cube is 1 in 3,041,280.

Wow.

 

[IsThatA4x4]

chances of pll skip with no auf? I just got one

1/288

 

[Imsoosm]

What are the chances of getting a PLL skip (including AUF) on a kilominx? Also LL skip?

 

[IsThatA4x4]

What are the chances of getting a PLL skip (including AUF) on a kilominx? Also LL skip?

I know nothing about kilo, but this should be simple. Somebody please correct me if I'm wrong about this because I've probably made a silly mistake somewhere!
There are 5 corners.
In PLL, we only need to worry about their permutation. Because the last corner's permutation is determined by the other 4, there are really only four corners, giving us
5 x 4 x 3 x 2 (x 1) aka 5! total PLLs (120).
Because a skip has a rotational symmetry of 5, the chance of a skip is 1/(120/5) = 1/24.

For LL, we also care about orientation too. Because the last corner's orientation is determined by the other 4, there are again really only 4 corners, so the amount of total OLLs is 3^4 = 81 (this does not account for symmetry ofc).
So the total number of LL cases (not reduced) is 81 x 120 = 9720.
The skip has a rotational symmetry of 5, so the chance of an LL skip is 1/1944.

 

[Abram Grimsley]

What is the probability of getting a last-layer skip on megaminx after orienting the last-layer edges?

 

[IsThatA4x4]

What is the probability of getting a last-layer skip on megaminx after orienting the last-layer edges?

Giving it a go (mega may work differently so take this with... I don't know, a ball of salt?)

We need to consider:
CO
CP
EP

For CO:
Discounting AUFs, there are 3^4 cases, and I believe counting AUFs is incorrect in this scenario, so I'm leaving it there (we'll see why)

For CP:
There should be 5! (5x4x3x2x1) CPs

For EP:
Again, 5!

Because a swap of only two pieces is impossible, half of these cases are impossible.
This leaves us with (3^4 x (5!)^2) / 2 LLs (discounting AUFs), however this is not our answer, as a skip has a rotational symmetry of 5, so the chance of a skip is:
1 / (3^4 x (5!)^2) / 10
= 1/116640

This is what I came to, however it seems too small (?). Other people correct me if this is not the chance.

 

[Abram Grimsley]

Thanks man. That means I got really lucky on that solve

 

[Jaym-er]

Somebody please correct me if I'm wrong about this because I've probably made a silly mistake somewhere!
5 x 4 x 3 x 2 (x 1) aka 5! total PLLs (120).
Because a skip has a rotational symmetry of 5, the chance of a skip is 1/(120/5) = 1/24.

For CP:
There should be 5! (5x4x3x2x1) CPs

For EP:
Again, 5!

Because a swap of only two pieces is impossible, half of these cases are impossible.
This leaves us with (3^4 x (5!)^2) / 2 LLs (discounting AUFs), however this is not our answer, as a skip has a rotational symmetry of 5, so the chance of a skip is...

You can't simply divide by 5 for a megaminx as some permutation cases can occur by a multiple other than 5. I've explained how to derive megaminx permutation in a post a month earlier in this thread.

See here:

So I was curious about how to mathematically answer the megaminx LL skip:

Using a mathematical derivation for probabilities by solving EO-CO-CP-EP:

Edge orientation:

2^4 - each edge has 2 possible orientations, when you know the orientation of 4, the last is predetermined, hence 2*2*2*2 = 2^4, so chance of EO skip = 1/16

Corner orientation:

3^4 - same logic as edge orientation, except 3 orientations possible for each corner. Chance of CO skip = 1/81

Chance of OLL skip: 1/81*1/16 = 1/1296

As permutation of corners and edges are dependent on each other, I will use a 4LLL approach calculating skip chance of first CP (ignoring EP) then EP (dependent on knowing CP)

Corner permutation:

4*3 - taking the first corner as fixed and then permuting the remaining 4 corners around this. The first corner has 4 to choose from, the next has 3 to choose from. Once you know these corners the remaining two are fixed (this also means you can determine megaminx cp from 3 corners) Chance of CP skip = 1/12

Edge permutation:

5*4*3 - once corner permutation is known, the 5 edges need to be permuted around this. First edge position has 5 to choose from, next has 4, third position has 3. Knowing all corner positions and first 3 edges means the position of remaining 2 edges is fixed. Chance of EP skip = 1/60

Chance of PLL skip = CPskip*EPskip = 1/12*1/60 = 1/720

Chance of LL skip = OLL skip* PLL skip = 1/1296*1/720 = 1/933120

A little wordy but this gives a few extra probabilities and info.

 

[ruffleduck]

I don't know, a ball of salt?

Here's my dodecahedron of salt : )

CO: 3^4
CP: 5! / 2
EP: 5! / 2
AUF: 5

AUF / (CO * CP * EP) = 5 / (3^4 * (5! / 2)^2) = 1 / 58320