TheAwesomeAlex

Member
whats the probability of getting a 2x2 LL skip?
whats the probability of getting a pyraminx LL skip?

Applejuice

Member
I'm really starting to enjoy this thread. ;D
What's the probability of having sune as OLL + A perm as pll?

Bob

I'm really starting to enjoy this thread. ;D
What's the probability of having sune as OLL + A perm as pll?
P(Sune) = 1/54
P(A perm) = 2/18 = 1/9

P(Sune -> A) = 1/54 * 2/18 = 1/486.

Do you count Sune and Antisune the same? In that case, it would be twice as likely, 1/243.

cubenut99

Member
On the 5x5x5, the probability that all the edges would be tripled up correctly is:

1/24! = 1/620,448,401,733,239,439,360,000

For the 4x4x4, the probability of all edges being paired up correctly is:
1/23!! (where !! is the double factorial operator), or 1/(23*21*19*17*15*13*11*9*7*5*3*1) = 1/316,234,143,225.
holy crap

TMOY

Member
whats the probability of getting a pyraminx LL skip?
It's 1/12 (3 possible permutations, '4 possible orientations for each one of them).

Smiles

Member
what is the probability of getting an x-cross (unintentionally) after doing a regular cross?

Stefan

Member
what is the probability of getting an x-cross (unintentionally) after doing a regular cross?
About 1.0%. Determined by 10000 computer simulations and this:

Chance of specific F2L pair solved: 1/24 * 1/16.
Chance of specific F2L pair *not* solved: 1 - 1/24 * 1/16.
Chance of all four F2L pairs *not* solved: (1 - 1/24 * 1/16) ^ 4.
Chance of at least one F2L pair solved: 1 - (1 - 1/24 * 1/16) ^ 4 = 1.038%.
Note that the last two aren't exact, because the probabilities of the four pairs aren't independent, because where pieces of one pair are affects where pieces of the other pairs can be. But they seem to be fairly independent, as the experiment confirms.

For a description of the experiment and more results, see page 35 and the pages around it here:
http://www.stefan-pochmann.info/hume/hume_diploma_thesis.pdf

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mDiPalma

Member
this may be a hard one.

what are the probabilities of the fewest amount of misoriented edges [R,L,U,D,F2,B2] in all orientations as opposed to in a fixed orientation?

for example F' U D2 L2 R U R' U' B F' L B2 R2 F L' F R2 F2 R' B' L' B F L U'
gives 6 bad edges in wg,wb,yg,and yb orientations
but gives 4 bad edges in wr,wo,yr, and yo orientations

so, if i compared the amount of misoriented edges in fixed orientation to the least amount of misoriented edges out of all orientations (y,x,z,&combinations thereof), all orientations should have a distribution that is skewed to the left of the fixed orientation data.

crider's site gives this data for a fixed orientation

2 - 66/2048
4 - 495/2048
6 - 924/2048
8 - 495/2048
10 - 66/2048
12 - 1/2048

so i guess im wondering if somebody somehow could fill out a chart like ^^that^^ counting the likelihoods of the FEWEST amount of misoriented edges in all orientations in all scrambles.

does that make sense?

Escher

Babby
Does anybody know of a decent book covering probability?

Just looking for a reasonable 'introductory' text, University level is fine.

Thanks :3

cuBerBruce

Member
what are the probabilities of the fewest amount of misoriented edges [R,L,U,D,F2,B2] in all orientations as opposed to in a fixed orientation?
There are a total of 12! * 2^11 = 980,995,276,800 configurations of the edges. For the purposes of calculating what mDiPalma wants, I believe we can consider the four E-layer edges indistinguishable, the four M-layer edges indistinguishable, and the four S-layer edges to be indistinguishable. This reduces the number of configurations we actually need to consider to 12! * 2^11 / 4!^3 = 70,963,200. Also, I believe three (appropriately chosen) cube orientations suffice for counting the misoriented edges.

Using GAP, I've come up with the following distribution:

Code:
bad edges   count
0      103741
2     6500978
4    35204527
6    26728948
8     2411347
10       13658
12           1
Note that for 12 bad edges, every edge must be in its proper inner layer. If an edge is in the wrong inner layer, there will be some cube orientation that will make it oriented. Hence, the above distribution has only 1 case with 12 bad edges.

mDiPalma

Member
Using GAP, I've come up with the following distribution:
i love you.

so a color neutral zz solver would have this:

2 - 9.161%
4- 49.610%
6 - 37.666%
8 - 3.398%
10 - .019%
12 - ~0%

as opposed to this:

2 - 3.223%
4- 24.170%
6 - 45.117%
8 - 24.170%
10 - 3.223%
12 - .049%

cool thanx man <3 <3 <3

theZcuber

Note that for 12 bad edges, every edge must be in its proper inner layer. If an edge is in the wrong inner layer, there will be some cube orientation that will make it oriented. Hence, the above distribution has only 1 case with 12 bad edges.
So basically the only position where all edges are bad in any orientation is superflip (corners could be scrambled of course)

cuBerBruce

Member
So basically the only position where all edges are bad in any orientation is superflip (corners could be scrambled of course)
Edges can also be scrambled, but only within their respective layers (E,M,S). So actually 13824 possible edge configurations.

kinch2002

Last 2 Center skip on 5x5?
Fairly simple just to think through logically.
+ and x centres will clearly behave the same in term of probability, so just calculate the probability of skipping a particular set and square the result.
Pick a piece of colour 1 and place it on a centre: 4/8 (1/2) chance of getting it right
Pick 2nd piece of colour 1 and place it in a free space: 3/7
3rd piece: 2/6 (1/3)
4th piece: 1/5
1/2 x 3/7 x 1/3 x 1/5 = 1/70
(1/70)^2=1/4900

Christopher Mowla

Last 2 Center skip on 5x5?
Just as kinch2002 said, it's 1/4900.

In general, the probability for getting a C center skip on a cube of size n is (with 6-C centers already solved) is:

$$\frac{1}{\left( \frac{\left( 4C \right)!}{\left( 4! \right)^{C}} \right)^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }}$$

Now, this doesn't take into account how many of the other pieces are in their solved positions.

Let me explain.
The 5x5x5 has 48 non-fixed center pieces (+ center pieces and X-center pieces), 8 corners, 12 middle edges, 24 wing edges and 6 fixed center pieces (on the xyz axis).

Since the 5x5x5 cube has fixed centers, then we can actually count how many pieces are not in their exact solved positions by comparing their locations with respect to the fixed centers.

If you use pure reduction, meaning that you do not make an effort to solve any corners, middle edges, or wing edges prior to completing all centers, then it is very unlikely that you will have:

4 or more corners in their correct locations (not to mention oriented correctly in their correct locations).
4 or more wing edges in their solved locations.
5 or more middle edges in their solved locations (not to mention oriented correctly in their correct locations).

And...not to mention that to have two or all three of the above on a cube at once.

These high improbabilities are true for the corners, wings, and middle edges on the 5x5x5 from the scramble to the point to where you intentionally just solve the first 4 centers.

Although this is all my opinion, you can give me a "more realistic" set of restrictions if you think that I should be more lenient.
So the probability of getting a last two center skip is indeed 1/4900 if you include the instances when even all corners, all middle edges, and all wing edges are in their correct locations (the most unlikely extreme case).

Just as a realistic guess of mine (you can request a different amount of wings, middle edges and corners to be in their correct locations), but ignoring the orientations of the corners and the middle edges (regardless whether or not they are in their correct locations or not), suppose that we consider the scenarios where:

No more than 3 wing edges are solved, no more than 3 corners are in their correct locations, and no more than 4 middle edges are in their correct locations.

Goal: We are going to calculate a percentage of the number of possible configurations of the 5x5x5 cube which meet the conditions in the above sentence (ignoring the centers completely...just the cage portion of the 5x5x5 cube). We will then multiply this percentage (in decimal form) by 1/4900 to get a smaller probability fraction...because it's less likely to get a last two centers skip than 1/4900.
The formula for the number of positions of the regular 6-color nxnxn cube is:

$$F\left( n \right)=\frac{8!\times 3^{7}}{24^{\left( n+1 \right)\bmod 2}}\left( 12!\times 2^{10} \right)^{n\bmod 2}24!^{\left\lfloor \frac{n-2}{2} \right\rfloor }\left( \frac{24!}{4!^{6}} \right)^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }$$

Since we are ignoring the permutations which the non-fixed centers contribute, we are left with:

$$F\left( n \right)=\frac{8!\times 3^{7}}{24^{\left( n+1 \right)\bmod 2}}\left( 12!\times 2^{10} \right)^{n\bmod 2}24!^{\left\lfloor \frac{n-2}{2} \right\rfloor }$$

Since the 5x5x5 is an odd cube, we have:

$$F\left( n \right)=\left( 8!\times 3^{7} \right)\left( 12!\times 2^{10} \right)24!^{\left\lfloor \frac{n-2}{2} \right\rfloor }$$

and substituting 5 for n, the denominator of the fraction which we will use to calculate a percentage is:

$$\left( 8!\times 3^{7} \right)\left( 12!\times 2^{10} \right)\left( 24! \right)$$

For the numerator of the fraction, we will use the same thing as the denominator, but we will substitute a (smaller) value for each of the factorials.

Let $$f\left( p,x \right)=\sum\limits_{k=p-x}^{p}{\left( \frac{p!\left\lfloor \frac{k\left( k-2 \right)!}{e}+\frac{1}{2}\cos ^{2}\left( \frac{k\pi }{2} \right) \right\rfloor }{k\left( k-2 \right)!\left( p-k \right)!} \right)}$$

(I derived the expression in the summation in this thread. I did remove the "round" operator and replaced it with the fldoor and cosine function, but it's still the same formula for the integers...which is all that counts with permutation puzzles).

This can be simplified to:

$$f\left( p,x \right)=\sum\limits_{k=p-x}^{p}{\left( \frac{p!\left( \frac{!k}{k-1} \right)}{k\left( k-2 \right)!\left( p-k \right)!} \right)=\sum\limits_{k=p-x}^{p}{\left( \frac{p!\left( k-1 \right)\left( \frac{!k}{k-1} \right)}{k\left( k-1 \right)\left( k-2 \right)!\left( p-k \right)!} \right)}}$$

$$=\sum\limits_{k=p-x}^{p}{\left( \frac{p!\left( !k \right)}{k!\left( p-k \right)!} \right)}=\sum\limits_{k=p-x}^{p}{\left( \begin{matrix} p \\ k \\ \end{matrix} \right)\left( !k \right)}$$

Where:
p = the number of the piece type. For wing edges, p = 24, for middle edges, p = 12, and for corners, p = 8.
(p is just the total number of each piece type).

No more than x of each piece type are solved.
$$0\le x\le p$$

(k is just the index of the summation).

Note that $$\sum\limits_{k=p-p}^{p}{\left( \begin{matrix} p \\ k \\ \end{matrix} \right)\left( !k \right)}=\sum\limits_{k=0}^{p}{\left( \begin{matrix} p \\ k \\ \end{matrix} \right)\left( !k \right)}=p!$$

So, here are the calculations which must be done to get a percentage:
We must do a calculation for the wing edges, middle edges, and corners separately.

The Wing Edges
p is always 24 (in cube sizes 4x4x4 and greater) because there 24 wings per orbit.
Since I gave the scenario that we allow no more than 3 wing edges to be solved, x = 3.

So we have:

$$\sum\limits_{k=24-3}^{24}{\left( \begin{matrix} 24 \\ k \\ \end{matrix} \right)\left( !k \right)}$$

The Middle Edges
p is always 12 (in cube sizes 3x3x3 and greater) because there are 12 middle edges on every odd cube greater than the 1x1x1.
Since I gave the scenario that we allow no more than 4 middle edges to be in their correct positions, x = 4.

So we have:

$$\sum\limits_{k=12-4}^{12}{\left( \begin{matrix} 12 \\ k \\ \end{matrix} \right)\left( !k \right)}$$

The Corners
p is always 8 (in cube sizes 2x2x2 and greater) because there 8 corners in cube sizes 2x2x2 and greater.
Since I gave the scenario that we allow no more than 3 corners to be solved, x = 3.

So we have:

$$\sum\limits_{k=8-3}^{8}{\left( \begin{matrix} 8 \\ k \\ \end{matrix} \right)\left( !k \right)}$$
And so our fraction which gives us our percentage is:

$$\frac{\left( \text{39549}\times 3^{7} \right)\left( \text{477248562}\times 2^{10} \right)\left( \text{608667230147569787984693} \right)}{\left( 8!\times 3^{7} \right)\left( 12!\times 2^{10} \right)\left( 24! \right)}\approx \text{0}\text{.958731}$$

And therefore, with the (realistic) assumption that
No more than 3 wing edges are solved, no more than 3 corners are in their correct locations, and no more than 4 middle edges are in their correct locations.
the probability of a last two centers skip on the 5x5x5 cube (using pure reduction) (ignoring the orientations of the corners and middle edges) is

$$\approx \frac{1}{4900}\left( \text{0}\text{.958731} \right)\approx \frac{1}{5111}$$

In other words, the official probability claim for a last two center skip on the 5x5x5 with reduction claimed that it is 4.13% more likely to occur than what it actually is (again, by the "realistic" assumptions/restrictions I presumed). And, I think I might have been too lenient at that. So it could well be more than 4.13% off in reality.

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ThomasJE

Skip on Guimond 1st step (3/4 of a face with opposite colours) and Ortega step 1 (face)?

cuBerBruce

Member
Skip on Guimond 1st step (3/4 of a face with opposite colours)
Assuming you're interested in the probability for the color neutral solving...
[post]147363[/post]

So probability is 3097152/3674160 or approximately 84.3%.

and Ortega step 1 (face)?
[post]152726[/post]
[post]737035[/post]

So probability (again, assuming color neutral) is 22654/3674160 (approximately 0.617% or about 1 in 162).

Pokerizer

Member
To give a comparison, the chance of a LL skip with no partial edge control, and with the possibility of AUF, is
1 / 15552

I thought it was pretty neat.

Chris
well that sux, I had a solve the other day with a LL skip without having to align edges.... So much for seeing that again any time soon lmao