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Probability Thread

IRNjuggle28

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a) What's the probability that a random 2x2x2 position can be completely identified by only seeing the U, F, R, and L faces?
b) Same as a, but the 6 faces are a random permutation of the 6 colors.
c) How about on a 3x3x3?
d) How about on a 4x4x4? :p

I lack the knowledge to answer these, sadly. My thoughts for A are that the ones where it could be identified with UFRL are the ones where the UFR and UFL corners don't share any colors. If, for example, the blue/yellow/orange corner and the white/green/red corner were the two UF corners, other corners could be determined from only seeing two colors because, for example, if the DFL corner had blue on L and yellow on F, the other color has to be red because there are only two pieces with blue and yellow, and one of them already has all 3 colors visible on UFRL. The less variety of UFR and UFL colors, the harder it'll be to identify. It wouldn't shock me if the answer is 100% and I simply can't see how it would be done, though.

I doubt I said anything you don't already know. :p

B is stated a bit unclearly. Can you clarify what you're asking?

I have no idea about C. I feel confident that D is 0%, since the centers on B and D are completely undetectable, and could be in any configuration. Even if the cube looked completely solved only looking at UFRL, it could still be not solved.
 

kinch2002

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whats the probability that the 4 cross edges are oriented but not permuted, ie all on D layers but not correctly permuted?

on one side

\( \frac{23}{12!*2^{11}} = \frac{23}{980995276800} \)
Just been reading through recent posts in this thread and I think for once Chris is wrong :p
23 = Number of ways the 4 cross pieces can be arranged to create the case we are interested in
12! * 2 ^ 11 = Number of different edge configurations
I believe for the second number you actually want 24*22*20*18 = 190080 = Number of different configurations for those 4 edges
This gives 23/190080 = 0.0121%
 

cmhardw

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Just been reading through recent posts in this thread and I think for once Chris is wrong :p
23 = Number of ways the 4 cross pieces can be arranged to create the case we are interested in
12! * 2 ^ 11 = Number of different edge configurations
I believe for the second number you actually want 24*22*20*18 = 190080 = Number of different configurations for those 4 edges
This gives 23/190080 = 0.0121%

That's true, I see that now.

Using the approach I used, my first calculation should have been:
\( \frac{23*8!*2^7}{12!*2^{11}}=\frac{23}{12*11*10*9*2^4}=\frac{23}{190080} \)

Good catch!
 

uberCuber

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If, for example, the blue/yellow/orange corner and the white/green/red corner were the two UF corners, other corners could be determined from only seeing two colors because, for example, if the DFL corner had blue on L and yellow on F, the other color has to be red because there are only two pieces with blue and yellow, and one of them already has all 3 colors visible on UFRL.

If you can see two of the stickers on a corner piece, knowing your color scheme will automatically tell you the third color; looking at other corners is unnecessary. DBL and DBR are the only potentially ambiguous pieces because you can only see one sticker of each.
 

IRNjuggle28

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If you can see two of the stickers on a corner piece, knowing your color scheme will automatically tell you the third color; looking at other corners is unnecessary. DBL and DBR are the only potentially ambiguous pieces because you can only see one sticker of each.

Very true. I'm thinking that with each DB corner, you have a 1/3 chance of the color showing that differentiates the corners, so if the two corners at DBL and DBR were a pair of corners that shared two colors with each other, there would be a 4/9 chance that neither corner would have that color showing. So, it may be that (odds of DBL and DBR corners sharing 2 colors)X(4/9) is the answer. I'm guessing it's more complex, though.

Well, the answer is not 100%. I have found two permutations that cannot be differentiated from each other by UFLR. A pure flip T case versus a T case with the two corners that are not oriented swapped. Algs below.

R U2 R' U' R U' R' L' U2 L U L' U L y' x
cannot be differentiated from
R U R' U' R' F R2 U' R' U' R U R' F' R U2 R' U' R U' R' L' U2 L U L' U L y' x
 
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Tempus

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a) What's the probability that a random 2x2x2 position can be completely identified by only seeing the U, F, R, and L faces?
I think that this is 16/21, or about 76.19%, but I'm not completely certain that I haven't missed some esoteric parity complication. Needs more thought...

b) Same as a, but the 6 faces are a random permutation of the 6 colors.
I've been thinking about this one for a while, and it seems that the more I think about it, the more complex the question turns out to be. Every time I think I'm beginning to get a handle on it, I discover some new level of complexity behind the previous one.

c) How about on a 3x3x3?
I'm not even going to touch this one.

d) How about on a 4x4x4? :p
I'm pretty sure that IRNjuggle28 is correct about this one being 0%.
 
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Phillip1847

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Well, the answer is not 100%. I have found two permutations that cannot be differentiated from each other by UFLR. A pure flip T case versus a T case with the two corners that are not oriented swapped. Algs below.
R U2 R' U' R U' R' L' U2 L U L' U L y' x
cannot be differentiated from
R U R' U' R' F R2 U' R' U' R U R' F' R U2 R' U' R U' R' L' U2 L U L' U L y' x
the red sticker on UB?
 

Tempus

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I think that this is 16/21, or about 76.19%, but I'm not completely certain that I haven't missed some esoteric parity complication. Needs more thought...
Okay, I've given the parity issue more thought, and have determined that I was right to doubt my first impression. Parity excludes some of the scenarios I was thinking were possible. If we assume that the observer is clever enough to calculate the corner parity of a scrambled cube in his head, then the result increases to 6/7, or about 85.714%.

Given a scrambled 2x2 that is stickered the standard way, one can see all three sides of the UFL and UFR pieces, so we know their permutation and orientation. Similarly, we see two sides each of the UBR, DFR, UBL, and DFL pieces, and since we know that the color scheme is standard, we can determine the permutation and orientation of them as well. This leaves only the DBL and DBR pieces, of which we only see one sticker each. Since we know by process of elimination what pieces occupy those two positions, we need only figure out their permutation (2 possibilities) and orientation (3 possibilities taking parity into account).

There are three main scenarios for these two semi-hidden pieces. The first is that they are corners that share no colors, i.e. opposite corners. Each piece only has one such counterpart out of 7 available, so this scenario has a 1/7th chance of occurring. In this case, knowing the parity state of the rest of the cube, seeing one sticker is enough to determine both the orientation and permutation of both pieces. We can see two stickers, so in this scenario there is zero chance of confusion.

The second scenario is that the two pieces share exactly one color, i.e. they are corners that belong on the same face but are otherwise opposite. Each piece has 3 such counterparts out of 7 available, so this scenario has a 3/7ths chance of occurring. In this case, any sticker that isn't the one color they share will instantly give the game away, so only the orientation where the common color is exposed on both pieces will work. In 2/3rds of these cases, parity will make this arrangement impossible, and in the remaining third, there is a 1/3rd chance that the common color will be the one exposed. (1/3)*(1/3)*(3/7)=1/21.

The third and final scenario is that the two pieces share exactly two colors, i.e. they belong adjacent to each other. Let us call the two colors they have in common color A and color B. As before, there are three parity cases. In one case, an orientation exists that reveals stickers A and B, but not one that reveals stickers B and A. In another parity case, an orientation exists that reveals stickers B and A, but not one that reveals stickers A and B. (This bit is what tripped me up the first time 'round.) It seems like you would be able to confuse them, but if you have perfect understanding of parity, it should be possible to determine which corner is which just by seeing A and B or B and A along with all the other stickers that are in view.

This leaves a third and final parity state where either A and A can be in view, or B and B can be in view, or the distinct colors can be in view, with equal probability. A and A will confuse, as will B and B, and so there is a 2/3rds chance of confusion, but only a 1/3rd chance of this parity state. Multiply in the probability of having exactly two colors in common, and you get (2/3)*(1/3)*(3/7)=2/21.

Add them all up and you get 0+1/21+2/21=1/7. A 1 in 7 chance of the cube state being confusable with another, or a 6/7 chance of it being uniquely identifiable merely by viewing the U, F, L, and R sides.

Mind you, this is just for question A. Question B, which I believe allows for non-standard color schemes, is one I am unprepared to answer.
 

elrog

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I'm pretty sure that IRNjuggle28 is correct about this one being 0%.

But you could have a single dedge flip on the DB edge? There are also different ways you can move the centers.

I think it would be interesting to see these stats for both regular cubes and super cubes.
 

goodatthis

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What is the probability of a one move XCross on white/yellow cross? Just white? Full color neutral?

I actually had a scramble like that on the third solve I ever did on TTW. Unfortunately I copied the scramble down but lost it.
 

guysensei1

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That's why LSE skips don't happen often :p If you make sure the centres are either solved or a M2 away from being solved before/during CMLL, like I do, then it goes to under 100,000 (92160 to be precise). LSE skips still aren't going to be happening very often.

If we defined an LSE skip to include the U/U'/U2 or M/M'/M2 away from solved cases, what's the probability then?
(This should go in the probability thread but meh)
 

Renslay

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How many L6E cases are there? 100s? 1000s?

Assuming you fix the corners and centers AUF (so you have to deal only with the edges):

Permutation: 6!/2 = 360
Orientation: 2^5 = 32

Which means 11520 cases.

Edit: okay, didn't see TDM's reply... Which is the same, but together with AUFs (which I think can be a bit misleading: the number of PLLs is also counted without AUFs).
 
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TDM

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Assuming you fix the corners and centers AUF (so you have to deal only with the edges):

Permutation: 6!/2 = 360
Orientation: 2^5 = 32

Which means 11520 cases.

Edit: okay, didn't see TDM's reply... Which is the same, but together with AUFs (which I think can be a bit misleading: the number of PLLs is also counted without AUFs).
Yeah, I should probably have thought to not include AUF. So with the M slice solved it's more likely to get a LSE skip than to get a LL skip in CFOP, but if you have it either solved or an M2 out then it's less likely than a LL skip.
 

cmhardw

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a) What's the probability that a random 2x2x2 position can be completely identified by only seeing the U, F, R, and L faces?

My attempt before reading other responses:
My answer: \( \frac{16}{21} \) or approximately 76.2% chance.

We can completely identify all corners of the U layer as well as DFL and DFR. We cannot identify the corners at DBL or DBR if:
1) Their stickers share exactly 2 colors, and either of those shared colors from both corners is showing at LDB and RDB
2) Their stickers share exactly 1 color, and that color is the one showing at LDB and RDB.


Let's compute the probability that the DBL and DBR corners are unidentifiable:

Case1:
There are 12 possible choices of a pair of corners that share 2 colors on their sets of stickers (i.e. corners that are adjacent on an edge of the cube). Call the corners A and B. Let's say that colors A_1 and B_1 are the same, A_2 and B_2 are the same, and A_3 and B_3 are opposite colors.

If either of A_3 or B_3 is visible at LDB or RDB then both corners are identifiable. If neither of A_3 or B_3 is visible then the corners are ambiguous.

There are 4*2=8 ways to permute those corners such that neither A_3 nor B_3 is visible. Since the orientations of the corners must be a legal state, then after permuting the first corner (4 ways) there is only a 2/3 probability that the corner orientation doesn't force the last corner to show the identifiable color. Also, the last sticker doesn't have two choices anymore, only 1 since it is forced.

The probability of such a case is:
\( \frac{12}{\binom{8}{2}}*\frac{4*\frac{2}{3}}{6}=\frac{4}{21} \)

Case 2:
There are 12 pairs of corners that share exactly one color in their sets of stickers. These are corners on a diagonal of a face (for example UBR and UFL).

Call the corners A and B. Stickers A_1 and B_1 are the same color. Stickers A_2 and B_2 are opposite colors, and so are stickers A_3 and B_3.

If any of the stickers A_2, B_2, A_3, B_3 are visible at LDB and RDB, then both corners are identifiable. If both stickers A_1 and B_1 are visible, then the corners are ambiguous.

There are 2 ways for both stickers to be visible. Since the corner orientations must be in a legal state, we can permute the first sticker in 2 ways, but then there is only a 1/3 probability that the other unidentifiable sticker can be legally permuted to the visible location.

The probability of such a case is:
\( \frac{12}{\binom{8}{2}}*\frac{2*\frac{1}{3}}{6}=\frac{1}{21} \)

Quick note:
If the two corners at DBL and DBR are corners that belong in diagonally opposite corners across the cube, then they are always identifiable at DBL and DBR

Adding the two case probabilities we get:
\( \frac{4}{21}+\frac{1}{21}=\frac{5}{21} \)

Therefore the probability that the 2x2x2 is identifiable is \( \frac{16}{21} \) or approximately 76.2% chance.
 
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