Probability Thread

[CLL Smooth]

Let's say we count mirrors and inverses as distinct cases, and assume all the edges are already solved. The only cases with rotational symmetry are the skip case, the two O PLLs and the two "star" PLLs. (Not sure what the common names for these PLLs are; I don't do megaminx much.) Thus there are ((5!/2)(3^4)−5)/5 + 5 = 976 CLL cases in total.

If we don't assume the edges are solved, then we may fix (say) the UFR corner as a reference; this gives (4!/2)(3^4) = 972 cases without accounting for rotationally-equivalent cases. In addition to the CLL-skip case, there's now one more case with rotational symmetry, so there're (972−2)/5 + 2 = 196 CLL cases if edges aren't solved.

If we do count mirrored cases as the same, then the numbers drop to 490 (edges solved) and 100 (edges not solved); if we also count inverses as the same, then they further drop to 252 (edges solved) and 52 (edges not solved).

Wow! That's a lot to digest.
If I understand you correctly, which I'm unsure of, this means there are 196 distinct LL cases on the Shengshou 2x2 megaminx (kilominx). Sound right?

 

[xyzzy]

If I understand you correctly, which I'm unsure of, this means there are 196 distinct LL cases on the Shengshou 2x2 megaminx (kilominx). Sound right?

Yup.

 

[EetuK00]

What's the probability for no cycle breaks in 3x3 blind memo?

And bonus: How many different memos are there?

 

[tseitsei]

What's the probability for no cycle breaks in 3x3 blind memo?

And bonus: How many different memos are there?

I'm working on the cycle break question but isn't the bonus question quite obvious or am I being stupid?
Obviously there must be as many different memos as there are scrambles. Otherwise it would be impossible to know which scramble you are currently solving...

So bonus question answer for 3x3 is
{\displaystyle (8!\cdot 3^{8-1})\cdot \left({\frac {12!}{2}}\cdot 2^{12-1}\right)=43\,252\,003\,274\,489\,856\,000\approx 43\cdot 10^{18}}

But then again it depends what you mean by memo. Of course you can artificially create memos that don't correspond to any possible 3x3 scramble but if that is the case there are infinite amount of different memos. But to be able to solve every 3x3 scramble you need as many different memos as there are scrambles.

 

[EetuK00]

I'm working on the cycle break question but isn't the bonus question quite obvious or am I being stupid?
Obviously there must be as many different memos as there are scrambles. Otherwise it would be impossible to know which scramble you are currently solving...

So bonus question answer for 3x3 is
{\displaystyle (8!\cdot 3^{8-1})\cdot \left({\frac {12!}{2}}\cdot 2^{12-1}\right)=43\,252\,003\,274\,489\,856\,000\approx 43\cdot 10^{18}}

Oh yeah the bonus question is actually quite obvious now that I think about it

 

[xyzzy]

What's the probability for no cycle breaks in 3x3 blind memo?

One could "break cycles" in order to flip edges or twist corners, but (I think?) most people just use algs for that, so assuming that, we can ignore the orientation of the pieces entirely. In other words, we're asking for the probability that we have only one corner cycle and only one edge cycle. (Please correct me if this is wrong and not what you're asking for, because I have no idea how people do BLD.)

Consider the two cases based on permutation parity. Suppose the corner/edge permutation parities are even. (They must be either both even or both odd.)

P(one corner cycle | even parity) = (P(corner 7-cycle) + P(corner 5-cycle) + P(corner 3-cycle) + P(corners solved)) / P(even parity)
= [(C(8,7)6! + C(8,5)4! + C(8,3)2! + C(8,1)0!)/8!] ⋅ 2
= [1/(7⋅1!) + 1/(5⋅3!) + 1/(3⋅5!) + 1/(1⋅7!)] ⋅ 2
= 43/120

By the same reasoning:
P(one edge cycle | even parity) = [1/(11⋅1!) + 1/(9⋅3!) + 1/(7⋅5!) + 1/(5⋅7!) + 1/(3⋅9!) + 1/(1⋅11!)] ⋅ 2
= 48187/217728

As the corners and edges are conditionally independent once we know the parity, we can conclude that the probability of not having to break cycles when the parities are even is:
P(one corner cycle, one edge cycle | even parity) = P(one corner cycle | even parity) P(one edge cycle | even parity)
= 2072041/26127360
≈ 7.93%

Then we can do the same computation for the odd parity case:
P(one corner cycle, one edge cycle | odd parity) = 39784637/326592000
≈ 12.18%

And conclude:
P(one corner cycle, one edge cycle) = (1/2)(2072041/26127360) + (1/2)(39784637/326592000)
= 131370299/1306368000
≈ 10.06%

 

[EetuK00]

Cool! That answers my question perfectly

 

[tseitsei]

No for the real question. I hope someone can verify my result but I think this should be correct

For edges I get 22.65% chance for no cycle breaks.
And for corners I get 33.98% chance for no cycle breaks.
And together 22.65% * 33.98% = 7.70% for no cycle breaks in whole solve
NOTE: I'M NOT COUNTING FLIPPED/TWISTED PIECES AS CYCLE BREAKS HERE!
Note2: Should I do something for the total propability because of parity? Not sure about this one...

How I got hose numbers:

Spoiler

for edges:
First I have 1/12 of a chance that the buffer piece is in its correct location to start with.
In this case the only way I will have no cycle breaks is if all other edges are in their correct positions already (flipped or not). And possibility for that is quite obviously 1/11!.

Then next case: I have 11/12 * 1/11 (which obviously =1/12 that I only noticed a while later) chance that the buffer piece is my second target (the one right after the buffer piece) since 11/12 that it is not in the buffer position and then 1/11 that it is in the next location we go. And now I have solved 2 edges already so in order to have only 1 cycle the remaining edges must be in their correct positions. Propability for that is 1/10!.

Once more to make the idea clear: I have 11/12 * 10/11 * 1/10 =1/12 chance that the buffer piece is my 3rd target. And again the remaining 9 edges need to have correct permutation for only 1 cycle to exist so propability is 1/9!.

We continue this until we have covered all cycle lengths so up to cycle length of 11 where we have 1/12 chance for the buffer to be there and 0 edges remaining after that so we get 1/0! for propability for our remaining edges.

so now we put it all together and get a neat sum (don't know the correct format for this forum but...):

sum(n goes 0...11)[ (1/12)*(1/n!) = 22.65%

and similarly for corners we get a sum:
sum(n goes 0...7)[(1/8)*(1/n!) = 33.98%

EDIT: ninja'd and he did it better So I should have done something with parity... Well my numbers were approximately correct anyway

 

[DGCubes]

Oh yeah the bonus question is actually quite obvious now that I think about it

Technically multiple memos could correspond to the same scramble. For example, doing cycles in different orders makes a difference, as does how you memorize flipped edges and twisted corners. I think a lot of it also depends on a person's particular memo method. So, I don't think you could really get a true answer to the bonus question.

 

[tseitsei]

Technically multiple memos could correspond to the same scramble. For example, doing cycles in different orders makes a difference, as does how you memorize flipped edges and twisted corners. I think a lot of it also depends on a person's particular memo method. So, I don't think you could really get a true answer to the bonus question.

Yep there are really 2 possible answers:
Q: How many memos do you NEED to be able to solve any 3x3 scramble?
A: The amount of different scrambled states possible.
Q: How many different memos could be made?
A: Infinite amount.

 

[EetuK00]

Yep there are really 2 possible answers:
Q: How many memos do you NEED to be able to solve any 3x3 scramble?
A: The amount of different scrambled states possible.
Q: How many different memos could be made?
A: Infinite amount.

Wait what? Infinite? If we give every sticker its own letter, you cant write out infinite amount of memos(?) Am I missing something here?

 

[tseitsei]

Wait what? Infinite? If we give every sticker its own letter, you cant write out infinite amount of memos(?) Am I missing something here?

Well depends on how we define memo. If I have to memo letters WR I could memo WalRus or WaR or World Record... would those be the same memo or different memo.

Also even if we define 'memo' to be just the string of letters we can always just lengthen the memo (there is no practical reason to do so but we CAN do it.). Trivial example: I have a 3 cycle buffer -> A -> B -> buffer. Normally you would memo AB But you could also memo BABA for the same outcome. Or something like IB IA BI. As I said stupid but possible...

 

[EetuK00]

Well depends on how we define memo. If I have to memo letters WR I could memo WalRus or WaR or World Record... would those be the same memo or different memo.

Also even if we define 'memo' to be just the string of letters we can always just lengthen the memo (there is no practical reason to do so but we CAN do it.). Trivial example: I have a 3 cycle buffer -> A -> B -> buffer. Normally you would memo AB But you could also memo BABA for the same outcome. Or something like IB IA BI. As I said stupid but possible...

Ahh I see

 

[EetuK00]

Well what's the probability for not having to orientate centers in 3x3 supercube?
(If you completely ignore center orientarion during the solve)

 

[xyzzy]

Well what's the probability for not having to orientate centers in 3x3 supercube?

There're 4 possible orientations for each centre and there're 6 centres, along with a parity constraint (you can't rotate a single centre 90°), so that's \( 4^6/2=2048 \) centre orientations and a 1/2048 chance of not having to orient centres.

 

[PhillipEspinoza]

Can anyone tell me:

1) What are the chances, after EO Line, that any F2L edge (4/12) will be paired up with a correctly oriented 3rd layer corner (4/8)?

2) What about the chances that a top layer edge piece (non EO Line, so 2/12) will be connected to a correctly oriented 3rd layer corner (as if it were a 1st layer corner)? [You have 4 correct positions with 4 3rd layer corners that can join either side of the 2 edge pieces. There are 8 corners and they each have 3 different ways to be oriented. Hope this helps.]

3) What is the chance that either 1 OR 2 will randomly occur after EO Line during block building?

4) We all know the probability of an OLL skip with ZZ is 1/27. But what is the probability of an OLL skip after ELS? In other words, if you just randomly insert the edge piece into LS, without regard to corner orientation of the corner it's paired with, does that decrease the chances of an OLL skip to 1/81?

5) If you do ELS with a correctly oriented LL corner, your chances of ending up with all pieces in LL oriented is 1/27 right? And if I do not get an orientation skip, and I do a CLS alg, my chances of skipping PLL are 1/72. Given this, what are my chances that I will either get an OLL skip or a PLL skip (for ZZ)? In other words, what are the chances of a LL skip after LS using these combined methods?

THANKS IN ADVANCE TO ANYONE WHO CAN MAKE SENSE OF WHAT I'M SAYING IN MY SLEEP DEPRIVED STATE. MUCH APPRECIATED!

 

[AlphaSheep]

Can anyone tell me:

1) What are the chances, after EO Line, that any F2L edge (4/12) will be paired up with a correctly oriented 3rd layer corner (4/8)?

2) What about the chances that a top layer edge piece (non EO Line, so 2/12) will be connected to a correctly oriented 3rd layer corner (as if it were a 1st layer corner)? [You have 4 correct positions with 4 3rd layer corners that can join either side of the 2 edge pieces. There are 8 corners and they each have 3 different ways to be oriented. Hope this helps.]

3) What is the chance that either 1 OR 2 will randomly occur after EO Line during block building?

4) We all know the probability of an OLL skip with ZZ is 1/27. But what is the probability of an OLL skip after ELS? In other words, if you just randomly insert the edge piece into LS, without regard to corner orientation of the corner it's paired with, does that decrease the chances of an OLL skip to 1/81?

5) If you do ELS with a correctly oriented LL corner, your chances of ending up with all pieces in LL oriented is 1/27 right? And if I do not get an orientation skip, and I do a CLS alg, my chances of skipping PLL are 1/72. Given this, what are my chances that I will either get an OLL skip or a PLL skip (for ZZ)? In other words, what are the chances of a LL skip after LS using these combined methods?

THANKS IN ADVANCE TO ANYONE WHO CAN MAKE SENSE OF WHAT I'M SAYING IN MY SLEEP DEPRIVED STATE. MUCH APPRECIATED!

1) I'd prefer someone else to confirm my answers but I think the chances of one specific edge being paired with any 3rd layer corner in a specific orientation is 1/6 (1/3 chance of being oriented and 1/2 chance of being from the last layer). That means that the odds of at least one out of two specific edges being like that are 11/36, and for at least one out of 3 is 91/216 and for at least one of four specific edges, its 671/1296 or 51.77%.

2) Not sure I understand. Can you give an example?

3) Not sure how to work this out, because the state after 1 move is not completely independent of the state before that move. I think that the answer depends on how inefficient your solution is. The more moves you take, the more likely it is to happen, I'm sure.

4) Yes 1/81 is correct.

5) Yes, if the corner you insert is oriented then the chance of the rest also being oriented is 1/27. And if you don't influence permutation of the other corners during CLS, then yes, theres a 1/72 chance of a PLL skip. That means there's a 49/972 (about 5%) chance of skipping at least one of OLL or PLL and 1/1944 chance of skipping both. Same as nomal OLL+PLL for ZZ.

 

[guysensei1]

but isn't the bonus question quite obvious or am I being stupid?

I think you're missing something here, there's more than 1 valid way to memorise scrambles with cycle breaks. You can choose any unsolved piece to shoot to after all, so there exist many many scrambles with multiple valid memoes.

Example: assume UB and UF are swapped, and your buffer is somewhere else. You can memo UB UF UB or UF UB UF or BU FU BU or FU BU FU (convert those to whatever scheme you're using.)

 

[xyzzy]

1) I'd prefer someone else to confirm my answers but I think the chances of one specific edge being paired with any 3rd layer corner in a specific orientation is 1/6 (1/3 chance of being oriented and 1/2 chance of being from the last layer). That means that the odds of at least one out of two specific edges being like that are 11/36, and for at least one out of 3 is 91/216 and for at least one of four specific edges, its 671/1296 or 51.77%

The probabilities aren't independent (e.g. if you have FR in the UB position and BR in the UR position, the matching corner position for both of those edges is UBR), so this is not exactly accurate, but it's pretty close to the correct value. I'm too tired to exert my brain, so I just did a simulation on a million random states and got this:

No pairs: 453642
1 pair: 433056
2 pairs: 106154
3 pairs: 7070
4 pairs: 78

 

[SiTeMaRo]

I had a really interesting one come up a few days ago. After solving the cross and the first two pairs, all the remaining 12 pieces were in their correct locations, only 6 of them were misoriented.

The probability of having the last 12 pieces of the cube in the correct location after solving the cross and first two pairs is:

1 / [6! * 6!/2] = 1 / 259200

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

And I still had 3 LL skips in just an year.