Probability Thread

[ottozing]

Yet another set of FMC related requests. Looking to adjust my time management strategy and these would likely help me a bit with determining what I end up doing in terms of checking EO's & utilizing NISS prediction

1 - What are the odds that after your EO is solved, the corner and edge state on at least one of the two axis' is 4C4E, 4C2E, or 3C2E collectively without adding any moves after the EO? These are the 3 triggers I sometimes NISS for since they can be quite easy (R for 4C4E, R U2 R for 4C2E, R U/U' R for 3C2E). Similarly, what are the odds that an EO case will be only one move away from these 3 pre mentioned states? (This may be trickier to calculate accurately since for an EO that ends with F there's 8 possible 1 move follow-ups, and for EO's ending in F B there are 16. It's also not as important as my first request)

2 - For 4C4E, every case has between 0 and 8 pairs formed (4 possible top pairs and 4 possible side pairs). For all of these cases (example: 3 top pairs + 1 side pair, or 2 top pairs 1 side pair etc.), what % of those cases are x moves or less with x being from 0-6 moves? For info on what a pair is in this context, see this document by Rodney Kinney > https://docs.google.com/document/d/...SJpMArtRrDT-p_CwM/edit#heading=h.o0b5b27es870

(I'm looking to basically confirm a suspicion of mine that top pairs matter much more than side pairs. As for interchangeable vs non interchangeable pairs, I'm already pretty convinced they don't matter, and even if they do they're quite a bit harder to NISS predict so I'm reluctant to implement that into my FMC attempts)

 

[Puzzlerr]

whats the odds of me getting 2 last layer skips? Not in a row, just in the span of a year
because that happened to me
and my last one was timed

 

[White KB]

@ottozing How do I get Platinum Member status? Or is it just a legacy feature?

But more on point:

whats the odds of me getting 2 last layer skips? Not in a row, just in the span of a year
because that happened to me
and my last one was timed

Well, it really depends on how many solves you did. If you did 3 solves in a year, it would be fairly low, but if you did 3 million in a year it would be higher. The number of LL possibilities are 4!*4!*2^4*3^4/(2*3*2*4), which simplifies to 62,208.
Then one would have to divide by 4 to get the proper denominator since an LL skip usually means the last layer is solved minus parity. This yields a result of 15,552 possible cases, disregarding AUF.

Thus, the chances on any given solve of getting an LL skip are 1/15,552, or 0.0064300412%. This does not take "forcing" an LL skip into account, as often when a cuber is solving the cube they may run into a situation wherein the cube needs U2 R U2 R' to be solved, and the cuber (who would typically solve the case using U R U' R') realizes that they can get an LL skip if they use U2 R U2 R' to solve it. If we pretend that LL is not predictable at all whatsoever while solving the final F2L pair, then we can proceed. It is also good to note that ZZ / Petrus solvers are advantaged in this respect because at some point in the solve EO is done, meaning that the chances of getting an LL skip on any given solve multiply by 8 because all the edges are oriented. (1 in 1944; 0.0514403292%) DR solvers are further advantaged because the DR stage of a solve requires the corners to be oriented as well, thus multiplying the chances of an LL skip by an additional factor of 27. (1 in 72; 1.3888888888...%) HTR users enjoy a 1 in 4 chance of an LL skip, if you don't count AUF. (A 25% chance!)

Putting all of this aside, we can look at the chances of a CFOP solver [who is unable to see the last layer] getting a last layer skip.
To find the chances of this person getting at least 2 LL skips, we have to first determine the chances of them not getting any. We can put the number of solves in any given time period in as x, and then the chances of getting an LL skip based on this will be a function of x. The best equation I could come up with is:
1-\left(\left(\frac{15561}{15562}\right)^{x}\right)^{2}
which when put into the Desmos Graphing Calculator yields a probability chart.
If you can't access Desmos, here's an image of the equation:

I hope that was informative, but not too wordy. And I hope I was correct!

 

[Imsoosm]

@ottozing How do I get Platinum Member status? Or is it just a legacy feature?

Not completely sure but it's probably just an old member thing. (Kirjava is called Colourful)

 

[Thom S.]

Kirjava is called Colourful

If you are a premium Member you can change what is says under your Username.
Kirjava used Colourful.

 

[SollsIsCool]

how common is a accidental x cross

 

[ruffleduck]

how common is a accidental x cross

I assume you are asking what the probability is of having xcross solved provided cross is solved.
(4 * 7! * 7! * 2^6 * 3*6) / (8! * 8! * 2^7 * 3^7 / 2) = 1/1944

 

[turtwig]

I assume you are asking what the probability is of having xcross solved provided cross is solved.
(4 * 7! * 7! * 2^6 * 3*6) / (8! * 8! * 2^7 * 3^7 / 2) = 1/1944

I don't know where you got this number, but it's much higher. The chance that a given pair is solved is 1/8 * 1/2 * 1/8 * 1/3 = 1/384 (edge perm, edge orientation, corner perm, corner orientation) and there are 4 pairs (which does not mean the probability is 4x, although it would probably be close to that).

EDIT: I'm going to try to calculate this.

So there are 8! * 2^7 * 8! * 3^7 / 2 positions with a given cross solved, which is (edge permutation) * (edge orientation) * (corner permutation) * (corner orientation) / (parity).

For a given pair, there are 7! * 2^6 * 7! * 3^6 / 2 positions. Then, there are 4 * 7! * 2^6 * 7! * 3^6 / 2 positions where at least one pair is solved.

However, this overcounts positions where there are at least 2 pairs solved. Given 2 pairs, there are 6! * 2^5 * 6! * 3^5 / 2 positions. There are (4 choose 2) = 6 combinations of 2 pairs, so we subtract 6 * 6! * 2^5 * 6! * 3^5 / 2 to correct for overcounting positions with 2 pairs.

Now, we need to add back positions with 3 pairs (which have been counted (3 choose 1) - (3 choose 2) = 0 times). We add (4 choose 3) * 5! * 2^4 * 5! * 3^4 / 2.

Finally, positions with 4 pairs have been counted (4 choose 1) - (4 choose 2) + (4 choose 3) = 2 times, so we subtract 4! * 2^3 * 4! * 3^3 / 2.

Thus, there are 4 * 7! * 2^6 * 7! * 3^6 / 2 - 6 * 6! * 2^5 * 6! * 3^5 / 2 + 4 * 5! * 2^4 * 5! * 3^4 / 2 - 4! * 2^3 * 4! * 3^3 / 2 = 2358218126592 positions with at least one pair solved. Thus, our probability is 2358218126592 / (8! * 2^7 * 8! * 3^7 / 2) = 37908599/3657830400, which is about 1.04%. I hope I did this right, I'm pretty rusty on this kind of math. This aligns with my intuition though, ending up with a pair solved accidentally is not too uncommon in my experience, although I sometimes try to preserve corners/edges during my cross solution without (consciously) solving the pair entirely, so the chance I get accidental pairs is probably higher because of that.

This doesn't include the minute chances of ending up with an xcross on another color, but that would be much harder to calculate and also is probably against the spirit of the question.

 

[ruffleduck]

I don't know where you got this number, but it's much higher.

I multiplied the number of cases of a specific slot being solved by 4, thinking that would cover cases for any slot. Good catch

 

[Bruce MacKenzie]

I assume you are asking what the probability is of having xcross solved provided cross is solved.
(4 * 7! * 7! * 2^6 * 3*6) / (8! * 8! * 2^7 * 3^7 / 2) = 1/1944

There are 24 ways to place the first corner and 16 ways to place the fifth edge. So:

24 * 16 / 4

since the xcross may be completed on any of the four corners of the face. 1 in 96 is the chance of a windfall xcross after forming the cross.

 

[ottozing]

Yet another set of FMC related requests. Looking to adjust my time management strategy and these would likely help me a bit with determining what I end up doing in terms of checking EO's & utilizing NISS prediction

1 - What are the odds that after your EO is solved, the corner and edge state on at least one of the two axis' is 4C4E, 4C2E, or 3C2E collectively without adding any moves after the EO? These are the 3 triggers I sometimes NISS for since they can be quite easy (R for 4C4E, R U2 R for 4C2E, R U/U' R for 3C2E). Similarly, what are the odds that an EO case will be only one move away from these 3 pre mentioned states? (This may be trickier to calculate accurately since for an EO that ends with F there's 8 possible 1 move follow-ups, and for EO's ending in F B there are 16. It's also not as important as my first request)

2 - For 4C4E, every case has between 0 and 8 pairs formed (4 possible top pairs and 4 possible side pairs). For all of these cases (example: 3 top pairs + 1 side pair, or 2 top pairs 1 side pair etc.), what % of those cases are x moves or less with x being from 0-6 moves? For info on what a pair is in this context, see this document by Rodney Kinney > https://docs.google.com/document/d/...SJpMArtRrDT-p_CwM/edit#heading=h.o0b5b27es870

(I'm looking to basically confirm a suspicion of mine that top pairs matter much more than side pairs. As for interchangeable vs non interchangeable pairs, I'm already pretty convinced they don't matter, and even if they do they're quite a bit harder to NISS predict so I'm reluctant to implement that into my FMC attempts)

Bumping this just to say that part one of question one has already been answered via some work Cale is currently doing - https://docs.google.com/spreadsheet...4TLW45M2ZkctQZAVxgQqEtlUs0/edit#gid=973034617

 

[SollsIsCool]

I assume you are asking what the probability is of having xcross solved provided cross is solved.
(4 * 7! * 7! * 2^6 * 3*6) / (8! * 8! * 2^7 * 3^7 / 2) = 1/1944

i got lucky then!!

 

[ottozing]

Yet another set of FMC related requests. Looking to adjust my time management strategy and these would likely help me a bit with determining what I end up doing in terms of checking EO's & utilizing NISS prediction

1 - What are the odds that after your EO is solved, the corner and edge state on at least one of the two axis' is 4C4E, 4C2E, or 3C2E collectively without adding any moves after the EO? These are the 3 triggers I sometimes NISS for since they can be quite easy (R for 4C4E, R U2 R for 4C2E, R U/U' R for 3C2E). Similarly, what are the odds that an EO case will be only one move away from these 3 pre mentioned states? (This may be trickier to calculate accurately since for an EO that ends with F there's 8 possible 1 move follow-ups, and for EO's ending in F B there are 16. It's also not as important as my first request)

2 - For 4C4E, every case has between 0 and 8 pairs formed (4 possible top pairs and 4 possible side pairs). For all of these cases (example: 3 top pairs + 1 side pair, or 2 top pairs 1 side pair etc.), what % of those cases are x moves or less with x being from 0-6 moves? For info on what a pair is in this context, see this document by Rodney Kinney > https://docs.google.com/document/d/...SJpMArtRrDT-p_CwM/edit#heading=h.o0b5b27es870

(I'm looking to basically confirm a suspicion of mine that top pairs matter much more than side pairs. As for interchangeable vs non interchangeable pairs, I'm already pretty convinced they don't matter, and even if they do they're quite a bit harder to NISS predict so I'm reluctant to implement that into my FMC attempts)

Bump again because I have a better 2nd question than the one I gave

Rather than knowing the % breakdowns, I think it would be better to know how often a 4C4E gives x number of pairs. My reason for this is that, hypothetically, if I know that 4 pairs is a 10% chance & 3 pairs is a 33% chance (random numbers for now since I don't actually know the odds), I could know that if I have 20 possible 5 moves to 4C4E to choose from that I'll likely have 2 which are 4 pairs & more likely to give a short finish when I NISS for those specifically

 

[Noob's Cubes]

What are the chances of a complete x-cross skip?

 

[Jorian Meeuse]

I dont have proof so if you dont believe me its fine, i probably also wouldnt belive it if someone said this but my first sub 6 (5.33) was with a premade x-cross, i think on yellow. As said in the easy/lucky/a lot of other things scrambles thread, i lost the scramble, but if i somehow retrieve it, ill make sure to post it here.

 

[Bruce MacKenzie]

What are the chances of a complete x-cross skip?

The number of permutations of one corner cubie and five edge cubies may be calculated:

24 • ( 24 • 22 • 20 • 18 • 16) = 72,990,720

The X-Cross pattern may be applied to the cube in 24 orientations, so the chance of a random scramble giving an instance of X-Cross is 24 times that or 1 in 3,041,280.

 

[ottozing]

Bump again because I have a better 2nd question than the one I gave

Rather than knowing the % breakdowns, I think it would be better to know how often a 4C4E gives x number of pairs. My reason for this is that, hypothetically, if I know that 4 pairs is a 10% chance & 3 pairs is a 33% chance (random numbers for now since I don't actually know the odds), I could know that if I have 20 possible 5 moves to 4C4E to choose from that I'll likely have 2 which are 4 pairs & more likely to give a short finish when I NISS for those specifically

bump

anyone?

 

[Savi349]

I have a quick question about the PLL sheet that was developed by Feliks Zemdegs. Does anyone know how the probabilities were calculated for the different algorithms? My solves do generally seem to follow the probabilities listed, but I was wondering how he came to these numbers – did he do his own solves and note down each PLL he encountered? Was it some sort of computer generation? Because I also added up the probabilities, however they do not add up to 1; where am I going wrong?

Also, does a "skip" count as a PLL case, and therefore the probability of a "skip" is also part of these listed probabilities? Or does the PLL sheet somehow omit the probability of a skip?
Any answers to any of my queries are greatly appreciated! My intense curiosity of this paired with my lack of confidence with maths is not a very pleasant combination.
Thank you!

 

[OreKehStrah]

I have a quick question about the PLL sheet that was developed by Feliks Zemdegs. Does anyone know how the probabilities were calculated for the different algorithms? My solves do generally seem to follow the probabilities listed, but I was wondering how he came to these numbers – did he do his own solves and note down each PLL he encountered? Was it some sort of computer generation? Because I also added up the probabilities, however they do not add up to 1; where am I going wrong?

Also, does a "skip" count as a PLL case, and therefore the probability of a "skip" is also part of these listed probabilities? Or does the PLL sheet somehow omit the probability of a skip?
Any answers to any of my queries are greatly appreciated! My intense curiosity of this paired with my lack of confidence with maths is not a very pleasant combination.
Thank you!

Sure!

Basically you can think of full PLL as 72 cases. However, you also get some cases that are fully rotational symmetric, and can AUF the non-fully rotationally symmetric cases. So the 4 angles of T perm can be treated as 1 unique case by AUFing.

As such, any fully rotational case including a skip case is 1/72, while any other case has a probability of 4/72, which is equal to 1/18, because of the fact it has 4 AUFs.

Edit: Z perm and E perm have a probability of 1/36 because for solved cp and diag cp, the cp swap is rotationally symmetric, and those 2 PLLs have U2 angle symmetry.

Hope this helps!

 

[Mastermind2368]

(I only recently got interested in probability, so feel free to call me out if I got anything wrong)

To add on to the answer above, the original 72 cases is calculated from doing:

(4!*4!)/(2*4)

Incase you haven't learnt about it yet, the "!" Means to make a number factorial. This means to multiple it by every smaller natural number. In other words, 4! = (4*3*2*1).

The two 4! Represent all the possibile corner and edge cases. Because each are independent, they're multiplied together, so that each corner case can be lined up with each edge case.

The total number produced by that is quite high though. But there are a couple of other things that need to be taken into account. Firstly, parity. You can never have an edge swapped with another edge, or a corner swapped with a corner. So this reduces the cases by half. Then secondly, remember that we can do any auf. Think of it like this, even though there's only one h perm alg, the aufs would turn it in to four different algs. If we were solving for 8 pieces of the cube that weren't on the same face, we couldn't do an auf like this, but since we can, it reduces the total amount of cases to an eighth of the original amount.

Therefore the total amount is (4!*4!)/(2*4)
Which simplifies to: 576/8
Which is: 72

If you're interested in learning more about this stuff, I'd recommend watching Jperm's video about the total probability for the cube. He explains the concepts much better than I did