Probability Thread

[cmhardw]

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

On any side (color neutral)? On either of 2 opposite sides (opposite color neutral)? On only 1 side (no color neutrality)?

 

[Deleted member 36445]

on one side

 

[cmhardw]

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

\( \frac{23}{12!*2^{11}} = \frac{23}{980995276800} \)

--edit--
This calculation is incorrect, and was corrected by Daniel Sheppard here.

 

[Deleted member 36445]

thanks! one more thing, how much percent is that, please?

 

[Rocky0701]

thanks! one more thing, how much percent is that, please?

You couldn't of just typed in: 23 divided by 980995276800 into Google? Anyway, it is 2.34455767E-11% of a chance. AKA a .00000000234455767% chance of happening.

 

[Deleted member 36445]

You couldn't of just typed in: 23 divided by 980995276800 into Google? Anyway, it is 2.34455767E-11% of a chance. AKA a .0000000000234455767% chance of happening.

thanks I didn't know how to calcute the percentage though

 

[10461394944000]

thanks I didn't know how to calcute the percentage though

...just multiply it by 100

 

[Carrot]

thanks I didn't know how to calcute the percentage though

jeez... percentage... derives from 'per cent' (kind of ish) which roughly means 'hundredths'. so basically multipy any number by 100 and you have the number represented as percentage. (0.25 = 25%, 0.01 = 1%, 25 = 2500%) It's THAT simple

EDIT: Ben, y u ninja?

 

[Deleted member 36445]

jeez... percentage... derives from 'per cent' (kind of ish) which roughly means 'hundredths'. so basically multipy any number by 100 and you have the number represented as percentage. (0.25 = 25%, 0.01 = 1%, 25 = 2500%) It's THAT simple

sorry, I'm dumb

 

[kinch2002]

What would the prob. be of getting 3 LL skips in about 17k solves? (this is my situation)

I believe this should just be a binomial probability, right?
(17000 choose 3) * (1/15552)^3 * (15551/15552)^16997 = 7.3% chance of 3 LL skips in 17000 solves.

Indeed, but your question is needlessly specific, resulting in an answer that exaggerates your true 'luckiness'. Less misleading would be the probability of getting at least 3 LL skips in 17k solves, which is 9.82%

 

[Robert-Y]

^People need to stop forgetting this

On a related note... I was wondering how unlucky you have to be to get x solves in row without a LL skip. Here are the probabilities that you will receive at least one LL skip using the standard CFOP method without any special tricks, for certain x number of solves in a row:

100 solves: 1 - (15551/15552)^100 ~= 0.64%
500 solves: 1 - (15551/15552)^500 ~= 3.16%
1000 solves: 1 - (15551/15552)^1000 ~= 6.23%
5000 solves: 1 - (15551/15552)^5000 ~= 27.5%
10000 solves: 1 - (15551/15552)^10000 ~= 47.43%
50000 solves: 1 - (15551/15552)^50000 ~= 95.98%
100000 solves: 1 - (15551/15552)^50000 ~= 99.84%

So you could say that if you've done at least 50000 solves in your lifetime, you have probably received at least one LL skip hopefully
But if not, it doesn't mean you're *that* unlucky since the chances that you haven't received one in 50000 solves is ~= 4% or 1/25.

 

[advincubing]

What is the probability of the last F2L corner + edge set being already paired in the top layer, without manipulating the earlier inserts to force pairing? (Relates to question of how often partial edge control -- sledgehammer vs. RU'R' and their mirrors -- can be applied through the last insert.) The calculation is easy enough if someone can point me to a summary of F2L case probabilities....

Thanks.

 

[Tempus]

What is the probability of the last F2L corner + edge set being already paired in the top layer, without manipulating the earlier inserts to force pairing?

Assuming I understand you correctly, there should be five corners and five edges in an indeterminate state. As for the corner piece, there's a 4/5 probability that the relevant corner is in the top layer and a 2/3 probability that the cross color on the relevant corner isn't facing up. As for the edge piece, there's now a 1/5 probability that it's against the correct face of the corner piece and a 1/2 probability that it's oriented correctly.

Multiply them all together and you get p=4/75, or about a 5.333% chance that they are matched up together in the top layer.

 

[FailCuber]

What's the probablity of getting a PLL skip on a 2x2?

 

[ryanj92]

What's the probablity of getting a PLL skip on a 2x2?

1 in 6, as there are only 6 possible permutations of the top layer (front/back/left/right adjacent swap, diagonal swap and the solved case)

 

[tseitsei]

What's the probablity of getting a PLL skip on a 2x2?

There are 4! = 24 possible permutations of LL pieces so

AUFless skip is only 1 permutation of those so 1/24
With AUF 4/24=1/6

 

[advincubing]

Assuming I understand you correctly, there should be five corners and five edges in an indeterminate state. As for the corner piece, there's a 4/5 probability that the relevant corner is in the top layer and a 2/3 probability that the cross color on the relevant corner isn't facing up. As for the edge piece, there's now a 1/5 probability that it's against the correct face of the corner piece and a 1/2 probability that it's oriented correctly.

Multiply them all together and you get p=4/75, or about a 5.333% chance that they are matched up together in the top layer.

Thank you for the response. That methodology makes perfect sense, but the result seems surprisingly low. I feel like I get a made pair in the top layer more like 25-40% of the time. I wonder if the difference is due to something I didn't think to address/describe in my question: Often the last F2L edge and corner are setup so that they can be paired before being inserted. This would be a case like F2L #5 where you would do (U' R U R') to pair the edge and corner in the U layer, AUF, then have the option to insert with R U' R' or a hedgeslammer. Those cases are in contrast to the ones that are solved and inserted in one fell swoop with, say, R U R'.

That complicates things quite a bit, I know. Not sure the best methodology for calculating with that nuance....

 

[goodatthis]

Okay, kind of a complicated question here, but is there any difference between the probability of getting a PLL skip when you use OLL (non edges oriented) vs OCLL? Basically what I mean is, if a PLL skip is just a 1LLL case (or ZBLL) that you happened to get as your OLL, and since OLLs permute pieces as well, do non edges oriented OLLs permute in a more common pattern than OCLLs? (Or less common?)

For example, let's say that a certain OLL permutes in a U perm like fashion. U perms are common, so another OLL that permutes in an H perm (less common) like fashion should be less likely to cause an PLL skip, right? I'm sorry if I'm a noob at this kind of thing, haha.

 

[ryanj92]

Okay, kind of a complicated question here, but is there any difference between the probability of getting a PLL skip when you use OLL (non edges oriented) vs OCLL? Basically what I mean is, if a PLL skip is just a 1LLL case (or ZBLL) that you happened to get as your OLL, and since OLLs permute pieces as well, do non edges oriented OLLs permute in a more common pattern than OCLLs? (Or less common?)

For example, let's say that a certain OLL permutes in a U perm like fashion. U perms are common, so another OLL that permutes in an H perm (less common) like fashion should be less likely to cause an PLL skip, right? I'm sorry if I'm a noob at this kind of thing, haha.

No, because during OLL the permutation of your LL pieces is generally random, so the chance of the OLL and OCLL actually causing a PLL skip is pretty slim
Maybe the nuances you mention have some minute effect on the probabilities - however, it will definitely be minute and calculating the probabilities won't give us any profound results

Also, not everybody uses the same OLL's or even OCLL's anyway, so the probabilities will vary from person to person

 

[qqwref]

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?