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Probability Thread

Sorry I meant formed, not necessarily solved.

Do you happen to how people arrived at that figure?

There are 24 possible corner/edge pairs, and each one has (exactly) 1/24 chance of being formed. Assuming that the probabilities are approximately independent, this means that we can approximate the distribution of the number of pairs formed as a Poisson distribution with mean 1; under this approximation, the probability that there are no pairs at all is approximately 1/e.

Or, if we don't want to use a Poisson approximation, then under the same simplifying assumption that the probabilities are approximately independent, P(no pairs) ~ \( (1-1/24)^{24}\approx36\% \).
 
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ok ishaan asked me what the probablity is that you get at least one flipped edge for 3bld (excluding the buffer piece). check my work if you want lol

i got 36.82 %

(11 choose 1)/24-(11 choose 2)/24/22+(11 choose 3)/24/22/20-(11 choose 4)/24/22/20/18+(11 choose 5)/24/22/20/18/16-(11 choose 6)/24/22/20/18/16/14+(11 choose 7)/24/22/20/18/16/14/12-(11 choose 8)/24/22/20/18/16/14/12/10+(11 choose 9)/24/22/20/18/16/14/12/10/8-(11 choose 10)/24/22/20/18/16/14/12/10/8/6+(11 choose 11)/24/22/20/18/16/14/12/10/8/6/4
 
I think this is the right thread for this, maybe?
Hoe many cases would there be for this Last Slot + Last Layer thing?:

- LL Corners are already solved (both oriented and permuted)
- FRD corner is already solved
- All the Last Layer edges on the U layer are oriented, but not necessarily the LL edge in FR

For a better visual, here's a setup for a case of what I'm talking about: F' U2 F U' F' U' F U2 F' U F U'

Also, how many cases would there be if the LL edge in FR was oriented?
 
I think this is the right thread for this, maybe?
Hoe many cases would there be for this Last Slot + Last Layer thing?:

- LL Corners are already solved (both oriented and permuted)
- FRD corner is already solved
- All the Last Layer edges on the U layer are oriented, but not necessarily the LL edge in FR

For a better visual, here's a setup for a case of what I'm talking about: F' U2 F U' F' U' F U2 F' U F U'

Also, how many cases would there be if the LL edge in FR was oriented?

12 cases for when the edge in FR is oriented, 12 more for when it is unoriented (just the mirrors)
 
I think this is the right thread for this, maybe?
Hoe many cases would there be for this Last Slot + Last Layer thing?:

- LL Corners are already solved (both oriented and permuted)
- FRD corner is already solved
- All the Last Layer edges on the U layer are oriented, but not necessarily the LL edge in FR

For a better visual, here's a setup for a case of what I'm talking about: F' U2 F U' F' U' F U2 F' U F U'

Also, how many cases would there be if the LL edge in FR was oriented?
I believe you are talking about OLS. If so, there are 17,712 cases I believe. (I didn't use math I just saw that stat somewhere)
 
Again, I don't know if this is the right thread or not, but here's another question:
What would the maximum movecount be for creating a solved "block" of all pieces except the last layer, FD and FR edge, and FDR corner?
 
Again, I don't know if this is the right thread or not, but here's another question:
What would the maximum movecount be for creating a solved "block" of all pieces except the last layer, FD and FR edge, and FDR corner?

Determining the exact value would require going through almost 28 billion cases, so I don't think you'll be getting an answer to this question from someone else, ever.

If you really want to know what the worst optimal move count is, there isn't really much of a choice but to write some code and determine how many moves each of those 28 billion cases need. See also: https://www.jaapsch.net/puzzles/compcube.htm
 
How many cases would there be if 6 edges on the 3x3 were to be scrambled (4 on U and 2 on E), and all the yellow edges on the U layer were oriented correctly (assuming the U side is yellow)?
Example setup for one of the cases, in case I'm not making sense:
D' F2 D2 B' R2 B D' U R' F2 R' U2 R2 U'
 
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How many cases would there be if 6 edges on the 3x3 were to be scrambled (4 on U and 2 on E), and all the yellow edges on the U layer were oriented correctly (assuming the U side is yellow)?
Example setup for one of the cases, in case I'm not making sense:
D' F2 D2 B' R2 B D' U R' F2 R' U2 R2 U'

And everything else is solved?
 
i asked this 6 years ago but i didnt get an answer, ive been looking for it and the answer seems to be 10.xx, but i cant trust myself
whats the average number of targets if you are using 2 fixed buffers? (dont mind orientations; passive pieces unoriented count as 0 targets, parity edge and corner count as 1 each and breaking into a new n cycle counts as n+1 targets)
also whats the average of passive pieces to twist? not counting the buffers i guess, or whatever is easier to calculate
 
How many cases would there be if 6 edges on the 3x3 were to be scrambled (4 on U and 2 on E), and all the yellow edges on the U layer were oriented correctly (assuming the U side is yellow)?
Example setup for one of the cases, in case I'm not making sense:
D' F2 D2 B' R2 B D' U R' F2 R' U2 R2 U'
Assuming the 2 F2L slots are adjacent, there's are 403 cases, excuding solved, I think.

Someone else please check my math :
Both E edges in E: (4 EPLLs + 5 parity EPLLs) * 2 orientations + 2-flip = 19 cases
First E edge in U: (4! ways to arrange U edges * 1 (E edges determined by parity)) * (2 possible orientations for edges^(3 edges that can be flipped - 1 last edge orientation determined by others)) = 96 cases
Mirrors of above cases: 96 cases (no symmetries possible)
Both E edges in U: (4! ways to arrange U edges * 1 (E edges determined by parity)) * (2 possible orientations for edges^(4 edges that can be flipped - 1 last edge orientation determined by others)) = 192 cases.
 
Assuming the 2 F2L slots are adjacent, there's are 403 cases, excuding solved, I think.

Someone else please check my math :
Both E edges in E: (4 EPLLs + 5 parity EPLLs) * 2 orientations + 2-flip = 19 cases
First E edge in U: (4! ways to arrange U edges * 1 (E edges determined by parity)) * (2 possible orientations for edges^(3 edges that can be flipped - 1 last edge orientation determined by others)) = 96 cases
Mirrors of above cases: 96 cases (no symmetries possible)
Both E edges in U: (4! ways to arrange U edges * 1 (E edges determined by parity)) * (2 possible orientations for edges^(4 edges that can be flipped - 1 last edge orientation determined by others)) = 192 cases.
Well, I was experimenting with a method idea but this would be too much for that :/
 
The standard likelihood of getting a PLL skip is 1/72, but what is it if you know full COLL?

There's a 1/8 chance to have EO done when arriving at LL. However, if you skip OLL you won't be able to use COLL. The chance of an OLL skip is 1/216. The chance for an EPLL skip is 1/12. Hence it's (1/8 - 1/216) * 1/12 + (1 - 1/8 + 1/216) * 1/72 = 173/7776 which is approximately 2.2% so around 1.6 times as likely as without COLL.
 
How many L8E cases are there if all the edges are oriented?

Which eight edges are you considering? (Also, regardless of which eight edges you choose, the answer is likely to be in the thousands, since 8!/2 = 20160.)
 
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