Stefan
Legend
Hmm, I'd say that depends on your personal F2L then. A beginner might take more moves for F2L and thus have more chances to build such a block and thus have a higher probability.
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Probability Thread
- Thread starter CubesOfTheWorld
- Start date
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- Tags
- cfop oll probability
Robert-Y
Legend
Yeah, it depends on how many times the cross has been restored during F2L I think.
kinch2002
Premium Member
Definitely too vague to calculate...as Stefan says, it totally depends on your F2L. On another note, I hope you took advantage of it and switched LL color for at least one of them 
penguinz7
Veteran
Just got 15 oll paritys in a row. Yes, 15. wut. 1/32768?
0.00003051757% Pretty sure thats a record for number of oll paritys in a row..
0.00003051757% Pretty sure thats a record for number of oll paritys in a row..
FJT97
Veteran
If I do a first face (not a layer), how many positions are possible then?
cmhardw
Premium Member
Permute the solid color layer:
(4!)^2
Permute and orient the rest of the cube:
(4!)*3^3 corners
(8!/2)*2^7 edges
For edge permutations you divide by 2 since the edge permutation parity must match the corner permutation parity.
Altogether:
(4!)^2*(4!)*3^3*(8!/2)*2^7 = 963158 999040
guysensei1
Legend
How many cases are there to solve the cube in 1 look if one has built a tripod already? Assume that cases that are simply a cube rotated version of another case can be considered the same.
TMOY
Veteran
First we count the total number of cases (considering cases which are identical up to rotation as different); we obtain 3^3 corner orientations, 4! corner permutations, 2^2 edge orientations and 3! edge permutations, dividing by 2 because of parity, we obtain 3^3*4!*2^2*3!/2=7776 positions.
Next we count the number of symmetrical positions. For a position to be symmetrical, the corner on top of the tripod must be in place (not necessarily oriented), and the other three must all have orientations compatible with the orientation of the top corner (only 1 possibility), and their permutation must be a 3-cycle. The edges have only one possible orientation, and their permutation must also be a 3-cycle. Hence 3*3*3=27 symmetrical positions, and 7776-27=7749 asymmetrical positions..
So the total number of cases, excluding the solved case which is of course symmetrical, are: 7749/3+26=2609 cases.
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cmhardw
Premium Member
That's a difficult question. In terms of number of combinations in total there are:
4!*3^3*3!/2*2^2=
I tried using the lemma that is not Burnside's (and I may have used it incorrectly) and got:
1/6 * (7776 + 3*(4(2)) + 1(3(2*2))) = 1302 unique cases, which is more than total LL combinations.
Since the tripod is essentially the last layer with 1 solved edge, I would think it would be less than the 1211 algs required for 1LLL, but maybe since the tripod has fewer symmetries (I count 8 for a layer and 6 for a tripod, but that may be wrong) then there would be more tripod cases?
Can someone with a better math background comment on this?
--edit--
Ninja'd by TMOY
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guysensei1
Legend
Thanks for the calculation! What if we used 2 look tripod? Which set will have less cases: solve corners in 1 look, then edges, or edges first then corners?
cmhardw
Premium Member
My number is off by almost exactly a factor of 2 from yours. Can you help me figure out where I went wrong? I can explain my calculation if you need. It's been so long since I've used the (not) Burnside's lemma and I'd like to make sure I'm not misusing it.
G2013
Veteran
How probable is this to happen?
https://www.youtube.com/watch?v=iaTUkd948JM
https://www.youtube.com/watch?v=iaTUkd948JM
Renslay
Legend
TMOY
Veteran
The difference between our two calculations is simply that you're counting two cases which are mirror images of each other as only one while I count them as two.
SirWaffle
SHJ
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7x7 Last 4 edge odds?
I really am unsure the correct search terms to use to find the answer so that's why I am creating a new thread. What are the odds of skipping one of the last 4 edges after realigning the centers on 7x7? This happened to me in a solve so I was curious.
I really am unsure the correct search terms to use to find the answer so that's why I am creating a new thread. What are the odds of skipping one of the last 4 edges after realigning the centers on 7x7? This happened to me in a solve so I was curious.
Stefan
Legend
You can also use the four middle edges as reference.
http://www.wolframalpha.com/input/?...)^(i+1)+*+(8-2*i)!^2+/+8!^2+for+i+from+1+to+4
http://www.wolframalpha.com/input/?...)^(i+1)+*+(8-2*i)!^2+/+8!^2+for+i+from+1+to+4
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cmhardw
Premium Member
I was treating the 16 wings as 1 orbit of 16 pieces. My correction was to make it two separate orbits, each with 8 pieces.
cmhardw
Premium Member
Wow, I understand what you did there using the edges as reference. I also just realized that in my expression those terms would cancel out, effectively using the edges as reference.
I would not have thought to use inclusion-exclusion like you did where you start by counting the number of states with 1 (really at least 1) edge group solved. That makes sense now that I look at it. For some reason I always want to use it to find derangements, when it's obviously a more powerful method with further uses.
Thanks for the math tips! As always, Stefan, you've got skills!
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Stefan
Legend
Oh crap, I did that, too
(I developed mine independently from yours, so I made the mistake myself, didn't copy it from you)
(fixed it now)
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