Discussion in 'Puzzle Theory' started by CubesOfTheWorld, Apr 8, 2010.

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1. ### jonlinMember

409
3
Jul 22, 2011
MDVA
WCA:
2012LINJ01
ForTheLulz
I'm guessing that an LL skip is 1/933120 chance of happening. Now we have to wait for Simon to report back to us to see when he gets an LL skip on megaminx.

2. ### RileyMember

May 14, 2011
California
WCA:
2007WOOR01
mylifeisriley
Are the chances of getting an LL skip with winter variation (as in you have winter variation in the current solve) 1/22? Because there are 21 PLL's you might get and 1 just solved last layer? (don't count AUF's) Thanks.

May 17, 2009
Ponyville
WCA:
2009WHIT01
ben1996123
1/72 probly.

4. ### IanTheCuberMember

215
0
Oct 28, 2011
In the clouds
WCA:
2013SCAH01
UCZVfZuvPSoBn3Ch2s2v
If you're just orienting the corners, then it would be 1/63.

Otherwise, no idea. You might also have to include half the F2L's, since mirrors make no extra contribution.

5. ### RileyMember

May 14, 2011
California
WCA:
2007WOOR01
mylifeisriley
I'm trying to say, if you have a R U' R' insertion with 3 edges on the top layer that are already correct (like yellow facing up, if white was your cross color), what is the chance of an LL skip, using WV, so that the corners orient? No mirrors. And if it's still 63, can you please explain how you get that? Thanks!

6. ### RpottsMember

1,859
26
Mar 23, 2008
KC
WCA:
2010POTT01
rpotts1
Same chance as a PLL skip - 1/72

7. ### IanTheCuberMember

215
0
Oct 28, 2011
In the clouds
WCA:
2013SCAH01
UCZVfZuvPSoBn3Ch2s2v
Here is a good one: What is the propability of getting a Sub-30 solve on 4x4, considering your average is about 37 seconds, and about 1/150 you get is Sub-35? I don't think there is enough info to make the answer out, but try. I am.

Aug 12, 2010
Aberdeen, Scotland
Godmil
Hmm, you know you could probably draw a probability distribution, like a gaussian or something with the 1/150 defining the width. Should be possible to get a rough estimate. But I don't know how to do it

9. ### MTGjumperMember

Jun 21, 2008
Bath/Notts
WCA:
2008CRAW01
MTGjumper
I doubt that someone who averaged 37 would only get a sub-35 one in 150 solves. But a normal distribution should work, providing you chose the variance such that 1/150 solves was sub-35.

10. ### ben1996123Banned

May 17, 2009
Ponyville
WCA:
2009WHIT01
ben1996123
Assuming the solve time, X is random and normally distributed:

X ~ N(37, σ)
P(X < 35) = 0.0067
P(Z < -2/σ) = 0.0067
P(Z < 2/σ) = 0.9933
2/σ = 2.48
σ = 2/2.48 = 0.806
X ~ N(37, 0.806)
P(X < 30) = P(Z < (30-37)/0.806) = P(Z < -8.685) = 0.0000000000000001894%

Last edited: Mar 29, 2012
11. ### ben1996123Banned

May 17, 2009
Ponyville
WCA:
2009WHIT01
ben1996123
After solving the centres on a 4x4, how many possible states are there?

Attempt:

Edge permutations(24!) * Corner permutations(8!) * Corner orientations(3^7) = 54711040793092776444454502400000

So the probability of a centres skip on 4x4 is 54711040793092776444454502400000/7401196841564901869874093974498574336000000000 = $\frac{1}{135277939046250}$

Or do I need to divide something by 24 somewhere?

12. ### TMOYMember

Jun 29, 2008
WCA:
2008COUR01
Seems correct to me. Once the centers are solved, the orientation of the cube is fixed, so you don't have to divide by 24.

13. ### CubeRootsMember

538
0
Mar 22, 2012
Leicester, UK
WCA:
2012LIVS01
CrazyMathCuber
I know this was 2 weeks ago but i realised mistake, I thought there were 2 ways to get Hperm but in reality there was only one so i ended up with 73 instead of 72 possibilities

14. ### Akash RupelaMember

472
1
May 9, 2011
New Delhi, India, India
WCA:
2012RUPE01
akashrupela1
probability that corners are correctly oriented after f2l=?
Probability that corners are correctly solved after f2l?
Probability that no corner is correctly oriented after f2l?
Probability that edges are correctly solved after f2l?
Probability that No edge is correctly oriented after f2l?

1,859
26
Mar 23, 2008
KC
WCA:
2010POTT01
rpotts1
1/27
1/162
6/27
1/96
1/8

16. ### cuBerBruceMember

912
4
Oct 8, 2006
Malden, MA, USA
WCA:
2006NORS01
cuBerBruce
1/162, allowing AUF (1/648 if not allowing AUF)
1/48, allowing AUF (1/192 if not allowing AUF)

17. ### ben1996123Banned

May 17, 2009
Ponyville
WCA:
2009WHIT01
ben1996123
What is the probability that after a random 3x3 scramble, one side will be solved (not necessarily a layer, just a face)? Just curious because I got this scramble about a week ago: R2 B2 R2 D' L2 D F2 U F2 L2 D2 F U2 F D' F2 R' F D L2 F D'

7 greens on F. I think I had one other scramble like this some time last year too, but I've never had one with 8 or 9.

Edit: Might as well have a go at calculating it myself:

edge permutations: 4! = 24
edge orientations: 2^4 = 16
corner permutations: 4! = 24
corner orientations: 3^4 = 81
centres: 6

1/(24*16*24*81*6) = $\frac{1}{4478976}$

not sure if its correct.

Last edited: Apr 19, 2012
18. ### cuBerBruceMember

912
4
Oct 8, 2006
Malden, MA, USA
WCA:
2006NORS01
cuBerBruce
Centers can be considered fixed. Corner and edge stickers of a given color must lie on the same face as the center of that color.

Consider a particular face such as the U face. Probability that the ULF sticker is on the U face is 4/24. The probability that the UFR sticker is on the U face, given that we know the ULF sticker already is, is 3/21. Similarly for the URB and UBL stickers, the probabilities can be given as 2/18 and 1/15. Similarly for the four U-color edge stickers, we get 4/24, 3/22, 2/20, and 1/18. Thus, the probability is (4/24)*(3/21)*(2/18)*(1/15)*(4/24)*(3/22)*(2/20)*(1/18) = 4!*4!/(24*21*18*15*24*22*20*18) = 1/(21*18*15*22*20*18) = 1/44906400.

The probability of at least one of the six faces being a solid color should be roughly (but not exactly) 6 times that or somewhere in the neighborhood of 1 in 7.5 million.

19. ### SweetSolverMember

355
4
Apr 13, 2012
Australia
WCA:
2014SWEE01
What is the probability of one FACE being solved after a random scramble on a 2x2?

20. ### cubingawsumnessMember

458
1
Jan 16, 2012
California
WCA:
2012CHIE01
Just curious, has anyone ever estimated the percentage of people in the US who can solve a 3x3?