Probability Thread

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

Welcome to the Speedsolving.com. You are currently viewing our boards as a guest which gives you limited access to join discussions and access our other features. By joining our free community of over 30,000 people, you will have access to post topics, communicate privately with other members (PM), respond to polls, upload content and access many other special features. Registration is fast, simple and absolutely free so please, join our community today!

If you have any problems with the registration process or your account login, please contact us and we'll help you get started. We look forward to seeing you on the forums!

Already a member? Login to stop seeing this message.
  1. jonlin

    jonlin Member

    408
    0
    Jul 22, 2011
    MDVA
    WCA:
    2012LINJ01
    YouTube:
    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. Riley

    Riley Member

    1,303
    0
    May 14, 2011
    California
    WCA:
    2007WOOR01
    YouTube:
    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.
     
  3. ben1996123

    ben1996123 Banned

    4,972
    2
    May 17, 2009
    Ponyville
    WCA:
    2009WHIT01
    YouTube:
    ben1996123
    1/72 probly.
     
  4. IanTheCuber

    IanTheCuber Member

    215
    0
    Oct 28, 2011
    In the clouds
    WCA:
    2013SCAH01
    YouTube:
    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. Riley

    Riley Member

    1,303
    0
    May 14, 2011
    California
    WCA:
    2007WOOR01
    YouTube:
    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. Rpotts

    Rpotts Member

    1,788
    1
    Mar 23, 2008
    KC
    WCA:
    2010POTT01
    YouTube:
    rpotts1
    Same chance as a PLL skip - 1/72
     
  7. IanTheCuber

    IanTheCuber Member

    215
    0
    Oct 28, 2011
    In the clouds
    WCA:
    2013SCAH01
    YouTube:
    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.
     
  8. Godmil

    Godmil Premium Member

    2,196
    1
    Aug 12, 2010
    Aberdeen, Scotland
    YouTube:
    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 :p
     
  9. MTGjumper

    MTGjumper Member

    2,213
    0
    Jun 21, 2008
    Bath/Notts
    WCA:
    2008CRAW01
    YouTube:
    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. ben1996123

    ben1996123 Banned

    4,972
    2
    May 17, 2009
    Ponyville
    WCA:
    2009WHIT01
    YouTube:
    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. ben1996123

    ben1996123 Banned

    4,972
    2
    May 17, 2009
    Ponyville
    WCA:
    2009WHIT01
    YouTube:
    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. TMOY

    TMOY Member

    1,802
    0
    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. CubeRoots

    CubeRoots Member

    538
    0
    Mar 22, 2012
    Leicester, UK
    WCA:
    2012LIVS01
    YouTube:
    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 Rupela

    Akash Rupela Member

    472
    1
    May 9, 2011
    New Delhi, India, India
    WCA:
    2012RUPE01
    YouTube:
    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?
     
  15. Rpotts

    Rpotts Member

    1,788
    1
    Mar 23, 2008
    KC
    WCA:
    2010POTT01
    YouTube:
    rpotts1
    1/27
    1/162
    6/27
    1/96
    1/8
     
  16. cuBerBruce

    cuBerBruce Member

    911
    1
    Oct 8, 2006
    Malden, MA, USA
    WCA:
    2006NORS01
    YouTube:
    cuBerBruce
    1/162, allowing AUF (1/648 if not allowing AUF)
    1/48, allowing AUF (1/192 if not allowing AUF)
     
  17. ben1996123

    ben1996123 Banned

    4,972
    2
    May 17, 2009
    Ponyville
    WCA:
    2009WHIT01
    YouTube:
    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. cuBerBruce

    cuBerBruce Member

    911
    1
    Oct 8, 2006
    Malden, MA, USA
    WCA:
    2006NORS01
    YouTube:
    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. SweetSolver

    SweetSolver Member

    355
    3
    Apr 13, 2012
    Australia
    WCA:
    2014SWEE01
    What is the probability of one FACE being solved after a random scramble on a 2x2?
     
  20. cubingawsumness

    cubingawsumness Member

    458
    0
    Jan 16, 2012
    California
    WCA:
    2012CHIE01
    Just curious, has anyone ever estimated the percentage of people in the US who can solve a 3x3?
     

Share This Page