Probability Thread

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

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  1. CubesOfTheWorld

    CubesOfTheWorld Member

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    This is just a thread for people to post interesting possibilities of a case. Like, it is a 1/72 chance for a PLL skip. Enjoy.
     
    Last edited: May 11, 2015
  2. chinesed00d

    chinesed00d Member

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    What are the chances of having a cross already done during a scramble?
     
  3. CubesOfTheWorld

    CubesOfTheWorld Member

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    heheh. that is why I made this thread. I don't know this kind of math. That was one I was thinking about.
     
  4. 4Chan

    4Chan Premium Member

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    At US Nationals 2008, during 3x3 qualification, there was a 1 move cross on white.

    True fact.
     
  5. cmhardw

    cmhardw Premium Member

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    I had a really interesting one come up a few days ago. After solving the cross and the first two pairs, all the remaining 12 pieces were in their correct locations, only 6 of them were misoriented.

    The probability of having the last 12 pieces of the cube in the correct location after solving the cross and first two pairs is:

    1 / [6! * 6!/2] = 1 / 259200

    To give a comparison, the chance of a LL skip with no partial edge control, and with the possibility of AUF, is
    1 / 15552

    I thought it was pretty neat.

    Chris
     
  6. Lucas Garron

    Lucas Garron Super-Duper Moderator Staff Member

    Saying that doesn't make your thread more legitimate.

    I changed the title, and let's see if we can salvage it:

    If you keep doing random alignments and twists on a Square-1, you will hit the square-square shape 2/1839 of the time.
     
  7. Tim Major

    Tim Major Platinum Member

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    Wow, I didn't know that, and am interested in how you worked that out. How? :)
     
  8. DT546

    DT546 Member

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    Feb 10, 2010
    York, UK
    not cubing related, but interesting

    if you buy a lottery ticket on a tuesday, you are more likely to die before the results are announced than you are to win on the wednesday
     
    Matthew Anderson likes this.
  9. riffz

    riffz Member

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    I just have to say this:

    WOLFRAM ALPHA IS AMAZING

    That is all.

    That's actually quite funny. But how exactly were the odds of dying calculated?
     
  10. Rinfiyks

    Rinfiyks Member

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    The odds of winning the lottery jackpot are almost 1 in 14 million.
    The population of the usa is 300 million.

    300/14=20~, so if more than 20 people die from non-natural causes every day in the usa, DT546's statement is correct.
     
  11. DaijoCube

    DaijoCube Member

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    My friend (who is not very good at cubing) was solving his 4x4x4 and he got a 3x3x3 skip NO JOKE. He finished pairing up and after that, he had to do a D' and a U2 THAT'S IT.

    This might be the luckiest solve you could have on a 4x4x4!
     
  12. TheMachanga

    TheMachanga Member

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    He scrambled it him self...most likely. It was most likely a crappy scramble.
     
  13. TheMachanga

    TheMachanga Member

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    Oh and I had OLL parity and when I finished that it was an LL skip. (qqtimer)
     
  14. DaijoCube

    DaijoCube Member

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    After solving centers then edges, you most likely can't keep track of the scramble. Scramble was legit.
     
  15. That70sShowDude

    That70sShowDude Member

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    What are the chances of having a CCT 4x4 scramble that requires 4 or less moves to solve the centers.

    I solved the centers in 4 moves on one of em ...
     
  16. I want to know the chances of COLL cases and ZBLL cases. Cride5 has tried several times (maybe only once?) to explain this to me with complicated language. I might know what is going on now and I will try to explain.

    So for OLL, the chance of Pi, Sune, Anti-Sune, T, U, and L is 1/54 while H is 1/108 and a skip occurs every 1/216 (these statistics are pretty well known). Just taking these cases for EO (setting 1/54 to be the standard):

    6 * chance + 1 * .5 chance + 1 * .25 chance = 1

    4/27 is the standard chance of occurance.

    So the chance of OCLL occurrences after EO:
    T: 4/27
    U: 4/27
    L: 4/27
    H: 2/27
    Pi: 4/27
    Sune: 4/27
    Anti-Sune: 4/27
    Skip: 1/27

    Now for COLLs. PLL probabilities are well known, so I can use those for my data.

    Corners correctly placed: 1/18 + 1/18 + 1/36 + 1/72 (U,U,Z,H) = 11/72
    Diagonal corner swap: 1/72 + 1/72 + 1/36 + 1/18 + 1/18 (N,N,E,V,Y) = 1/6

    Now the others should be divided evenly amongst themselves, if my reasoning is correct (no reason you should trust this). Which means every other case should have a nice 49/432 chance of occurrence. What? Hmmm. Weird number. I guess I'll post it with PLL chances:

    T
    Correct: 11/486
    Diagonal: 2/81
    Left: 49/2916
    Right: 49/2916
    Top: 49/2916
    Bottom: 49/2916

    U
    Correct: 11/486
    Diagonal: 2/81
    Left: 49/2916
    Right: 49/2916
    Top: 49/2916
    Bottom: 49/2916

    L
    Correct: 11/486
    Diagonal: 2/81
    Left: 49/2916
    Right: 49/2916
    Top: 49/2916
    Bottom: 49/2916

    H
    Correct: 11/972
    Diagonal: 1/81
    Left: 49/5832
    Right: 49/5832
    Top: 49/5832
    Bottom: 49/5832

    Pi
    Correct: 11/486
    Diagonal: 2/81
    Left: 49/2916
    Right: 49/2916
    Top: 49/2916
    Bottom: 49/2916

    Sune
    Correct: 11/486
    Diagonal: 2/81
    Left: 49/2916
    Right: 49/2916
    Top: 49/2916
    Bottom: 49/2916

    Anti-Sune
    Correct: 11/486
    Diagonal: 2/81
    Left: 49/2916
    Right: 49/2916
    Top: 49/2916
    Bottom: 49/2916

    Skip
    Aa: 1/486
    Ab: 1/486
    E: 1/972
    F: 1/486
    Ga: 1/486
    Gb: 1/486
    Gc: 1/486
    Gd: 1/486
    H: 1/1944
    Ja: 1/486
    Jb: 1/486
    Na: 1/1944
    Nb: 1/1944
    Ra: 1/486
    Rb: 1/486
    T: 1/486
    Ua: 1/486
    Ub: 1/486
    V: 1/486
    Y: 1/486
    Z: 1/972

    I postulate that ZBLL has the same chance for every edge permutation. There are 12 possible edge permutations and COLL + EPLL generates each edge permutation equally often. Therefore the chance of each ZBLL case is that of the chance of the respective COLL case divided by 12. I don't want to have to post all of those probabilities.

    One last thing: am I right?
     
  17. 4Chan

    4Chan Premium Member

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    Mister eastamazonantidote, I have wondered the same thing. o:
    I made a thread about COLL probability, and I'm quite curious myself about ZBLL.
     
  18. ben1996123

    ben1996123 Banned

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    Is that including the middle layer flip or not?
     
  19. Owen

    Owen Member

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    Nov 2, 2009
    What about a edges skip? (on 3x3)
     
  20. kinch2002

    kinch2002 Premium Member

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    This is pretty easy when you think about it for a moment. You need the first edge in place - that's 1/12. Second edge must now go in place - that's 1/11. Etc...so you have 1/12*1/11*1/10...*1/2=1/12!. Then they all have to be orientated correctly (1/2 for each) but the last edge orientation is defined by the other 11, so you get (1/2)^11 for orientation

    Final answer: 1/(12!*2^11) = 9.81*10^11

    EDIT: I hope I got this right...I shouldn't be doing maths at uni otherwise...
     

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