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chi-square test

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  • h_kociemba
    Ron posted the results of random scrambles a few days ago, looking ... The theoretical average numbers for 20000 cubes then are 0:9.77, 2:644.53, 4:4834, 6:
    Message 1 of 2 , Apr 12 4:08 AM
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      Ron posted the results of random scrambles a few days ago, looking
      how many edges were flipped:

      > Here are the results of 20,000 random scrambles sequences of 25
      > moves (1 in 3 moves in a half turn).
      > 0: 28 0,14%
      > 2: 701 3,51%
      > 4: 5095 25,48%
      > 6: 8950 44,75%
      > 8: 4590 22,95%
      > 10: 627 3,14%
      > 12: 9 0,05%
      > Average: 5.929

      Jaap gave the theoretical distribution:

      > With perfect random scrambles, the results should in theory have the
      >following distribution:
      > 0: 1 0.05%
      > 2: 66 3.22%
      > 4: 495 24.17%
      > 6: 924 45.12%
      > 8: 495 24.17%
      > 10: 66 3.22%
      >12: 1 0.05%
      > Total: 2048

      The theoretical average numbers for 20000 cubes then are

      0:9.77, 2:644.53, 4:4834, 6: 9023.44, 8:4834, 10:644.53, 12:9.77


      The chi-square test
      http://en.wikipedia.org/wiki/Pearson's_chi-square_test
      is the appropriate tool to test if 25 move scrambles are "good":

      (28-9.77)^2/9.77 + (701-644.53)^2/644.53 + .....gives a result of 66.

      With 6 degrees of freedom the tables show, that the chance that this
      value is >18.55 is less then 0.005. So if I did not do anything
      false, the 25 move scramble is a very bad procedure. Especially the
      high occurrence of 0 and 4 flipped edges contribute to the high value
      of 66.

      Herbert
    • d_funny007
      That seems to be the right test to perform from what I remember of AP Stats class. So I was right! 25-turn scrmables are biased in some way. -Doug ... the ...
      Message 2 of 2 , Apr 12 7:14 AM
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        That seems to be the right test to perform from what I remember of
        AP Stats class. So I was right! 25-turn scrmables are biased in some
        way.


        -Doug


        --- In speedsolvingrubikscube@yahoogroups.com, h_kociemba
        <no_reply@...> wrote:
        >
        > Ron posted the results of random scrambles a few days ago, looking
        > how many edges were flipped:
        >
        > > Here are the results of 20,000 random scrambles sequences of 25
        > > moves (1 in 3 moves in a half turn).
        > > 0: 28 0,14%
        > > 2: 701 3,51%
        > > 4: 5095 25,48%
        > > 6: 8950 44,75%
        > > 8: 4590 22,95%
        > > 10: 627 3,14%
        > > 12: 9 0,05%
        > > Average: 5.929
        >
        > Jaap gave the theoretical distribution:
        >
        > > With perfect random scrambles, the results should in theory have
        the
        > >following distribution:
        > > 0: 1 0.05%
        > > 2: 66 3.22%
        > > 4: 495 24.17%
        > > 6: 924 45.12%
        > > 8: 495 24.17%
        > > 10: 66 3.22%
        > >12: 1 0.05%
        > > Total: 2048
        >
        > The theoretical average numbers for 20000 cubes then are
        >
        > 0:9.77, 2:644.53, 4:4834, 6: 9023.44, 8:4834, 10:644.53, 12:9.77
        >
        >
        > The chi-square test
        > http://en.wikipedia.org/wiki/Pearson's_chi-square_test
        > is the appropriate tool to test if 25 move scrambles are "good":
        >
        > (28-9.77)^2/9.77 + (701-644.53)^2/644.53 + .....gives a result of
        66.
        >
        > With 6 degrees of freedom the tables show, that the chance that
        this
        > value is >18.55 is less then 0.005. So if I did not do anything
        > false, the 25 move scramble is a very bad procedure. Especially
        the
        > high occurrence of 0 and 4 flipped edges contribute to the high
        value
        > of 66.
        >
        > Herbert
        >
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