[next] [prev] [up] Date: Sun, 11 Nov 90 15:34:23 -0500 (EST)
[next] [prev] [up] From: Haym Hirsh <hirsh@cs.rutgers.edu >
[next] [prev] [up] Subject: Re: Rubik's Cube reassembly problem and solution

I just caught a bug in my reasoning. The restickering need not yield
something equivalent to the original undestickered cube, but rather
just one that can be solved to obtain solid colors on each face.
Since there are 5*3*2 different distinguishable cubes (i.e., 30
different ways to label a die with the numbers 1-6) (6! labelings, but
rotational symmetry removes 24 -- six faces can be brought to the top,
and for each it can be rotated around the axis perpendicular to that
face in one of 4 ways), the numerator should be multiplied by 30, and
hence the probability is actually


= 3.0*10^54/2.3*10^71 = 1.3*10^-17


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