From hirsh@cs.rutgers.edu Sun Nov 11 15:34:28 1990 Return-Path: Received: from pei.rutgers.edu by life.ai.mit.edu (4.1/AI-4.10) id AA23245; Sun, 11 Nov 90 15:34:28 EST Received: by pei.rutgers.edu (5.59/SMI4.0/RU1.2/3.05) id AA00310; Sun, 11 Nov 90 15:34:25 EST Sender: Haym Hirsh Date: Sun, 11 Nov 90 15:34:23 EST From: Haym Hirsh Reply-To: Haym Hirsh To: Cube-Lovers@life.ai.mit.edu Subject: Re: Rubik's Cube reassembly problem and solution In-Reply-To: Your message of Sat, 10 Nov 90 18:49:48 EST Cc: Haym Hirsh Message-Id: 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 2962822807718827976654851209938335398226821120000000000 ------------------------------------------------------------------------ 230843697339241380472092742683027581083278564571807941132288000000000000 = 3.0*10^54/2.3*10^71 = 1.3*10^-17 Haym