Twenty-seven identical white cubes are assembled into a single cube, and the outside of that cube is painted black.

The cube is then disassembled and rebuilt randomly.

What is the probability that the outside of this cube is again completely black?

For positions:

There are 12 edge pieces to get back to an edge. The probability of each edge being filled by an edge is:

(12/27)*(11/26)*...*(1/16)

Given that that was done, the probability of filling the 6 face-center places with face-center pieces is

(6/15)*(5/14)*...*(1/10)

Given that was done, the probability that the rest of the positions are filled with the correct type of cube can be expressed in either of two ways, the probability that the center would find its way tot he center, which is 1/9, or that the 8 vertex places would be filled by the 8 vertex pieces:

(8/9)*(7/8)*...*(1/2) = 1/9

For orientation:

Each cube must be placed in the correct orientation:

The 12 edge pieces must be oriented to the edge with the painted pair of faces is along the edge of the big cube, with probability, for all 12:

(1/12)^12

For the faces:

(1/6)^6

and for the vertices:

(1/8)^8

In all the probability is

12! * 6! * 8! / (27! * 12^12 * 6^6 * 8^8)

This comes out to 1/5465062811999459151238583897240371200

The denominator is a 37-digit number, its common logarithm being  36.7375951578224093408522212147919334423910293248693654862854768504973912368...

In all as there are 27! * 24^27 ways of placing and orienting the cubes (including cases that look identical but are different cubes or different orientations that are however painted the same), which is 200764220619687053469776705897730625915702926996021687877632000000, the number of orientations that count as a match are this times the probability, or 36735940194296694948495360000.

 10   P=1
 20   for Edge=12 to 1 step -1
 30     P=P*Edge//((Edge+15)*12)
 40   next
 50   for Face=6 to 1 step -1
 60     P=P*Face//((Face+9)*6)
 70   next
 80   P=P//(9*8^8)
 90   print P
100   print 1//P:print log(1/P)/log(10)
110   print !(27)*24^27
120   print !(27)*24^27*P

point 20
Words for fractionals 20 (Decimals for display 96)
OK
run
 1//5465062811999459151238583897240371200
 5465062811999459151238583897240371200
 36.7375951578224093408522212147919334423910293248693654862854768504973912368071
96054483385864924917
 200764220619687053469776705897730625915702926996021687877632000000
 36735940194296694948495360000
OK

Posted by Charlie on 2007-04-02 12:11:14