Date modified: Thu 2023-05-11 . 11:51 AM
Date created: Sat 2023-10-07 . 09:02 PM
# Reassembling a cube. Consider a $3\times 3\times 3$ cube made of 27 smaller $1\times 1\times 1$ cubes. The outer 6 faces of the $3\times 3\times 3$ cube is painted black. Now shuffle all the tiny cubes and assemble them back randomly into a $3\times 3\times 3$ cube again. What is the probability that the surface is all black again? ![[---images/Reassembling a cube 2023-05-04 13.08.12.excalidraw.svg]] As a warm up, try intstead a $2\times 2 \times 2$ cube made of $8$ smaller $1\times 1\times1$ cubes. ///solution./// Let us do the $2\times 2\times 2$ case as a warm up. When the outer faces of this larger cube is painted, we see that all eight tiny cubes are all corner pieces. Each tiny cube has 24 possible orientations, of which 3 of them will be appropriate for a corner. So the probability that a reassembled $2\times 2\times 2$ has outer face all black again is just $(3/24)^8 \approx 5.96\times 10^{-8}$ . Now when the outer faces of the $3\times3\times3$ cube are painted black, there are four kinds of tiny cube pieces: Corner, edge, (middle) face, and central cube. And some of these types must be in the right spot, in addition to having correct orientation. Of the 27 spots, the 1 center tiny cube must be at the 1 central position. Of the 27 spots, the 8 corner tiny cubes must be at the 8 corner positions. Of the 27 spots, the 12 edge tiny cubes must be at the 12 edge positions. Of the 27 spots, the 6 central face tiny cubes must be at the 6 face positions. So, not accounting for orientations yet, for the 27 cubes to be at their right spots has probability $$\frac{1}{27 \choose1,8,12,6}=\frac{1!8!12!6!}{27!}$$ Now let us account for orientation. The 1 center tiny cube can bear any orientation. The 8 corner tiny cubes each has (3/24) chance of being in the right orientation. The 12 edge tiny cubes each has (2/24) chance of being in the right orientation. The 6 central face tiny cubes each has (4/24) chance of being in the right orientation. So given tha they are all in the right spots, the probability that the orientation give a full outer black coloring is $$(3/24)^8 (2/24)^{12}(4/24)^6$$ So all together, the probability of a reassembled $3\times3\times3$ cube will have outer face all black is $$\frac{(3/24)^8 (2/24)^{12}(4/24)^6}{27 \choose1,8,12,6}\approx 1.83\times 10^{-37}$$ In other words, **very unlikely**. Yet curiously, still more likely than a deck of 52 cards being reshffuled into a perfect deck order, a probability of $1/52!\approx 1.24\times 10^{-68}$. /// --- Related, what if this $3\times 3\times 3$ cube is a rubiks cube, with each of the 6 larger faces colored with one of six different colors? And somewhat related, if you disassemble a rubiks cube, and put it back randomly, it may not be solvable again. What is the probability of it being solvable? What about peeling off the stickers of the rubiks cube, and putting them back randomly, what is the probability of the rubiks cube is solvable? #puzzle #counting #probability