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(My previous question was maybe a bit too easy. This one is just slightly harder...)

Twenty-seven white blocks are assembled into a single cube and its outside is painted blue. Foolishly placed on the the floor, the cube is one day stepped on by a blind man, and it breaks into pieces (all twenty-seven are separated and spread across the floor completely at random). He tries to reassemble the pieces.

What is the probability that the cube which the blind man assembles is blue?

Note: the blind man cannot feel the blue paint in any way, it has identical structure to the white paint. No lateral thinking tag.

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    $\begingroup$ This seems like a possible duplicate of math.stackexchange.com/questions/1159899 $\endgroup$ – JonTheMon Dec 9 '15 at 15:27
  • $\begingroup$ Oh, you're right. What would be the proper thing to do? Should I remove the question? $\endgroup$ – Petriarch Dec 9 '15 at 15:36
  • $\begingroup$ I cannot possibly imagine how a puzzling.stackexchange question can be a duplicate of a math.stackexchange question... I can't even close it as belonging on that site, let alone as a duplicate! $\endgroup$ – Mr Lister Dec 9 '15 at 17:00
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    $\begingroup$ @JonTheMon that question is about a red cube, ;) $\endgroup$ – DrunkWolf Dec 9 '15 at 18:14
  • $\begingroup$ @Petriarch one question Is the bottom also painted blue?? by bottom i mean the bottom surface of the cubes $\endgroup$ – Roshan Rimal Sep 26 '17 at 5:55
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The center cube must be placed in the middle, the centers of faces must be placed in their proper places, and the edges must be placed in their proper places.This guarantees that the corners are in their places. The number of ways to place each cube in its proper place (ignoring rotation) is...

$${27\choose1}{26\choose6}{20\choose8}$$

Then we have to multiply by the probability that each cube is in its proper alignment. For the corners, the correct corner has to be pointing out (1/8); for the edges, the correct edge has to be pointing out (1/12); for the face centers, the correct face has to be pointing out (1/6). This gives a combined probability of...

$$\frac{1}{{27\choose1}{26\choose6}{20\choose8}8^812^{12}6^6}$$

which is about $1.83\times10^{-37}$.

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The cubes have 0, 1, 2, or 3 sides painted blue. The final assembly is correct if all the pieces are in the correct groups, and oriented so the blue faces are out.

There are 8 cubes with 3 blue faces. 12 with 2 blue faces, 6 with 1 blue face, and 1 with 0 blue faces.

Each cube can be oriented in 24 ways (pick any face to be "up", and then one of the four adjoining faces to be "forward"). The 3-blue cubes can be oriented with all blue sides out in 3 different ways. The 2-blue cubes in 2 ways. The 1-blue cube in 4 ways, and the 0-cube in 24 ways.

The odds of getting the cubes in the right groups are:

$$P_1 = {8! \times 12! \times 6! \times 1! \over 27!} \approx 1.27 \times 10^{-12}$$

Once the cubes are put into the right places, the odds that they are oriented correctly is:

$$P_2 = {3^8 \times 2^{12} \times 4^6\times 24^1\over 24^{27}} = 24^{-18}\approx 1.43\times 10^{-25}$$

The total probability is $P_3 = P_1 \times P_2 \approx 1.82 \times 10^{-37}$.

I think it's interesting that $P_2$ is so much smaller than $P_1$.

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  • $\begingroup$ I was too slow. :-( $\endgroup$ – user3294068 Dec 9 '15 at 18:04
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We must consider both position and orientation.

Consider a single cube like a die. There are 6 faces, so that means there are 6 different faces a cube can have up. For each of these ways, it can be rotated 4 ways. Thus, there are 24 ways to position a single cube.

The total number of possibilities for the positions of pieces of the cube are $27!$. The number of orientations of each cube is $24$, and their are $27$ of them. So the total number of possible composite cubes is:

$$27! \times 24^{27}$$

A $3\times3\times3$ composite cube has:

  • $1$ centre piece
  • $8$ corner pieces
  • $12$ edge pieces
  • $6$ middle pieces (centre of each face)

The centre piece must be placed in the middle, but orientation doesn't matter. Thus, if you have a valid composite cube with the centre piece in the right location, three are $24$ variants that are also valid simply by rotating this piece through all possible rotations.

The middle pieces can be placed in any of the middle places. They need the blue face outward, but other than that, can be in any of the $4$ possible rotations. So, if a composite cube is valid, you could make $6!$ valid composite cubes simply be re-positioning the middle pieces. For each of these valid composite cubes, you could take a middle piece and rotate it $4$ ways. Thus, there are $6! \times 4^6$ valid cubes that can be made by repositioning and rotating the middle pieces.

The edge pieces must be placed in the edge. The piece can be in two possible orientations where both outward sides are blue. Thus, given a valid cube, you could re-position the 12 edge pieces in $12!$ different ways. For each positioning, you could orient each edge piece in $2$ ways. Thus, there are $12! \times 2^{12}$ valid cubes that can be made by repositioning and rotating the edge pieces.

Lastly, the corner pieces must be placed at the corner. There are three valid ways that this can be done by rotating them so that the vertex of the three faces remains the vertex of the composite cube. Thus, there are $8!$ ways of positioning these corners in a valid cube, and $8^3$ ways of rotating the corners. So, there are $8! \times 3^8$ valid cubes made from repositioning and rotating the corners.

So, putting this all together, if we divide the number of valid composite cubes by the total number of composite cubes, we will get our answer.

$$\frac{24 \times 8! \times 3^8 \times 12! \times 2^{12} \times 6! \times 4^6}{27! \times 24^{27}}=1.8298051 \times 10^{-37}$$

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  • $\begingroup$ I added this answer when I thought the accepted answer was wrong. After I corrected my mistakes, I realize that it is correct. I will leave this here for now, since I think it shows a different way of thinking about it. $\endgroup$ – Trenin Dec 9 '15 at 18:30

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