How to count - probability puzzle

Question

A $3 \times 3 \times 3$ big cube consists of $1 \times 1 \times 1$ smaller cubes. The big cube is painted black on the outside. Suppose we disassemble the cube and randomly put it back together. What is the probability of perfectly assembling the cube i.e. all faces are black again?

Do you think my partial progress is correct? If not, could you provide a hint/clue (without giving away the entire solution)?

Solution (My Attempt).

There are $8$ corner(vertex) cubes, having $3$ faces black - label them $V$

There are $12$ edge cubes, having $2$ faces black - label them $E$

There are $6$ center of face cubes, having $1$ face black - label it $C$

There is $1$ origin cube, that is not black - label it $O$

For the $1 \times 1 \times 1$ cube, there are $6$ choices for the top face, and for each such choice, $4$ choices for an adjacent face. This completely determines the orientation. So, there are $24$ distinguishable orientations of the unit cube.

The probability that a $V$ cube is correctly oriented = $\frac{1}{24}$.

The probability that a $E$ cube is correctly oriented = $\frac{1}{24}$.

The probability that a $C$ cube is correctly oriented = $\frac{4}{24}$.

The probability that a $O$ cube is correctly oriented = $1$.

Furthermore, we can swap one corner cube for another, one edge cube for another and so forth leaving the outer appearance unchanged. So, there are $\frac{27!}{8!12!6!1!}$ favourable arrangements of the cubes.

Putting it all together:

$$P \{\text{Perfectly assembling the cube} \} = \frac{1}{24^8} \times \frac{1}{24^{12}} \times \left(\frac{4}{24}\right)^6 \times \frac{\frac{27!}{8!12!6!1!}}{27!}$$

Answer 1

Mostly good, but there are three mistakes. (Thanks to @obscurans for spotting the third.) The probabilities for $E$ and $V$ cubes to be correctly oriented are wrong. And $\frac{27!}{8!12!6!1!}$ counts the distinguishable configurations if you consider the four types of cubes as indistinguishable; whereas what you want is the number of equivalent configurations corresponding to a particular one of those distinguishable configurations (namely the one where all the types are in the right places). Note that your count decreases the more equivalent pieces there are (and would be $1$ if all pieces were of one type), whereas the correct count should increase the more equivalent pieces there are (and would be $27!$ if all pieces were of one type).

Answer 2

The good configurations place center, outer center, corner and remaining ones correctly in $1, 6!, 8!, 12!$ of the total configurations ($27!$). The probability that all the center, outer center, corner and remaining ones are correctly oriented is 1, $(4/24)^6$, $(3/24)^8$ and $(2/24)^{12}$. The probability of correctly assembling the cube is therefore $$\frac{6!8!12!}{27!}\frac{1}{2^{54}3^{18}} = \frac{1}{27\cdot 26 \cdot25^{1/2}\cdot 24^{18}\cdot 223092870}.$$

where $223092870$ is the 10th primorial number, the product of all primes smaller than 27.

Numerical approximation

_{The $6^6, 8^8, 12^{12}$ factors are familiar from the factorial approximation and $\frac{6!8!12!}{27!}\approx 2\pi e \sqrt{6\cdot 8\cdot 12/27}\frac{6^68^812^{12}}{27^{27}}$ which curiously simplifies the probability to $$ \frac{2\pi e}{\sqrt{27}}\frac{24}{\cdot27^{27}}=\pi e\sqrt{\frac{2^8}{3}}\frac{1}{3^{81}} \approx\frac{81}{3^{81}} = 3\frac{27}{27^{27}}= \frac{3^4}{3^{3^4}}=3^{-77}. $$}