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| − | ==Solution 1.5==
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| − | (The above solution explains the solution very well; however the thought process is not explicitly addressed)
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| − | We want P(two cubes can be rotated to match after they are painted); i.e. P(cubes are rotationally identical). By Polya's enumeration theorem (from Burnside's Lemma), there are <math>n^6 + 3n^4 +12n^3 +8n^2</math> distinct ways to color a cube using <math>n</math> colors. Plugging in <math>n = 2</math>, we find that there are 10 distinct ways to color a cube with 2 colors (meaning none of these 10 colorations can be rotated to match another). However, each coloration does not have the same probability of occuring (the cubes are painted randomly); for example, there is only 1 way to paint both cubes all black, as any rotations will not yield a visually different result. There are multiple ways to paint each cube with 1 white face and 5 black faces, as one cube can simply be rotated to match the other. Therefore, we have to do casework for 10 cases, where each case is a distinct way to color a cube. (The above solution sometimes addresses more than one case at once)
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| − | Case 1: Both cubes are completely black <math>\rightarrow</math> 1 way, P(Case 1) = <math>\frac{1^2}{4096}</math> (There are 2^12 or 4096 total colorations of the cubes)
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| − | Case 2: Exactly one face is white, the rest are black <math>\rightarrow</math> 6 ways, P(Case 2) = <math>\frac{6^2}{4096}</math>
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| − | Case 3: 2 white faces share an edge, the rest are black <math>\rightarrow</math> <math>\frac{\text{6 choices for the first white face * 4 choices for the second adjacent white face}}{2!\text{(because the white faces are identical)}}</math> = 12 ways, P(Case 3) = <math>\frac{12^2}{4096}</math>
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| − | Case 4: 2 white faces opposite each other, the rest are black <math>\rightarrow</math> <math>\frac{\text{6 choices for the first white face * 1 choice for the second white face}}{2!}</math> = 3 ways, P(Case 4) = <math>\frac{3^2}{4096}</math>
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| − | Case 5: 3 white faces share a vertex, the rest are black <math>\rightarrow</math> <math>\frac{\text{6 choices for the first * 4 choices for the second * 3 choices for the third}}{3!}</math> = 12 ways, P(Case 5) = <math>\frac{12^2}{4096}</math>
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| − | Case 6: 3 white faces in a "row", the rest are black <math>\rightarrow</math> <math>\frac{\text{6 choices for the first * 4 choices for the second * 2 choice for the third}}{3!}</math> = 8 ways, P(Case 5) = <math>\frac{8^2}{4096}</math>
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| − | Cases 7, 8, 9, and 10: These cases are the same as cases 1, 2, 3, and 4, albeit with a color swap. Doubling the probabilities of cases 1-4 will cover these cases.
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| − | Therefore we have P(cubes are rotationally identical) = <math>\frac{1^2 + 6^2 + 12^2 + 3^2 + 12^2 + 8^2 + 3^2 + 12^2 + 6^2 + 1^2}{4096}</math> = <math>\frac{588}{4096} =\boxed{\textbf{(D) } \frac{147}{1024}}.</math>
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| − | P.S. Sorry for the bad formatting. My latex isn't very good =(
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| | ==Video Solution1== | | ==Video Solution1== |
Revision as of 12:59, 8 November 2022
Problem
Two different cubes of the same size are to be painted, with the color of each face being chosen independently and at random to be either black or white. What is the probability that after they are painted, the cubes can be rotated to be identical in appearance?
Solution
Define two ways of painting to be in the same 
if one can be rotated to form the other.
We can count the number of ways of painting for each specific .
Case 1: Black-white color distribution is 0-6 (out of 6 total faces)
Trivially 
way to paint the cubes.
Case 2: Black-white color distribution is 1-5
Trivially all 
ways belong to the same , so 
ways to paint the cubes.
Case 3: Black-white color distribution is 2-4
There are two 
for this case: the where the two red faces are touching and the other where the two red faces are on opposite faces. There are 
members of the latter since there are unordered pairs of 
opposite faces of a cube. Thus, there are 
members of the former . Thus, 
ways to paint the cubes for this case.
Case 4: Black-white color distribution is 3-3
By simple intuition, there are also two for this case, the where the three red faces meet at a single vertex, and the other class where the three red faces are in a "straight line" along the edges of the cube. Note that since there are 
vertices in a cube, there are members of the former class and 
members of the latter class. Thus, 
ways to paint the cubes for this case.
Note that by symmetry (since we are only switching the colors), the number of ways to paint the cubes for black-white color distributions 4-2, 5-1, and 6-0 is 2-4, 1-5, and 0-6 (respectively).
Thus, our total answer is![\[\frac{2(6^2 + 1^2 + 12^2 + 3^2) + 12^2 + 8^2}{2^{12}} = \frac{588}{4096} = \boxed{\textbf{(D) } \frac{147}{1024}}.\]](//latex.artofproblemsolving.com/4/0/9/4090b7116c16c14b3f455aea87e8b6520c02044a.png)
-fidgetboss_4000
Video Solution1
https://youtu.be/n9zwaCZnDPg
~Education, the Study of Everything
Video Solution by TheBeautyofMath
https://youtu.be/V5qI12icrm8
~IceMatrix
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
if one can be rotated to form the other.
way to paint the cubes.
ways belong to the same 
ways to paint the cubes.
for this case: the 
members of the latter 
opposite faces of a cube. Thus, there are 
members of the former 
ways to paint the cubes for this case.
vertices in a cube, there are 
members of the latter class. Thus, 
ways to paint the cubes for this case.
![\[\frac{2(6^2 + 1^2 + 12^2 + 3^2) + 12^2 + 8^2}{2^{12}} = \frac{588}{4096} = \boxed{\textbf{(D) } \frac{147}{1024}}.\]](http://latex.artofproblemsolving.com/4/0/9/4090b7116c16c14b3f455aea87e8b6520c02044a.png)
