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Each face of a cube is painted either red or blue, each with probability $\frac{1}{2}$. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?

The probability of each instance is $(1/2)^6$. So for $4$ colors the same, there are $2$ scenarios and for $5$ colors the same, there is also $2$ scenarios, for $6$ colors the same it is $1$ scenario. My result is $\frac{5}{64}\cdot 2$ since red and blue can be the same color. Thus I get $5/32$.

Is my solution correct and why don't rotations add to the scenario. I don't grasp this fully quite yet.

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    $\begingroup$ Here is the official solution: artofproblemsolving.com/wiki/…. Is there something you don't understand there? $\endgroup$ – Isaac Browne Dec 27 '17 at 0:54
  • $\begingroup$ @IsaacBrowne why are the scenarios in which the the vertical faces do not have the same color counted in the desirable outcomes $\endgroup$ – jamarcus tyrone Dec 27 '17 at 0:58
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Note that the question asks for the probability that the cube can be placed. That means we don't start with 4 designated vertical faces; we have freedom to choose which faces to place vertical.

There are 3 scenarios in which four vertical faces are red and two blue: pick any pair of opposing faces of the cube to be blue, and the rest red. Then we can choose to place the cube blue-face-down, so the red faces come out vertical. There are 3 such pairs (since there are 6 faces).

There are 6 scenarios in which five faces are red and one blue: pick any particular face to be blue, then we can choose to place the cube blue-face-down (or blue-face-up), and the red faces come out vertical.

There is 1 scenario in which six faces are red. The red faces will come out vertical no matter how we place the cube.

So there should be a total of 10 scenarios in which we can place the cube with four vertical red faces. As you alluded to, the same argument goes through for blue faces, giving 20 scenarios in total. So the probability should be $\frac{20}{64}$

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  • $\begingroup$ why are the scenarios in which the the vertical faces do not have the same color counted in the desirable outcomes $\endgroup$ – jamarcus tyrone Dec 27 '17 at 1:05
  • $\begingroup$ @jamarcus can you explain what you mean? $\endgroup$ – greenturtle3141 Dec 27 '17 at 1:26
  • $\begingroup$ @greenturtle3141 The scenarios in which the vertical faces are not the same color are chosen why is this so and isnt the whole problem under the assumption that the cube is fixed $\endgroup$ – jamarcus tyrone Dec 27 '17 at 1:33
  • $\begingroup$ @jamarcustyrone Precisely which scenarios are being counted which you think should not be? $\endgroup$ – Y. Forman Dec 27 '17 at 1:35
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    $\begingroup$ @jamarcustyrone What do you mean by "fixed"? My interpretation of the problem is that we can rotate the cube after we paint it and before we place it on the horizontal surface. $\endgroup$ – Y. Forman Dec 27 '17 at 1:36

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