# Probability that a square of a Rubik's cube has a painted face

The outer surface of a 3x3 Rubik's cube is painted red. You break the Rubik's cube into 27 smaller cubes. You select 1 at random and put it on the table. You see that the 5 visible spaces have no red painting. What is the probability that the bottom face is painted red?

So I am given that the selected cube has 5 faces that are not painted red. This eliminates 20 of the cubes. The remaining cubes are the 6 centered cubes on each of the 6 faces, and the very center cube.

So isn't the probability simply 6/7? There is a hint that says to use Bayes theorem, but I don't see how that comes in. All that is needed seems to be just the definition of conditional probability. Specifically we want $$P(A\mid B)$$ where $$A$$ is the event that the selected cube has 1 painted face and $$B$$ is the event that the selected cube has 5 un-painted faces.

If it’s one of the cubes in the centres of the faces, there is only one chance in six that the red face is on the bottom, but if it’s the central cube, any of the six faces could be on the bottom. To put it a bit differently, the $$7$$ cubes have a total of $$42$$ faces, all of which are equally likely to be on the bottom. If you don’t see a red face, you know that the face on the bottom is either one of the $$6$$ faces of the central cube or the one red face of one of the other $$6$$ cubes. Thus, it’s one of $$12$$ faces, and your sample space actually has size $$12$$. In $$6$$ of those $$12$$ cases the bottom face is red, so the probability is actually $$\frac12$$.

• ahhhhh I got tricked. Where does Bayes rule come in to play here? Still seems that conditional probability is sufficient. Commented Oct 5, 2020 at 18:52
• Never thought about it that way, nice explanation Commented Oct 5, 2020 at 18:54
• @BrianM.Scott Yeah, I just deleted that comment for the reason you stated. I got 0.5 now. Commented Oct 5, 2020 at 19:50
• @student010101: Let $A$ be the event that the bottom face is red, and let $B$ be the event that you see no red face. There are $162$ faces, $54$ of which are red, so $P(A)=\frac{54}{162}=\frac13$. As noted in my answer, there are $12$ possible bottom faces that result in $B$, so $P(B)=\frac{12}{162}=\frac2{27}$. And $P(B\mid A)=\frac6{54}=\frac19$, so Bayes gives us $$P(A\mid B)=\frac{\frac19\cdot\frac13}{\frac2{27}}=\frac12\,.$$ Commented Oct 5, 2020 at 20:07

In fact, you only need Laplace Definition of probability:

• Possible Cases: Cubes with AT LEAST 5 non-painted faces. They are 7 (All the 6 centers of de faces and the center of teh Rubik cube).
• Favorable Cases: Cubes with 5 non-painted faces and 1 painted face. They are the 6 centers of the RubikCube-faces.

And that's all.

• Your cases are not equally likely. The $27$ minicubes have a total of $162$ faces, of which $12$ placed face-down would show five unpainted faces. $6$ of those $12$ have the face-down face painted and $6$ out of $12$ have the face-down face unpainted. So the probability is $\frac{6}{12}=\frac12$ Commented Jul 21, 2021 at 12:33