### Probabilities in a Painted Cube

A wooden cube - after being painted all over - has been cut into 3\times 3\times 3 smaller cubes. These were thoroughly mixed in a bag, from which one was produced and tossed. What is the probability that a painted side turned up?

On impulse, one would approach the problem in a more or less standard way. There are 8 corner cubes with 3 painted sides, 12 mid-edge cube with 2 painted sides, and 6 face-central cubes with only 1 side painted. The total probability is then

\displaystyle\frac{8}{27}\cdot\frac{3}{6}+\frac{12}{27}\cdot\frac{2}{6}+\frac{6}{27}\cdot\frac{1}{6}=\frac{1}{3}.

Now generalize: cut the cube into n\times n\times n, n\gt 1 smaller cubes and ask the same question. The problem is not awfully difficult but needs some figuring out. Following the foregoing pattern, we eventually arrive at

\displaystyle\frac{8}{n^3}\cdot\frac{3}{6}+\frac{12\cdot (n-2)}{n^3}\cdot\frac{2}{6}+\frac{6\cdot (n-2)^2}{n^3}\cdot\frac{1}{6}

This expression simplifies, as you can verify, to \displaystyle\frac{1}{n}, which, by the way, confirms the answer for n=3.

The above is a useful and not too difficult exercise, but there is a delightful shortcut that avoids most of the counting of cubes and their sides.

n^3 cubes have a total of 6\cdot n^3 sides. Of these, 6\cdot n^2 are painted. All sides have the same probability of turning up, therefore, a painted side will turn up with the probability \displaystyle\frac{6n^2}{6n^3}=\frac{1}{n}.

### References

- R. Honsberger,
*Mathematical Delights*, MAA, 2004, pp. 77-78

Even easier: As you look at the cube from any side at any location, you see one painted face out of the n painted faces aiming toward you along that column. So there's everywhere 1/n faces painted.

May 20th, 2013 at 6:35 pmJoshua, I am very much sorry but I do not understand that. Am I at any location on the cube looking along a column? If so, I probably see only a part of that column - this if making an effort not to see anything else, even a little sideways. In any event, what I see very much depends on my location. This is according to my interpretation of what you wrote. I do want to understand that.

May 21st, 2013 at 8:55 amI'm imagining that I can see the "top" face of all n little cubelets in one column stacked up as I look down on the top. My use of "see" was probably too metaphorical as you certainly can't have transparent painted faces! Anyway in each column of n small cubes, and in each of the six directions of the faces, we'll have exactly one painted side, so 1/n of all the small faces are painted.

May 23rd, 2013 at 4:57 pmNow I see! Standing on a tall (even semi)transparent column and looking all the way down may make one dizzy - depending on n of course. Thank you. That's a transparent way to think of the problem.

May 25th, 2013 at 10:22 am[...] Probabilities in a Painted Cube, Cut the Knot examines solutions to a problem about painting and cutting a larger cube into unit [...]

June 14th, 2013 at 9:28 amMy tip: As you look at it from any side at any location, you see the painted face out of the n painted faces aiming toward you along that column. So there's everywhere 1/n faces painted.

January 13th, 2015 at 8:41 am