AMC 10 Step by Step

2004 AMC 10B Problem 23

P21-25~6 minPrint

Each face of a cube is painted either red or blue, each with probability 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?

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Key insight

It works when the less common color is on no faces, one face, or two opposite faces (top and bottom). That is 20 of 64.

Solution

1. Count all colorings. Each of the 66 faces is red or blue, so there are 26=642^6 = 64 equally likely colorings.
2. What does "placed so the four sides match" mean? Setting the cube down picks a top and a bottom, which are opposite faces. The other four faces go around the sides. So we need: some pair of opposite faces can take all the faces of the less common color, leaving four faces of one color around the sides.
3. Sort by how many faces have the less common color.
- 00 faces (all six the same): 22 colorings (all red or all blue). These work.
- 11 face is the odd color: put it on top. The four sides are all the other color. It works. Choose the odd face (66 ways) and its color (22 ways): 6×2=126 \times 2 = 12 colorings.
- 22 faces are the odd color: both must go on top and bottom, so they must be opposite each other. A cube has 33 pairs of opposite faces, and the odd color can be red or blue: 3×2=63 \times 2 = 6 colorings. If the two odd faces are next to each other, one of them always ends up on a side, so those fail.
- 33 and 33: any four side faces include both colors. These all fail.
4. Add up. 2+12+6=202 + 12 + 6 = 20 colorings work. The probability is
2064=516.\frac{20}{64} = \frac{5}{16}.

The answer is (B) 516\boxed{\textbf{(B)}\ \frac{5}{16}}.

Why this works

First turn the physical question into a counting question: "four sides match" means "the odd-colored faces all fit on top and bottom." Then sort the colorings by how many faces have the less common color; the cube's shape only matters in the case with two odd faces, where they must be opposite.

Alternative approach

Count each possible "band" of four side faces. There are 33 bands (one for each choice of top and bottom). For one band to be all one color: 22 choices for the band's color and 22 each for top and bottom, so 88 colorings. Three bands give 3×8=243 \times 8 = 24, but some colorings are counted more than once. Two bands together cover all six faces and share faces, so if both are one color the whole cube is one color: just 22 colorings, the same for any two bands and for all three. Those all-one-color cubes were counted 33 times instead of once, so subtract 2×2=42 \times 2 = 4: 24−4=2024 - 4 = 20.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

There are 26=642^6 = 64 equally likely colorings. Placing the cube on a surface chooses a top-bottom pair of opposite faces; the four vertical faces are the other four. So the cube is "good" if some pair of opposite faces can be removed leaving four faces of one color.

Sort colorings by how many faces carry the less common color (call it the minority color).

  • 00 minority faces (all six alike): 22 colorings, always good.
  • 11 minority face: put that face on top; the vertical faces are all majority. Choose the odd face (66 ways) and its color (22 ways): 1212 colorings.
  • 22 minority faces: the vertical band must avoid both, so both must be the top and bottom, i.e. an opposite pair. There are 33 opposite pairs and 22 colors: 66 colorings. If the two minority faces are adjacent, no band avoids both, so those fail.
  • 33 and 33: any four faces include both colors. Fail.

Good colorings: 2+12+6=202 + 12 + 6 = 20, so the probability is
2064=516.\frac{20}{64} = \frac{5}{16}.

The answer is (B) 516\boxed{\textbf{(B)}\ \frac{5}{16}}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.