2004 AMC 10B Problem 23

1. Count all colorings. Each of the $6$ faces is red or blue, so there are $2^6 = 64$ 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.
- $0$ faces (all six the same): $2$ colorings (all red or all blue). These work.
- $1$ face is the odd color: put it on top. The four sides are all the other color. It works. Choose the odd face ($6$ ways) and its color ($2$ ways): $6 \times 2 = 12$ colorings.
- $2$ faces are the odd color: both must go on top and bottom, so they must be opposite each other. A cube has $3$ pairs of opposite faces, and the odd color can be red or blue: $3 \times 2 = 6$ colorings. If the two odd faces are next to each other, one of them always ends up on a side, so those fail.
- $3$ and $3$: any four side faces include both colors. These all fail.
4. Add up. $2 + 12 + 6 = 20$ colorings work. The probability is
$$\frac{20}{64} = \frac{5}{16}.$$

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

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.

Count each possible "band" of four side faces. There are $3$ bands (one for each choice of top and bottom). For one band to be all one color: $2$ choices for the band's color and $2$ each for top and bottom, so $8$ colorings. Three bands give $3 \times 8 = 24$, 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 $2$ colorings, the same for any two bands and for all three. Those all-one-color cubes were counted $3$ times instead of once, so subtract $2 \times 2 = 4$: $24 - 4 = 20$.

There are $2^6 = 64$ 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).

• $0$ minority faces (all six alike): $2$ colorings, always good.
• $1$ minority face: put that face on top; the vertical faces are all majority. Choose the odd face ($6$ ways) and its color ($2$ ways): $12$ colorings.
• $2$ minority faces: the vertical band must avoid both, so both must be the top and bottom, i.e. an opposite pair. There are $3$ opposite pairs and $2$ colors: $6$ colorings. If the two minority faces are adjacent, no band avoids both, so those fail.
• $3$ and $3$: any four faces include both colors. Fail.

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

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