2004 AMC 10B Problem 23

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

Translate the physical condition into a combinatorial one: "some placement has a monochrome band" means "some opposite pair contains all faces of the other color." Then organize by the size of the minority color class; the geometry of the cube (three opposite pairs, adjacency) enters only in the two-face case.

Inclusion-exclusion over the three bands. Each band is monochrome in $2 \cdot 2 \cdot 2 = 8$ colorings (band color, two free faces). Any two bands overlap in $4$ faces and together cover all six, so both monochrome forces a solid cube: $2$ colorings, the same for all three bands at once. Union: $3 \cdot 8 - 3 \cdot 2 + 2 = 20$.

• Counting the 4-2 split with any two faces of the minority color (15 pairs) instead of only the 3 opposite pairs.
• Adding the three bands' $8$ colorings each ($24$) without subtracting the solid-color cubes counted three times.