2024 AMC 10A Problem 24

P21-25~8 minPrint

Basic Probability Paths & Grids Split into exhaustive cases and handle each Set up the equation/formula and compute; no special trick needed

A bee is moving in three-dimensional space. A fair six-sided die with faces labeled $A^+, A^-, B^+, B^-, C^+,$ and $C^-$ is rolled. Suppose the bee occupies the point $(a,b,c).$ If the die shows $A^+$ , then the bee moves to the point $(a+1,b,c)$ and if the die shows $A^-,$ then the bee moves to the point $(a-1,b,c).$ Analogous moves are made with the other four outcomes. Suppose the bee starts at the point $(0,0,0)$ and the die is rolled four times. What is the probability that the bee traverses four distinct edges of some unit cube?

Show solution without answering

Key insight

Each roll must use a new direction, and a direction used before must be reversed. Counting roll by roll gives 6 · 4 · 7 = 168.

Solution

There are $6^4 = 1296$ equally likely ways to roll the die four times. Count the good ones roll by roll.

Call the three directions the $A$ , $B$ and $C$ directions. A path stays on one unit cube as long as, in each direction, the bee only moves between two neighbouring values. So if the bee moves again in a direction it already used, it must go the opposite way (after $A^+$ , only $A^-$ is allowed, and the reverse).

1. Roll 1: anything. $6$ choices. Say it is $A^+$ (the other cases are the same by symmetry).
2. Roll 2: it cannot be $A^+$ (that leaves the cube) or $A^-$ (that goes back over the same edge). So it must be a new direction: $B^\pm$ or $C^\pm$ . $4$ choices. Say it is $B^+$ .
3. Roll 3: it cannot use the $B$ direction (same reasons as roll 2). Two kinds of choice remain:
- Case 1: go back the other way in direction $A$ ( $A^-$ ). $1$ choice. The bee has now walked three sides of a square.
- Case 2: a new direction ( $C^+$ or $C^-$ ). $2$ choices.
4. Roll 4.
- In Case 1 (moves $A^+, B^+, A^-$ ): not $A$ (just used). $B^-$ closes the square, which is fine: $1$ choice. Or a $C$ move, either way: $2$ choices. Total $3$ choices.
- In Case 2 (moves $A^+, B^+, C^\pm$ ): not $C$ (just used). $A^-$ or $B^-$ are allowed; $A^+$ or $B^+$ would leave the cube. Total $2$ choices.
5. Count. After the first two rolls, the last two rolls can go $1 \cdot 3 + 2 \cdot 2 = 7$ ways. So the number of good sequences is
$6 \cdot 4 \cdot 7 = 168.$
6. Probability: $\dfrac{168}{1296} = \dfrac{7}{54}$ (divide top and bottom by $24$ ).

In every good case the four edges are different, because the bee never goes straight back along the edge it just used, and in four moves that is the only way to repeat an edge.

The answer is $\boxed{\textbf{(B)}\ \frac{7}{54}}$ .

Why this works

The words "four distinct edges of some unit cube" turn into two simple rules for the rolls: never use the same direction twice in a row, and when you return to a direction, go the other way. With rules like that, counting one roll at a time and splitting into cases only where the choices differ is quick and safe.

Alternative approach

Count by the pattern of directions. The directions used by the four rolls, with no direction twice in a row, can be chosen in $3 \cdot 2 \cdot 2 \cdot 2 = 24$ ways. In each direction, the first move can be $+$ or $-$ , and every later move in that direction is forced to be the opposite of the one before.

Patterns that use only $2$ directions must alternate, like $ABAB$ . There are $3 \cdot 2 = 6$ of them, each with $2 \cdot 2 = 4$ sign choices: $24$ sequences.
The other $18$ patterns use all $3$ directions, each with $2 \cdot 2 \cdot 2 = 8$ sign choices: $144$ sequences.

That is $24 + 144 = 168$ good sequences again, and $\frac{168}{1296} = \frac{7}{54}$ .

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 $6^4 = 1296$ equally likely roll sequences. Translate the condition.

Staying on one cube. The four edges lie on a common unit cube exactly when the path stays inside a $1\times1\times1$ box, i.e. each coordinate only ever takes two adjacent values. For a given axis this means its steps alternate in sign: after a $+$ step the next step on that axis must be $-$ , and vice versa.

Distinct edges. Two consecutive steps on the same axis would have to be opposite (to stay in the box), which retraces the same edge, so consecutive steps must use different axes. With only four steps, any other repeated edge would also force such an immediate backtrack, so this is the only extra restriction.

So a sequence is good iff (i) consecutive steps are on different axes, and (ii) the steps on each axis alternate in sign.

Count. Axis patterns of length $4$ with no two consecutive equal: $3\cdot 2\cdot 2\cdot 2 = 24$ . For a given pattern, the first step on each axis has a free sign and every later step on that axis is forced, so there are $2^k$ sign choices where $k$ is the number of distinct axes used.

Patterns using only $2$ axes must alternate, $XYXY$ : choose the two axes and which goes first, $3\cdot 2 = 6$ patterns, each with $2^2 = 4$ sign choices: $24$ sequences.
The remaining $24 - 6 = 18$ patterns use all $3$ axes: $2^3 = 8$ sign choices each: $144$ sequences.

Total $24 + 144 = 168$ , and $\dfrac{168}{1296} = \dfrac{7}{54}$ .

The answer is $\boxed{\textbf{(B)}\ \frac{7}{54}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Counting & Probability, Chapter 5: More With Combinations (§5.2)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques