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2024 AMC 10A Problem 24

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2024 AMC 10A Problem 24AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Probability · difficulty 4 of 5 · about 8 min · position P21-25

A bee is moving in three-dimensional space. A fair six-sided die with faces labeled A+,A−,B+,B−,C+,A^+, A^-, B^+, B^-, C^+, and C−C^- is rolled. Suppose the bee occupies the point (a,b,c).(a,b,c). If the die shows A+A^+ , then the bee moves to the point (a+1,b,c)(a+1,b,c) and if the die shows A−,A^-, then the bee moves to the point (a−1,b,c).(a-1,b,c). Analogous moves are made with the other four outcomes. Suppose the bee starts at the point (0,0,0)(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?

  1. A)

    154\frac{1}{54}

  2. B)

    754\frac{7}{54}

  3. C)

    16\frac{1}{6}

  4. D)

    518\frac{5}{18}

  5. E)

    25\frac{2}{5}

Answer

B

Key insight

Four steps trace distinct edges of one cube iff consecutive steps use different axes and repeated axes alternate sign: 24 axis patterns, 2^(axes used) sign choices, total 168/1296.

Solution

There are 64=12966^4 = 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 1×1×11\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 44 with no two consecutive equal: 3⋅2⋅2⋅2=243\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 2k2^k sign choices where kk is the number of distinct axes used.

  • Patterns using only 22 axes must alternate, XYXYXYXY: choose the two axes and which goes first, 3⋅2=63\cdot 2 = 6 patterns, each with 22=42^2 = 4 sign choices: 2424 sequences.
  • The remaining 24−6=1824 - 6 = 18 patterns use all 33 axes: 23=82^3 = 8 sign choices each: 144144 sequences.

Total 24+144=16824 + 144 = 168, and 1681296=754\dfrac{168}{1296} = \dfrac{7}{54}.

The answer is (B) 754\boxed{\textbf{(B)}\ \frac{7}{54}}.

Why this works

The geometric phrase "edges of some unit cube" is really two bookkeeping conditions on the roll sequence: bounded range per coordinate (alternating signs) and no repeated edge (no consecutive same axis). Once the event is restated combinatorially, the count splits naturally by how many axes appear, because the number of free sign choices depends only on that.

The trap

Only forbidding immediate backtracking (probability 125/216) without requiring the path to stay in a unit cube, i.e. that a repeated axis be traversed back the other way.

Common mistakes

  • Only forbidding immediate backtracking (probability 125/216) without requiring the path to stay in a unit cube, i.e. that a repeated axis be traversed back the other way.
  • Forgetting that the XYXYXYXY patterns have only 44 sign choices, not 88, and getting 24⋅8=19224\cdot 8 = 192.

Techniques

Split into exhaustive cases and handle each · Set up the equation/formula and compute; no special trick needed