AMC 10 · 2025 · #24

Grade 8 probability

Archetype Branching State Enumeration └ Sub-archetype Markov State Probability

probability-basicrecursive-sequencesystems-of-equations convert-to-algebra ↑ Prerequisites: probability-basic

📏 Medium solution 💡 4 insights

Problem

A frog starts at 0 on the number line. From 0 it hops to 1 with probability

\frac{1}{2}

and otherwise disappears. From any of 1, 2, or 3 it hops to the next spot up with probability

\frac{1}{4}

, back one spot with probability

\frac{1}{4}

, and disappears with probability

\frac{1}{2}

. Find the probability that the frog ever reaches 4.

Pick an answer.

(A)

$\frac1{101}$

(B)

$\frac 1{100}$

(C)

$\frac1{99}$

(D)

$\frac1{98}$

(E)

$\frac1{97}$

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

The frog can wander forever, so chasing individual paths is hopeless. Tool #4 (Introduce a Variable) fixes this: attach one unknown P_n to each spot, standing for "my chance of ever reaching 4 from here." Tool #11 (Work Backwards) anchors everything on the one chance we already know, P₄ = 1, and pushes that certainty back toward the start P₀. Tool #7 (Identify Subproblems) treats each spot as its own tiny equation — one hop's worth of reasoning — so the tangled wandering becomes five clean equations in five unknowns that solve by simple substitution.

1STEP 1

Name the chance at each spot

Let P_n be the chance of eventually reaching 4 from spot n; we want P₀, and standing on 4 already wins, so P₄ = 1.

P_n = Pr(reach 4 ∣ now at n), P₄ = 1

2STEP 2

Turn one hop into one equation

Weighting one hop from n = 1, 2, 3 gives P_n = $\frac{1}{4}$ P_n+1 + $\frac{1}{4}$ P_n-1, since vanishing contributes 0.

P_n = 1/4 P_n+1 + 1/4 P_n-1 (n = 1, 2, 3)

3STEP 3

Handle the two special ends

The ends differ: 4 is already done, and from 0 the only move up carries chance $\frac{1}{2}$ , so P₀ = $\frac{1}{2}$ P₁.

P₄ = 1, P₀ = 1/2 P₁

4STEP 4

Chase every chance back to P₀

Chain the equations upward as multiples of P₀: P₁ = 2P₀, then P₂ = 7P₀, then P₃ = 26P₀.

P₁ = 2P₀, P₂ = 7P₀, P₃ = 26P₀

5STEP 5

Close the loop and solve

The unused spot-3 equation becomes 104P₀ = 1 + 7P₀, so 97P₀ = 1 and P₀ = $\frac{1}{97}$ — choice (E).

104P₀ = 1 + 7P₀ → 97P₀ = 1 → P₀ = 1/97 (E)

Answer

1/97

The chance should be tiny: the frog must survive at least four hops (each move keeps it alive with probability only 1/2) and it can also drift backward and waste moves, so a value near 1/100 is expected, and 1/97 ≈ 0.0103 fits. A floor check: the straight path 0 → 1 → 2 → 3 → 4 alone has probability 1/2 · 1/4 · 1/4 · 1/4 = 1/128, and adding all the meandering paths should lift the total slightly above that; indeed 1/97 ＞ 1/128. Each of the five equations was used exactly once and every substitution is reversible, so no information was lost.

💡Key takeaway

Label each spot with its chance of reaching 4, turn each hop into one weighted-average equation, then chain them all back to the start: 97P₀ = 1, so the answer is 1/97.

Name the chance at each spot
Turn one hop into one equation
Handle the two special ends
Chase every chance back to P₀
Close the loop and solve

Name the chance at each spot

Turn one hop into one equation

Handle the two special ends

Chase every chance back to P₀

Close the loop and solve

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