AMC 10 · 2014 · #22

Grade 8 probability

Archetype Branching State Enumeration └ Sub-archetype Markov State Probability

probability-basicrecursive-sequencesystems-of-equationssymmetry-argument convert-to-algebra ↑ Prerequisites: probability-basicsystems-of-equations

📏 Long solution 💡 4 insights

Problem

Each step moves one way or the other with a position-dependent chance, ending at one of two edges. Find the chance of reaching the far edge.

Pick an answer.

(A)

$\frac{32}{79}$

(B)

$\frac{161}{384}$

(C)

$\frac{63}{146}$

(D)

$\frac{7}{16}$

(E)

$\frac{1}{2}$

How to solve

Strategy Introduce a Variable

There are eleven pads and one wandering frog, so instead of tracking every possible path we name one unknown per pad: s_N, the probability of escaping when currently on pad N (Tool #4, Introduce a Variable). One jump then links each s_N to its neighbors, turning the puzzle into a chain of equations. The board is mirror-symmetric about pad 5, so a change of focus (Tool #16) hands us s₅=1/2 for free. Solving the chain one pad at a time (Tool #7, Identify Subproblems) and then back-substituting (Tool #11, Work Backwards) pins down s₁.

1STEP 1

Name the escape chance per pad

Each position gets its own escape chance.

s₀ = 0, s₁₀ = 1, find s₁

2STEP 2

Link each pad to its neighbors

Every position links to its two neighbours.

10 s_N = N s_N-1 + (10-N) s_N+1 (0 ＜ N ＜ 10)

3STEP 3

Use symmetry at the middle pad

Symmetry gives the middle value free.

s₅ = 1/2

4STEP 4

Climb from the bottom

Climbing from the bottom is straightforward.

s₁ = 9/10s₂, s₃ = 41/40s₂

5STEP 5

Reach the known middle value

That chain reaches the known middle value.

219/35 s₂ = 3 → s₂ = 35/73

6STEP 6

Back-substitute for pad 1

Back-substituting gives 63/146, choice (C).

s₁ = 9/10·35/73 = 63/146 → (C)

Answer

63/146

63/146≈ 0.431 is exactly choice (C), and it is just under 1/2. That fits the story: the frog starts on pad 1, right next to the snake, so escape should be a bit less than even — yet the strong upward push there (up-probability 9/10) keeps it close. The relation 10 s₁ = 9 s₂ checks out: 9·35/73=315/73 equals 10·63/146=630/146=315/73. The symmetric value s₅=1/2 also lands between the endpoints as it must.

💡Key takeaway

Give every pad its own escape chance, use the fair 1/2 at the middle pad, then walk the equations up from the bottom to pin down pad 1.

Name the escape chance per pad
Link each pad to its neighbors
Use symmetry at the middle pad
Climb from the bottom
Reach the known middle value
Back-substitute for pad 1