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2014 AMC 10B Problem 25

2014 AMC 10B Problem 25AMC 10 Step by Step

Name: Date: September 13, 2026

Conditional Probability & States · difficulty 5 of 5 · about 10 min · position P21-25

In a small pond there are eleven lily pads in a row labeled $0$ through $10$ . A frog is sitting on pad $1$ . When the frog is on pad $N$ , $0<N<10$ , it will jump to pad $N-1$ with probability $\frac{N}{10}$ and to pad $N+1$ with probability $1-\frac{N}{10}$ . Each jump is independent of the previous jumps. If the frog reaches pad $0$ it will be eaten by a patiently waiting snake. If the frog reaches pad $10$ it will exit the pond, never to return. What is the probability that the frog will escape without being eaten by the snake?

A)
$\frac{32}{79}$
B)
$\frac{161}{384}$
C)
$\frac{63}{146}$
D)
$\frac{7}{16}$
E)
$\frac{1}{2}$

Answer

Key insight

Consecutive differences of the escape probabilities satisfy d_N = N/(10-N) d_{N-1}, so d_N = d_0 / C(9,N), and the differences sum to 1.

Solution

Let $p_N$ be the probability of escaping from pad $N$ , so $p_0 = 0$ and $p_{10} = 1$ . One jump from pad $N$ gives
$p_N = \frac{N}{10}\,p_{N-1} + \frac{10 - N}{10}\,p_{N+1}.$
Move $p_N$ across: $N(p_N - p_{N-1}) = (10 - N)(p_{N+1} - p_N)$ . Writing $d_N = p_{N+1} - p_N$ , this is
$d_N = \frac{N}{10 - N}\, d_{N-1}.$
So $d_1 = \frac{1}{9} d_0$ , $d_2 = \frac{2}{8} \cdot \frac{1}{9} d_0 = \frac{d_0}{36}$ , $d_3 = \frac{3}{7} \cdot \frac{d_0}{36} = \frac{d_0}{84}$ , $d_4 = \frac{4}{6} \cdot \frac{d_0}{84} = \frac{d_0}{126}$ ; in general $d_N = \frac{d_0}{\binom{9}{N}}$ , which is symmetric ( $d_{9-N} = d_N$ ).

The differences telescope: $d_0 + d_1 + \cdots + d_9 = p_{10} - p_0 = 1$ . Hence
$1 = 2 d_0\left(1 + \frac19 + \frac{1}{36} + \frac{1}{84} + \frac{1}{126}\right) = 2 d_0 \cdot \frac{252 + 28 + 7 + 3 + 2}{252} = d_0 \cdot \frac{292}{126} = d_0 \cdot \frac{146}{63}.$
So $d_0 = \frac{63}{146}$ , and since $p_1 = p_0 + d_0$ , the frog escapes with probability $\frac{63}{146}$ .

The answer is $\boxed{\textbf{(C)}\ \frac{63}{146}}$ .

Why this works

Absorbing random walks are solved by first-step equations, but ten coupled equations are painful. The trick that recurs in every such problem: rewrite the equation as a relation between consecutive differences, which makes the recurrence first-order and multiplicative, then use the known endpoints to normalize because the differences telescope to $p_{10} - p_0$ .

Alternative approach

By the symmetry $N \leftrightarrow 10 - N$ (which swaps snake and exit), $p_5 = \frac12$ . Set $p_1 = a$ and march forward: $p_2 = \frac{10a}{9}$ , $p_3 = \frac{41a}{36}$ , $p_4 = \frac{145a}{126}$ , $p_5 = \frac{73a}{63}$ . Then $\frac{73a}{63} = \frac12$ gives $a = \frac{63}{146}$ .

The trap

Solving the ten linear equations by brute substitution and making a fraction error; the difference recurrence turns it into one clean sum.

Common mistakes

Solving the ten linear equations by brute substitution and making a fraction error; the difference recurrence turns it into one clean sum.
Guessing $\frac12$ from the symmetry of the jump rule; the symmetry gives $p_5 = \frac12$ , but the frog starts at pad $1$ , far from the middle.

Techniques

Define states/recurrence and iterate · Exploit symmetry to reduce work or pair up objects · Collapse a sum or product by cancellation