2025 AMC 10B Problem 24

For $n = 0,1,2,3,4$ let $p_n$ be the probability that the frog ever reaches $4$, given that it is sitting at $n$. Then $p_4 = 1$, and disappearing contributes $0$. Each hop has probability at least $\tfrac12$ of ending the walk, so the frog is certain to disappear or reach $4$ eventually and these probabilities are well defined.

Condition on the next hop:
$$p_0 = \tfrac12 p_1, \qquad p_1 = \tfrac14\left(p_0 + p_2\right), \qquad p_2 = \tfrac14\left(p_1 + p_3\right), \qquad p_3 = \tfrac14\left(p_2 + 1\right).$$

Rather than eliminate blindly, express everything in terms of $p_1$, working up the line.

From the second equation, $4p_1 = p_0 + p_2 = \tfrac12 p_1 + p_2$, so
$$p_2 = \tfrac72 p_1 .$$
From the third, $4p_2 = p_1 + p_3$, so
$$p_3 = 4 \cdot \tfrac72 p_1 - p_1 = 13 p_1 .$$
From the fourth, $4p_3 = p_2 + 1$, so
$$52 p_1 = \tfrac72 p_1 + 1 \implies \tfrac{97}{2} p_1 = 1 \implies p_1 = \frac{2}{97}.$$

Finally $p_0 = \tfrac12 p_1 = \dfrac{1}{97}$.

The answer is $\boxed{\textbf{(E)}\ \frac{1}{97}}$.

First-step analysis converts an infinite collection of hopping sequences — which may revisit $0, 1, 2, 3$ any number of times — into four linear equations, because the future depends only on where the frog currently is. The useful trick is the direction of the substitution: starting from $p_0 = \tfrac12 p_1$ and pushing upwards writes each $p_n$ as an explicit multiple of $p_1$, so the last equation is a single equation in one unknown, with no fractions to clear until the end. The same method handles any walk on a finite set of states with killing, and the denominators it produces ($97$ here) are exactly the kind of unusual numbers that answer choices like these are testing.

Work one rung at a time. Let $u_n$ be the probability that a frog at $n$ ever reaches $n+1$. From $0$ there is no way back, so $u_0 = \tfrac12$. From $n \ge 1$, the frog either steps up at once (probability $\tfrac14$), or steps down to $n-1$ (probability $\tfrac14$) and must then climb back to $n$ and start again, or disappears:
$$u_n = \tfrac14 + \tfrac14 u_{n-1}u_n \implies u_n = \frac{1}{4 - u_{n-1}} .$$
So $u_0 = \tfrac12$, $u_1 = \tfrac{2}{7}$, $u_2 = \tfrac{7}{26}$, $u_3 = \tfrac{26}{97}$, and the answer is the product
$$u_0u_1u_2u_3 = \frac12 \cdot \frac27 \cdot \frac{7}{26} \cdot \frac{26}{97} = \frac{1}{97},$$
which telescopes because each numerator cancels the previous denominator.

• Adding up only the straight-up path $0 \to 1 \to 2 \to 3 \to 4$, giving $\left(\tfrac12\right)\left(\tfrac14\right)^3 = \tfrac1{128}$ and a guess at the smallest choice (A); the backtracking paths are exactly what turn $128$ into $97$.
• Treating $0$ as fatal, i.e. setting $p_0 = 0$ in the equation for $p_1$; the frog that falls back to $0$ still has a fresh chance to climb.
• Solving the system correctly but reporting $p_1 = \tfrac{2}{97}$; the frog starts at $0$, not at $1$.