AMC 10 · 2025 · #24
Grade 8 probabilityA frog hops along the number line according to the following rules:
It starts at 0.
If it is at 0, then it moves to 1 with probability 21 and disappears with probability 21.
For n=1,2,3, if it is at n, then it moves to n+1 with probability 41, to n−1 with probability 41, and disappears with probability 21.
What is the probability that the frog reaches 4?
Pick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A frog starts at $0$ on the number line. From $0$ it hops to $1$ with probability $\tfrac12$ and otherwise disappears. From any of $1$, $2$, or $3$ it hops to the next spot up with probability $\tfrac14$, back one spot with probability $\tfrac14$, and disappears with probability $\tfrac12$. Find the probability that the frog ever reaches $4$.
Givens: The frog starts at position $0$; From $0$: hop to $1$ with probability $\tfrac12$, disappear with probability $\tfrac12$; From $n = 1, 2, 3$: hop to $n+1$ with probability $\tfrac14$, to $n-1$ with probability $\tfrac14$, disappear with probability $\tfrac12$; Answer choices: (A) $\tfrac1{101}$, (B) $\tfrac1{100}$, (C) $\tfrac1{99}$, (D) $\tfrac1{98}$, (E) $\tfrac1{97}$
Unknowns: The probability that the frog, starting at $0$, eventually lands on $4$
Understand
Restated: A frog starts at $0$ on the number line. From $0$ it hops to $1$ with probability $\tfrac12$ and otherwise disappears. From any of $1$, $2$, or $3$ it hops to the next spot up with probability $\tfrac14$, back one spot with probability $\tfrac14$, and disappears with probability $\tfrac12$. Find the probability that the frog ever reaches $4$.
Givens: The frog starts at position $0$; From $0$: hop to $1$ with probability $\tfrac12$, disappear with probability $\tfrac12$; From $n = 1, 2, 3$: hop to $n+1$ with probability $\tfrac14$, to $n-1$ with probability $\tfrac14$, disappear with probability $\tfrac12$; Answer choices: (A) $\tfrac1{101}$, (B) $\tfrac1{100}$, (C) $\tfrac1{99}$, (D) $\tfrac1{98}$, (E) $\tfrac1{97}$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #11 Work Backwards, #7 Identify Subproblems
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, $P_4 = 1$, and pushes that certainty back toward the start $P_0$. 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.
Execute — Answer: E
6.EE.B.6 Step 1 Name the chance at each spot
- Give the thing we want a name at every position.
- Let $P_n$ be the probability that the frog eventually reaches $4$ once it is standing on spot $n$.
- The frog begins at $0$, so the answer we are after is $P_0$.
- Reaching $4$ from $4$ needs no hops at all, so $P_4 = 1$.
- Now the wandering-frog question becomes: find the five numbers $P_0, P_1, P_2, P_3, P_4$.
💡 Labeling each spot with its own chance of success turns the frog's choices into things we can write as equations.
7.SP.C.7 Step 2 Turn one hop into one equation
- Stand on an interior spot $n = 1, 2, 3$.
- The very next hop goes to $n+1$ (chance $\tfrac14$), to $n-1$ (chance $\tfrac14$), or vanishes (chance $\tfrac12$).
- The frog's overall chance of reaching $4$ from $n$ is the weighted average of the chances from wherever it lands next, and a disappearance contributes $0$.
- That gives $P_n = \tfrac14 P_{n+1} + \tfrac14 P_{n-1}$.
- Writing it for $n = 1, 2, 3$ yields three equations tying each spot to its neighbors.
💡 A future chance of success is just the average of the chances from each place the next hop could take you.
7.SP.C.7 Step 3 Handle the two special ends
- The endpoints follow their own rules.
- At $4$ the frog is already finished, so $P_4 = 1$.
- At $0$ the frog has only two moves — to $1$ with chance $\tfrac12$ or gone with chance $\tfrac12$ — so $P_0 = \tfrac12 P_1$.
- Together with the three interior equations, that is five equations for the five unknowns.
💡 The start and the finish behave unlike the middle spots, so each earns its own equation.
8.EE.C.8 Step 4 Chase every chance back to $P_0$
- Rewrite each $P_n$ as a multiple of $P_0$.
- From $P_0 = \tfrac12 P_1$ we get $P_1 = 2P_0$.
- Plug that into $P_1 = \tfrac14 P_2 + \tfrac14 P_0$ and multiply by $4$: $8P_0 = P_2 + P_0$, so $P_2 = 7P_0$.
- Plug both into $P_2 = \tfrac14 P_3 + \tfrac14 P_1$ and multiply by $4$: $28P_0 = P_3 + 2P_0$, so $P_3 = 26P_0$.
💡 Because each spot's chance is chained to its neighbor, a single starting value $P_0$ locks in all the rest.
8.EE.C.7 Step 5 Close the loop and solve
- One equation is still unused: the one at spot $3$, namely $P_3 = \tfrac14 P_4 + \tfrac14 P_2 = \tfrac14 + \tfrac14 P_2$.
- Substitute $P_3 = 26P_0$ and $P_2 = 7P_0$, then multiply by $4$: $104P_0 = 1 + 7P_0$, so $97P_0 = 1$ and $P_0 = \tfrac{1}{97}$.
- The frog reaches $4$ with probability $\tfrac{1}{97}$, which is choice (E).
💡 The last leftover equation pins down the one free value, and every other chance collapses along with it.
6.EE.B.6 Give the thing we want a name at every position. Let $P_n$ be the probability th 7.SP.C.7 Stand on an interior spot $n = 1, 2, 3$. The very next hop goes to $n+1$ (chance 7.SP.C.7 The endpoints follow their own rules. At $4$ the frog is already finished, so $P 8.EE.C.8 Rewrite each $P_n$ as a multiple of $P_0$. From $P_0 = \tfrac12 P_1$ we get $P_1 8.EE.C.7 One equation is still unused: the one at spot $3$, namely $P_3 = \tfrac14 P_4 + Review
Reasonableness: The chance should be tiny: the frog must survive at least four hops (each move keeps it alive with probability only $\tfrac12$) and it can also drift backward and waste moves, so a value near $\tfrac{1}{100}$ is expected, and $\tfrac{1}{97} \approx 0.0103$ fits. A floor check: the straight path $0 \to 1 \to 2 \to 3 \to 4$ alone has probability $\tfrac12 \cdot \tfrac14 \cdot \tfrac14 \cdot \tfrac14 = \tfrac{1}{128}$, and adding all the meandering paths should lift the total slightly above that; indeed $\tfrac{1}{97} > \tfrac{1}{128}$. Each of the five equations was used exactly once and every substitution is reversible, so no information was lost.
Alternative: Track instead the first-passage chance $q_k$ of ever climbing from spot $k-1$ up to $k$. Then $q_1 = \tfrac12$, and each later one satisfies $q_k = \tfrac14 + \tfrac14 q_{k-1} q_k$, giving $q_2 = \tfrac27$, $q_3 = \tfrac{7}{26}$, $q_4 = \tfrac{26}{97}$; the product $q_1 q_2 q_3 q_4$ telescopes to $\tfrac{1}{97}$. A third route notices the coefficients $1, 2, 7, 26$ obey $a_{n+1} = 4a_n - a_{n-1}$ (characteristic roots $2 \pm \sqrt{3}$), again landing on $97$.
CCSS standards used (min grade 8)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Attaching an unknown $P_n$ to each spot to stand for its chance of eventually reaching $4$.)7.SP.C.7Develop probability models and use them to find probabilities of events (Writing each hop as a weighted-average equation $P_n = \tfrac14 P_{n+1} + \tfrac14 P_{n-1}$ and handling the boundary chances at $0$ and $4$.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Substituting the boundary and interior equations to express $P_1, P_2, P_3$ as multiples of $P_0$.)8.EE.C.7Solve linear equations in one variable (Reducing the last equation to $97P_0 = 1$ and solving for $P_0 = \tfrac{1}{97}$.)
⭐ 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: $97P_0 = 1$, so the answer is $\tfrac{1}{97}$.
⭐ 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: $97P_0 = 1$, so the answer is $\tfrac{1}{97}$.
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