AMC 10 · 2019 · #22

Grade 7 probability

Archetype Branching State Enumeration └ Sub-archetype Markov State Probability

probability-basicsymmetry-argumentconditional-probabilityinvariant-monovariant systematic-enumerationpattern-recognition ↑ Prerequisites: probability-basicconditional-probability

📏 Medium solution 💡 3 insights

Problem

Three players (Raashan, Sylvia, Ted) start with 1 each. Every bell, everyone who still has money gives 1 to one of the other two players, chosen independently and uniformly at random. After 2019 bell rings, what is the probability all three again have exactly $1?

Pick an answer.

(A)

$\frac{1}{7}$

(B)

$\frac{1}{4}$

(C)

$\frac{1}{3}$

(D)

$\frac{1}{2}$

(E)

$\frac{2}{3}$

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

Tool #9 (Easier Problem): 2019 is intimidating; replace with 1 ring, then 2, then ask whether the answer depends on the ring count. Tool #2 (Systematic List): list all 2 × 2 × 2 = 8 outcomes from state (1,1,1) to see which return to (1,1,1). Tool #15 (Reorganize): group the 6 ordered (2,1,0)-type states into one bucket by symmetry. Tool #5 (Pattern): once we compute the transition probabilities, the answer is the same regardless of starting state — so it is the same for all n ≥ 1.

1STEP 1

Swap 2019 for one ring: from (1,1,1) each player passes to one of two neighbors around a circle, giving 8 equally likely outcomes.

outcomes from (1,1,1) = 2 × 2 × 2 = 8

2STEP 2

Only all passing the same circular direction keeps (1,1,1): 2 of 8, so return probability is $\frac{1}{4}$ and shifting to (2,1,0) is $\frac{3}{4}$ .

P((1,1,1) → (1,1,1)) =

\frac{2}{8}

\frac{1}{4}

3STEP 3

From a (2,1,0) start only two players give: 2×2 = 4 outcomes, and just one (A→B, B→C) restores (1,1,1) — probability $\frac{1}{4}$ .

P((2,1,0) → (1,1,1)) =

\frac{1}{4}

4STEP 4

Key pattern: from (1,1,1) or any (2,1,0)-type state, the next ring returns to (1,1,1) with the same probability $\frac{1}{4}$ .

P(next = (1,1,1) ∣ any current state) =

\frac{1}{4}

5STEP 5

Because the per-ring return probability is $\frac{1}{4}$ whatever the state now, it holds for rings 1 through 2019 — the count is irrelevant.

P(after 2019 rings) =

\frac{1}{4}

6STEP 6

The answer is (B) $\frac{1}{4}$ .

\frac{1}{4}

Answer

\frac{1}{4}

Spot check at n = 1: 2 of 8 outcomes return to (1,1,1), so P =

\frac{1}{4}

✓. At n = 2: starting from (1,1,1), we either stay at (1,1,1) with prob

\frac{1}{4}

(and then go to (1,1,1) with prob

\frac{1}{4}

) or move to a (2,1,0)-type state with prob

\frac{3}{4}

(and then go to (1,1,1) with prob

\frac{1}{4}

). Total:

\frac{1}{4}

\frac{1}{4}

\frac{3}{4}

\frac{1}{4}

\frac{1}{4}

✓. The answer choices include

\frac{1}{4}

and there is no n-dependence — fits AMC style of 'big number is decoy'.

💡Key takeaway

This AMC 10 problem only needs Grade 7 probability lists — list the 8 possible exchanges from (1,1,1) and the 4 from (2,1,0), see that each lands back in (1,1,1) with probability $\frac{1}{4}$ , so 2019 doesn't matter — the answer is $\frac{1}{4}$ .

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