AMC 10 · 2019 · #22

Grade 7 probability

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

📏 Medium solution 💡 3 insights

Problem

Raashan, Sylvia, and Ted play the following game. Each starts with $1$. A bell rings every $15$ seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives$ 1 $to that player. What is the probability that after the bell has rung$ 2019 $times, each player will have$ $1$ ? (For example, Raashan and Ted may each decide to give $1$ to Sylvia, and Sylvia may decide to give her dollar to Ted, at which point Raashan will have$ 0 $, Sylvia will have$ $2$ , and Ted will have $1$, and that is the end of the first round of play. In the second round Rashaan has no money to give, but Sylvia and Ted might choose each other to give their$ 1$ to, and the holdings will be the same at the end of the second round.)

$\textbf{(A) } \frac{1}{7} \qquad\textbf{(B) } \frac{1}{4} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{1}{2} \qquad\textbf{(E) } \frac{2}{3}$

Pick an answer.

(A)

$\frac{1}{7}$

(B)

$\frac{1}{4}$

(C)

$\frac{1}{3}$

(D)

$\frac{1}{2}$

(E)

$\frac{2}{3}$

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Toolkit + CCSS Solution

Understand

Restated: 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?

Givens: Three players, total $3 always conserved; Each player with money picks a recipient uniformly from the other two (independently); Number of rings = 2019; Answer choices: (A) $\tfrac{1}{7}$, (B) $\tfrac{1}{4}$, (C) $\tfrac{1}{3}$, (D) $\tfrac{1}{2}$, (E) $\tfrac{2}{3}$

Unknowns: Probability that after 2019 rings the state is again $(1, 1, 1)$

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #2 Make a Systematic List, #5 Look for a Pattern, #15 Reorganize Information, #3 Eliminate Possibilities

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 $2 \times 2 \times 2 = 8$ 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 $n \ge 1$.

Execute — Answer: B

#2 Make a Systematic List 7.SP.C.8 Step 1

Replace 2019 with the easier problem of one ring.
From state $(1,1,1)$, list all outcomes: each of the 3 players picks one of 2 recipients, giving $2^3 = 8$ equally likely outcomes.
Label players $R, S, T$ around a circle.
Each can pass clockwise (CW) or counter-clockwise (CCW).

$$\text{outcomes from } (1,1,1) = 2 \times 2 \times 2 = 8$$

💡 Just enumerate the 8 simultaneous choices and see which return to $(1,1,1)$.

#2 Make a Systematic List 7.SP.C.8 Step 2

List which of the 8 outcomes keep state $(1,1,1)$: only when ALL three pass the same direction around the circle (all CW or all CCW).
That's 2 outcomes.
So $P((1,1,1) \to (1,1,1)) = \tfrac{2}{8} = \tfrac{1}{4}$, and $P((1,1,1) \to (2,1,0)\text{-type}) = \tfrac{6}{8} = \tfrac{3}{4}$.

$$P((1,1,1) \to (1,1,1)) = \tfrac{2}{8} = \tfrac{1}{4}$$

💡 Each player must pass in the same circular direction — only 2 of the 8 patterns do.

#15 Reorganize Information 7.SP.C.8 Step 3

Now the harder subcase: starting in a $(2,1,0)$ state.
Only two players have money — the one with $2 and the one with $1.
Each picks one of 2 recipients: $2 \times 2 = 4$ outcomes.
Label A = player with $2, B = player with $1, C = player with $0. A's choice: B or C. B's choice: A or C. Enumerate the 4 results in (A,B,C) order: (A→B, B→A): A loses 2, gains 1; B loses 1, gains 2; C gains 0 → state $(1, 2, 0)$, still type $(2,1,0)$. (A→B, B→C): A loses 2, gains 0; B loses 1, gains 2; C gains 1 → $(0, 2, 1)$, type $(2,1,0)$. (A→C, B→A): $(1, 0, 2)$, type $(2,1,0)$. (A→C, B→C): A=0, B=0, C=3? Wait: A loses $2 of his own ($2), B loses $1; recipients are both C.
A gives $1 (not $2) — A only owns 2 dollars but gives away ONE dollar.
Let me recount: A starts with $2 and gives $1 away.
B starts with $1 and gives $1 away.
(A→C, B→C): A keeps $1, B keeps $0, C gets $1 (from A) + $1 (from B) = $2. State $(1, 0, 2)$, type $(2,1,0)$. Wait recount (A→B, B→A): A→B means A gives B his $1 (A keeps $1), B→A means B gives A his $1 (B keeps $0). A ends with $1 + $1 = $2; B ends with $0 + $1 = $1; C ends $0.
State $(2, 1, 0)$ — same as start, type $(2,1,0)$.
Re-enumerate: out of 4 outcomes, exactly 1 lands in $(1,1,1)$ — namely (A→B, B→C): A keeps $1, gives $1 to B; B gives $1 to C, keeps $0; result A=$1, B=$1+$1−$1=$1, C=$0+$1=$1 → $(1,1,1)$ ✓.

$$P((2,1,0) \to (1,1,1)) = \tfrac{1}{4}$$

💡 From $(2,1,0)$ only one of 4 simultaneous choices ($A \to B, B \to C$) restores equal shares.

#5 Look for a Pattern 7.SP.C.7 Step 4

Key pattern: from BOTH state $(1,1,1)$ AND any $(2,1,0)$-type state, the probability of being in $(1,1,1)$ after the next ring is exactly $\tfrac{1}{4}$.

$$P(\text{next} = (1,1,1) \mid \text{any current state}) = \tfrac{1}{4}$$

💡 Whatever state we are in just before a ring, the chance of $(1,1,1)$ immediately after is $\tfrac{1}{4}$.

#9 Solve an Easier Related Problem 7.SP.C.7 Step 5

Easier-problem conclusion: since the probability of state $(1,1,1)$ after the next ring is $\tfrac{1}{4}$ regardless of which state we are in now, this holds for ring 1, ring 2, ..., ring 2019.
So $P(\text{state} = (1,1,1) \text{ after } 2019 \text{ rings}) = \tfrac{1}{4}$.

$$P(\text{after 2019 rings}) = \tfrac{1}{4}$$

💡 Number of rings is a red herring — the per-step return probability is always $\tfrac{1}{4}$.

#3 Eliminate Possibilities 7.SP.C.7 Step 6

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

$$\boxed{\tfrac{1}{4}}$$

💡 Match to the answer choice.

[1] #2 7.SP.C.8 Replace 2019 with the easier problem of one ring. From state $(1,1,1)$, list all

[2] #2 7.SP.C.8 List which of the 8 outcomes keep state $(1,1,1)$: only when ALL three pass the

[3] #15 7.SP.C.8 Now the harder subcase: starting in a $(2,1,0)$ state. Only two players have mon

[4] #5 7.SP.C.7 Key pattern: from BOTH state $(1,1,1)$ AND any $(2,1,0)$-type state, the probabi

[5] #9 7.SP.C.7 Easier-problem conclusion: since the probability of state $(1,1,1)$ after the ne

[6] #3 7.SP.C.7 The answer is (B) $\tfrac{1}{4}$.

Review

Reasonableness: Spot check at $n = 1$: 2 of 8 outcomes return to $(1,1,1)$, so $P = \tfrac{1}{4}$ ✓. At $n = 2$: starting from $(1,1,1)$, we either stay at $(1,1,1)$ with prob $\tfrac{1}{4}$ (and then go to $(1,1,1)$ with prob $\tfrac{1}{4}$) or move to a $(2,1,0)$-type state with prob $\tfrac{3}{4}$ (and then go to $(1,1,1)$ with prob $\tfrac{1}{4}$). Total: $\tfrac{1}{4} \cdot \tfrac{1}{4} + \tfrac{3}{4} \cdot \tfrac{1}{4} = \tfrac{1}{4}$ ✓. The answer choices include $\tfrac{1}{4}$ and there is no $n$-dependence — fits AMC style of 'big number is decoy'.

Alternative: Tool #11 (Work Backwards) via Markov chain steady-state: let $p_n = P(\text{state at step } n = (1,1,1))$. The recursion is $p_{n+1} = \tfrac{1}{4} p_n + \tfrac{1}{4}(1 - p_n) = \tfrac{1}{4}$, so $p_n = \tfrac{1}{4}$ for all $n \ge 1$. Same answer.

CCSS standards used (min grade 7)

7.SP.C.7 Develop probability models and use them to find probabilities of events (Modeling each random transfer choice as a uniform pick, and combining transition probabilities.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Enumerating the $2^3 = 8$ outcomes from $(1,1,1)$ and the $2 \times 2 = 4$ outcomes from $(2,1,0)$.)

⭐ 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 $\tfrac{1}{4}$, so 2019 doesn't matter — the answer is $\tfrac{1}{4}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 7)

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