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2019 AMC 10B Problem 22

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2019 AMC 10B Problem 22AMC 10 Step by Step
Name:                             Date: September 13, 2026
Conditional Probability & States · difficulty 4 of 5 · about 7 min · position P21-25

Raashan, Sylvia, and Ted play the following game. Each starts with $1\$1 . A bell rings every 1515 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\$1 to that player. What is the probability that after the bell has rung 20192019 times, each player will have $1\$1 ? (For example, Raashan and Ted may each decide to give $1\$1 to Sylvia, and Sylvia may decide to give her dollar to Ted, at which point Raashan will have $0\$0 , Sylvia will have $2\$2 , and Ted will have $1\$1 , and that is the end of the first round of play. In the second round Raashan has no money to give, but Sylvia and Ted might choose each other to give their $1\$1 to, and the holdings will be the same at the end of the second round.)

  1. A)

    17\frac{1}{7}

  2. B)

    14\frac{1}{4}

  3. C)

    13\frac{1}{3}

  4. D)

    12\frac{1}{2}

  5. E)

    23\frac{2}{3}

Answer

B

Key insight

Only two kinds of holdings occur, (1,1,1) and a permutation of (2,1,0), and from either one the next state is (1,1,1) with probability exactly 1/4.

Solution

The total is always $3\$3. A player can gain at most $2\$2 in a round and must give away $1\$1 if they have any, so nobody ever reaches $3\$3. The only possible holdings are (1,1,1)(1,1,1) or some arrangement of (2,1,0)(2,1,0).

From (1,1,1)(1,1,1): each of the three players picks one of two recipients, 23=82^3 = 8 equally likely outcomes. Everyone ends with $1\$1 exactly when each player receives one dollar, i.e. the gifts form a cycle: R→S→T→RR \to S \to T \to R or R→T→S→RR \to T \to S \to R. That is 22 of 88, probability 14\frac14.

From (2,1,0)(2,1,0), say Raashan 22, Sylvia 11, Ted 00: only Raashan and Sylvia give, 2⋅2=42 \cdot 2 = 4 equally likely outcomes. For everyone to end with $1\$1, Ted must receive exactly one dollar and Raashan must receive none: Raashan gives to Sylvia and Sylvia gives to Ted. That is 11 of 44, probability 14\frac14.

So no matter what the holdings are before the final bell, the probability that the 20192019th ring produces (1,1,1)(1,1,1) is 14\frac14.

The answer is (B) 14\boxed{\textbf{(B)}\ \frac{1}{4}}.

Why this works

A process with many steps looks intimidating until you list its states; here there are effectively two. Because the transition probability into the target state is identical from every state, the distribution before the last step is irrelevant and the huge number 20192019 is a red herring. Always compute one-step transition probabilities first and look for such a coincidence.

The trap

Trying to track 2019 rounds or assuming the answer depends on the parity of 2019; the transition probability into (1,1,1) is the same from every state.

Common mistakes

  • Trying to track 2019 rounds or assuming the answer depends on the parity of 2019; the transition probability into (1,1,1) is the same from every state.
  • Miscounting the (1,1,1)(1,1,1) case as 26\frac{2}{6} or 38\frac{3}{8} by treating the three choices as a permutation instead of 232^3 independent choices.

Techniques

Split into exhaustive cases and handle each · Define states/recurrence and iterate