2017 AMC 12B Problem 22

Question

Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn—one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?

\dfrac{7}{576}

\dfrac{5}{192}

\dfrac{1}{36}

\dfrac{5}{144}

\dfrac{7}{48}

Answer: B

Concepts:basic probability casework

Difficulty rating: 2330

Small hint:

Each round has

4 \cdot 3 = 12

equally likely (giver, receiver) outcomes, for

12^4

total

Big hint:

Count outcome sequences that return everyone to

4

coins: each player’s net transfers must cancel; split into cases by the pattern of exchanges

Solution:

Each round has

4 \cdot 3 = 12

equally likely (giver, receiver) pairs, so there are

12^4

outcome sequences. Everyone ends with four coins exactly when the four transfers cancel. The favorable patterns are: a

4

-cycle of gifts (

24 \cdot 6 = 144

ways), two disjoint mutual exchanges (

24 \cdot 3 = 72

), one pair exchanging twice (

6 \cdot 6 = 36

), and one player both giving to and receiving from each of two others (

4 \cdot 3 \cdot 24 = 288

). These total

144 + 72 + 36 + 288 = 540.

The probability is

\dfrac{540}{12^4} = \dfrac{540}{20736} = \dfrac{5}{192}.

Thus, the correct answer is B.