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?

$\dfrac{7}{576}$

$\dfrac{5}{192}$

$\dfrac{1}{36}$

$\dfrac{5}{144}$

$\dfrac{7}{48}$

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.

2017 AMC 12B Problem 22

Solution:

Problem 22 in Other Years