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2017 AMC 12B Problem 22

Problem 22 of 25HarderCounting & Probability

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?

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Solution

Each round has 4⋅3=124 \cdot 3 = 12 equally likely (giver, receiver) pairs, so there are 12412^4 outcome sequences. Everyone ends with four coins exactly when the four transfers cancel. The favorable patterns are: a 44-cycle of gifts (24⋅6=14424 \cdot 6 = 144 ways), two disjoint mutual exchanges (24⋅3=7224 \cdot 3 = 72), one pair exchanging twice (6⋅6=366 \cdot 6 = 36), and one player both giving to and receiving from each of two others (4⋅3⋅24=2884 \cdot 3 \cdot 24 = 288). These total 144+72+36+288=540.144 + 72 + 36 + 288 = 540. The probability is 540124=54020736=5192.\dfrac{540}{12^4} = \dfrac{540}{20736} = \dfrac{5}{192}. Thus, the correct answer is B.

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Concepts: basic probability · casework

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.