Problem:
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?
Answer Choices:
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B.
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E.
Solution:
There are outcomes for each set of draws and therefore outcomes in all. To count the number of outcomes in which each player will end up with four coins, note that this can happen in four ways:
For some permutation of Abby, Bernardo, Carl, Debra , the outcomes of the four draws are that gives a coin to gives a coin to gives a coin to , and gives a coin to , in one of orders. There are 3 ways to choose whom Abby gives her coin to and 2 ways to choose whom that person gives his or her coin to, which makes 6 ways to choose the givers and receivers for these transaction. Therefore there are ways for this to happen.
One pair of the players exchange coins, and the other two players also exchange coins, in one of orders. There are 3 ways to choose the pairings. Therefore there are ways for this to happen.
Two of the players exchange coins twice. There are ways to choose those players and ways to choose the orders of the exchanges, for a total of ways for this to happen.
One of the players is involved in all four transactions, giving and receiving a coin from each of two others. There are 4 ways to choose this player, 3 ways to choose the other two players, and ways to choose the order in which the transactions will take place. Therefore there are ways for this to happen.
In all, there are outcomes that will result in each player having four coins. The requested probability is .
The problems on this page are the property of the MAA's American Mathematics Competitions