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2019 AMC 12B Problem 19

Problem 19 of 25HarderCounting & Probability

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.)

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Solution

From (1,1,1),(1,1,1), each of the three players gives to one of two others, so there are 88 equally likely outcomes; only the 22 cyclic gift patterns return to (1,1,1),(1,1,1), a probability of 14.\dfrac14. From a (2,1,0)(2,1,0) state the broke player gives nothing, and checking the 44 equally likely choices of the other two shows exactly one yields (1,1,1),(1,1,1), again probability 14.\dfrac14. So after any ring the probability of (1,1,1)(1,1,1) is 14,\dfrac14, including after 20192019 rings. Thus, B is the correct answer.

More practice

Concepts: recursive probability · symmetry · invariant

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