Question

Raashan, Sylvia, and Ted play the following game. Each starts with $\$1.$ A bell rings every $15$ 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 $\$1$ to that player. What is the probability that after the bell has rung $2019$ times, each player will have $\$1?$

(For example, Raashan and Ted may each decide to give $\$1$ to Sylvia, and Sylvia may decide to give her dollar to Ted, at which point Raashan will have $\$0,$ Sylvia will have $\$2,$ and Ted will have $\$1,$ 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 $\$1$ to, and the holdings will be the same at the end of the second round.)

$\dfrac{1}{7}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

Solution:

There are only two possible money configurations up to order: $(1,1,1)$ and $(2,1,0)$ . From $(1,1,1)$ , the next state is again $(1,1,1)$ exactly when all three players pass dollars in the same cyclic direction, which has probability $2\left(\dfrac12\right)^3=\dfrac14$ .

From $(2,1,0)$ , only the players with money give a dollar. The next state is $(1,1,1)$ exactly when the player with $2$ dollars gives to the player with $0$ dollars and the player with $1$ dollar gives to the player with $2$ dollars, also with probability $\dfrac14$ .

Therefore, regardless of the state after $2018$ rings, the probability that the state after the next ring is $(1,1,1)$ is $\dfrac14$ . Thus, B is the correct answer.

2019 AMC 10B Problem 22

Solution:

Problem 22 in Other Years