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 has 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.)

$\dfrac{1}{7}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

Answer: B

Concepts:recursive probability symmetry

Difficulty rating: 1950

Solution:

The only reachable money configurations up to order are $(1,1,1)$ and $(2,1,0)$ . A player cannot finish a round with all $3$ dollars: anyone who begins with money must give a dollar to someone else, and no one can give to themselves. 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 $(2,1,0)$ , label the players' holdings $A=2,B=1,C=0$ . The next state is $(1,1,1)$ exactly when $A$ gives to $B$ and $B$ gives to $C$ , one of the four equally likely pairs of choices. This also has 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

Answer: B

Difficulty rating: 1950

Solution:

Problem 22 in Other Years