2019 AMC 10B Problems/Problem 22

The following problem is from both the 2019 AMC 10B #22 and 2019 AMC 12B #19, so both problems redirect to this page.

Problem

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 her dollar to Ted, at which point Raashan will have $$0$ , Sylvia will have $$2$ , and Ted will have $$1$ , and that is the end of the first round of play. In the second round Rashaan has no money to give, but Sylvia and Ted might choose each other to give their $$1$ to, and the holdings will be the same at the end of the second round.)

$\textbf{(A) } \frac{1}{7} \qquad\textbf{(B) } \frac{1}{4} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{1}{2} \qquad\textbf{(E) } \frac{2}{3}$

Solution 1

On the first turn, each player starts off with $\text{$1}$ . Each turn after that, there are only two possibilities: either everyone stays at $\text{$1}$ $\text{(1-1-1)}$ , or the distribution of money becomes $\text{$2-$1-$0}$ in some order, which we write as $\text{(2-1-0)}$ .

We also write $\text{S-T-R}$ (with letters, as distinct from e.g. $\text{(1-1-1)}$ ) which means that each person has $\text{$1}$ ) to mean that $\text{R}$ gives his money to $\text{S}$ , $\text{S}$ gives her money to $\text{T}$ , and $\text{T}$ gives his money to $\text{R}$ . We also write e.g. $\text{S-T-0}$ to mean that $\text{S}$ gives her money to $\text{T}$ , $\text{T}$ gives his money to $\text{R}$ , but $\text{R}$ has no money, so gives nothing to $\text{S}$ .

From the $\text{1-1-1}$ state, there are two ways to distribute the money so that it stays in a $\text{1-1-1}$ state: $\text{S-T-R}$ and $\text{T-R-S}$ . There are 6 ways to change the state to $\text{2-1-0}$ : $\text{S-R-R}$ , $\text{T-R-R}$ , $\text{S-R-S}$ , $\text{S-T-S}$ , $\text{T-T-R}$ , and $\text{T-T-S}$ . This means that the probability that the state stays $\text{1-1-1}$ is $\frac{2}{2+6}=\frac{1}{4}$ , and the probability that the state changes to $\text{2-1-0}$ is $\frac{6}{2+6}=\frac{3}{4}$ .

From the $\text{2-1-0}$ state, there is one way to change the state back to $\text{1-1-1}$ : $\text{S-T-0}$ . (We can assume that $\text{R}$ has $\text{$2}$ , $\text{S}$ has $\text{$1}$ , and $\text{T}$ has $\text{$0}$ since only the distribution of money matters, not the specific people.) There are three ways to keep the $\text{2-1-0}$ state: $\text{S-R-0}$ , $\text{T-R-0}$ , $\text{T-T-0}$ . This means that the probability that the state changes to $\text{1-1-1}$ is $\frac{1}{1+3} = \frac{1}{4}$ , and the probability that the state stays $\text{2-1-0}$ is $\frac{3}{1+3} = \frac{3}{4}$ .

We can see that there will always be a $\frac{1}{4}$ chance that the money is distributed $\text{1-1-1}$ (as long as the bell rings once), since the probabilities are the same in both cases, so the answer is $\boxed{\textbf{(B) }\frac{1}{4}}$ . Humpty Dumpty sat on a wall.

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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All AMC 10 Problems and Solutions

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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2019 AMC 10B Problems/Problem 22

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Solution 1

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