2019 AMC 10B Problem 22

$2019$ rounds is far too many to follow. So first ask which situations can even occur, and what happens from each on a single ring.

Step 1: the possible situations.

1. There is always $\$3$ in total.
2. Could one player have all $\$3$? A player with money must give $\$1$ away, and each of the other two can give at most $\$1$. If the others each give $\$1$ to one player, that player also gives $\$1$ away, so they end with at most $\$2$. Nobody ever reaches $\$3$.
3. So the money is always split $1, 1, 1$ or $2, 1, 0$ (in some order).

Step 2: from $1, 1, 1$.

1. Each of the three players picks one of two others: $2 \cdot 2 \cdot 2 = 8$ equally likely ways.
2. Everyone keeps $\$1$ only if each player receives exactly one dollar. That happens when the gifts go around in a circle: Raashan to Sylvia to Ted to Raashan, or the reverse. That is $2$ of the $8$ ways.
3. Probability: $\frac{2}{8} = \frac14$.

Step 3: from $2, 1, 0$. Say Raashan has $\$2$, Sylvia $\$1$, Ted $\$0$.

1. Only Raashan and Sylvia give, each choosing one of two players: $4$ equally likely ways.
2. To end at $1, 1, 1$, Raashan must receive nothing and Ted must receive exactly $\$1$. So Sylvia must give to Ted, and Raashan must give to Sylvia. That is $1$ of the $4$ ways.
3. Probability: $\frac14$.

Step 4: conclude. Whatever the situation just before the $2019$th ring, the chance that ring leaves everyone with $\$1$ is $\frac14$. So the answer is $\frac14$.

The answer is $\boxed{\textbf{(B)}\ \frac{1}{4}}$.

A long random process becomes simple when you list its possible situations; here there are only two kinds. Then look at a single step from each one. Because both give the same chance of reaching $1, 1, 1$, the history before the last ring does not matter, and the number $2019$ is only there to scare you.

The total is always $\$3$. A player can gain at most $\$2$ in a round and must give away $\$1$ if they have any, so nobody ever reaches $\$3$. The only possible holdings are $(1,1,1)$ or some arrangement of $(2,1,0)$.

From $(1,1,1)$: each of the three players picks one of two recipients, $2^3 = 8$ equally likely outcomes. Everyone ends with $\$1$ exactly when each player receives one dollar, i.e. the gifts form a cycle: $R \to S \to T \to R$ or $R \to T \to S \to R$. That is $2$ of $8$, probability $\frac14$.

From $(2,1,0)$, say Raashan $2$, Sylvia $1$, Ted $0$: only Raashan and Sylvia give, $2 \cdot 2 = 4$ equally likely outcomes. For everyone to end with $\$1$, Ted must receive exactly one dollar and Raashan must receive none: Raashan gives to Sylvia and Sylvia gives to Ted. That is $1$ of $4$, probability $\frac14$.

So no matter what the holdings are before the final bell, the probability that the $2019$th ring produces $(1,1,1)$ is $\frac14$.

The answer is $\boxed{\textbf{(B)}\ \frac{1}{4}}$.