2023 AMC 10A Problem 7

1. Why only the first three rolls matter. Each roll is at least $1$, so the running total goes up every time. After three rolls it is at least $3$. If it has not hit $3$ by then, it has already jumped past $3$ and can never come back. So the fourth roll does not matter.
2. List every way to hit exactly 3. The rolls before the total reaches $3$ must add to exactly $3$. Order matters, because the rolls happen one at a time.
- One roll: roll a $3$ first.
- Two rolls: $1$ then $2$, or $2$ then $1$.
- Three rolls: $1$, $1$, $1$.
3. Find each probability. One particular roll has probability $\frac16$, two particular rolls $\frac{1}{36}$, three particular rolls $\frac{1}{216}$.
- Roll $3$ first: $\frac16 = \frac{36}{216}$.
- $1$ then $2$, or $2$ then $1$: $2 \times \frac{1}{36} = \frac{12}{216}$.
- $1, 1, 1$: $\frac{1}{216}$.
4. These cases never overlap. For example, if the first roll is $3$, the first two rolls cannot also add to $3$. So we may add the probabilities:
$$\frac{36}{216} + \frac{12}{216} + \frac{1}{216} = \frac{49}{216}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{49}{216}}$.

A running total that always goes up can pass each number at most once. So "the total equals $3$ at some point" means "the first few rolls add to exactly $3$," and there are only four such lists of rolls. Whenever a quantity only moves in one direction, "at some point" questions break into separate cases that do not overlap.