2023 AMC 10A Problem 7

Every roll is at least $1$, so the running total reaches $3$ only within the first three rolls, and once it passes $3$ it never returns. The fourth roll is irrelevant.

The ways to reach a total of exactly $3$ are the ordered sequences of positive rolls summing to $3$:

• one roll: $3$, probability $\frac16$;
• two rolls: $1, 2$ or $2, 1$, probability $2 \cdot \frac{1}{36} = \frac{2}{36}$;
• three rolls: $1, 1, 1$, probability $\frac{1}{216}$.

These events are disjoint (a total of exactly $3$ occurs at a single moment). Over a common denominator:
$$\frac{36}{216} + \frac{12}{216} + \frac{1}{216} = \frac{49}{216}.$$

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

A running total is monotone, so "hits $3$ at some point" is the same as "some initial segment of rolls sums to exactly $3$." That converts a four-roll problem into listing the compositions of $3$, each weighted by $6^{-k}$ for its length $k$. Look for monotone quantities: they make "at some point" events into disjoint cases.

• Forgetting the order matters (counting 1+2 once instead of twice) or double-counting outcomes that hit 3 in two different ways, which is impossible.
• Using the fourth roll in the count, or normalizing over $6^4$ without multiplying each case by the correct number of continuations.