2022 AMC 10B Problem 12

"At least once" is hard to count directly: it could be on the first roll, the third, or several. The opposite, "never", is easy. So work with "never".

1. One roll. Two dice give $6 \times 6 = 36$ equally likely results. The ones that sum to $7$ are $1{+}6$, $2{+}5$, $3{+}4$, $4{+}3$, $5{+}2$, $6{+}1$: six of them. So the chance of a $7$ is $\frac{6}{36} = \frac16$, and the chance of no $7$ is $\frac56$.
2. Several rolls. The rolls do not affect each other, so the chance of no $7$ in $n$ rolls is $\frac56$ multiplied by itself $n$ times, $\left(\frac56\right)^n$.
3. What we need. The chance of at least one $7$ is $1 - \left(\frac56\right)^n$. For this to be more than $\frac12$, the chance of "never" must be less than $\frac12$.
4. Try values of $n$.

To compare a fraction with $\frac12$, compare its top with half its bottom.
5. The first $n$ that works is $4$.

The answer is $\boxed{\textbf{(C)}\ 4}$.

For "at least once" questions, find the chance that it never happens and subtract from $1$. When the tries do not affect each other, "never" is one fraction multiplied by itself, so the question becomes a short table. Adding the chances ($\frac16 + \frac16 + \cdots$) is wrong because it counts rolls with two sevens twice; with enough rolls it would even pass $1$.

On a single roll of two dice there are $36$ outcomes, and six of them ($1{+}6, 2{+}5, 3{+}4, 4{+}3, 5{+}2, 6{+}1$) sum to $7$. So the probability of a $7$ is $\frac16$, and of no $7$ is $\frac56$.

The rolls are independent, so the probability of never seeing a $7$ in $n$ rolls is $\left(\frac56\right)^n$. We need
$$1 - \left(\tfrac56\right)^n > \tfrac12 \quad\Longleftrightarrow\quad \left(\tfrac56\right)^n < \tfrac12.$$

Compute: $\left(\frac56\right)^2 = \frac{25}{36} \approx 0.69$; $\left(\frac56\right)^3 = \frac{125}{216} \approx 0.58$; $\left(\frac56\right)^4 = \frac{625}{1296} \approx 0.48$.

The first power below $\frac12$ is $n = 4$.

The answer is $\boxed{\textbf{(C)}\ 4}$.