A pair of standard $6$-sided dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?
For the circumference to be greater than the area, we must have $\pi d > \pi \left( \frac{d}{2} \right) ^2$, or $d<4$. Now since $d$ is determined by a sum of two dice, the only possibilities for $d$ are thus $2$ and $3$. In order for two dice to sum to $2$, they most both show a value of $1$. The probability of this happening is $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$. In order for two dice to sum to $3$, one must show a $1$ and the other must show a $2$. Since this can happen in two ways, the probability of this event occurring is $2 \times \frac{1}{6} \times \frac{1}{6} = \frac{2}{36}$. The sum of these two probabilities now gives the final answer: $\frac{1}{36} + \frac{2}{36} = \frac{3}{36} = \frac{1}{12} \rightarrow \fbox{\frac{1}{12}}$