Problem (Combinatorics):
A seventeen-sided die has faces numbered 1 through 17, but it is not fair: 17 comes up with probability $1 / 2$, and each of the numbers 1 through 16 comes up with probability $1 / 32$. Compute the probability that the sum of two rolls is either 20 or 12.

Solution:
The rolls that add up to 20 are $17+3,16+4,15+5,14+6,13+7,12+8,11+9$, and $10+10$. Accounting for order, the probability of $17+3$ is $\frac{1}{2} \cdot \frac{1}{32}+\frac{1}{32} \cdot \frac{1}{2}=2 \cdot \frac{1}{2} \cdot \frac{1}{32}=\frac{32}{1024}$. The combination $10+10$ has probability $\frac{1}{32} \cdot \frac{1}{32}=\frac{1}{1024}$; the other six combinations have probability $2 \cdot \frac{1}{32} \cdot \frac{1}{32}=\frac{2}{1024}$, for a total of $\frac{32+1+6 \cdot 2}{1024}=\frac{45}{1024}$ (again, accounting for two possible orders per combination). The rolls that add up to 12 are $1+11,2+10,3+9,4+8,5+7,6+6$, all



of which have probability $2 \cdot \frac{1}{32} \cdot \frac{1}{32}=\frac{2}{1024}$ except the last, which has probability $\left(\frac{1}{32}\right)^{2}$, for a total of $\frac{11}{1024}$. Thus the probability of either sum appearing is $\frac{45}{1024}+\frac{11}{1024}=\frac{56}{1024}=\frac{\mathbf{7}}{\mathbf{1 2 8}}$.

Final answer: $\frac{7}{128}$