Problem 33

Note that the probability of Diana rolling a number larger than Apollo's is the same as the probability of Apollo's being more than Diana's. If we denote this common probability $D$, then $2D+P($Apollo=Diana$)=1$. Now all we need to do is find $P($Apollo=Diana$)$. There are $6(6)=36$ possibilities total, and 6 of those have Apollo=Diana, so $P($Apollo=Diana$)=\frac{6}{36}=\frac{1}{6}$. Going back to our first equation and solving for D, we get $$2D+\frac{1}{6}=1$$ $$2D=\frac{5}{6}$$ $$D=\frac{5}{12} \Rightarrow \mathrm{(B)}$$