No. 11!

Let us first consider the chances of calvin winning it : The probability of getting a six is $\frac{1}{6}$ and that of getting another no. is $\frac{5}{6}$ $\quad \quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad$ . Chances of Calvin winning on the first throw : $\frac{1}{6}$
$$ Second throw : $\frac{5}{6}*\frac{5}{6}*\frac{1}{6}$ (Aaron must also not get a six) $\quad\quad\quad\quad\quad\quad\quad\quad\quad$ Third throw : $\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}$ and so on. Hence chances of Calvin winning are $\frac{1}{6}+\frac{5^{2}}{6^{3}}+\frac{5^{4}}{6^{5}}+...=\frac{\frac{1}{6}}{1-\frac{5^{2}}{6^{2}}}=\frac{6}{11}$ . Now chances of Aaron winning =1- $\frac{6}{11}=\frac{5}{11}$ . Hence our answer is $6!-5!=600$