If we roll a die three times, the probability of the maximum value being equal to $x$ is equal to the probability that all three die rolls are less than or equal to $x$, minus the probability that all die rolls are less than $x$ $$ P(\text{max} = x) = P(X_1 \leq x) P(X_2 \leq x) P(X_3 \leq x)\\ \hspace{20mm}- P(X_1 < x)P(X_2 < x)P(X_3 < x). $$ It is straightforward to see that $$ P(X \leq x) = \frac{x}{6} $$ and $$ P(X < x) = \frac{x-1}{6}. $$ Therefore, $$ P(\max = x) = \left( \frac{x}{6} \right)^3 - \left( \frac{x-1}{6} \right)^3. $$ If we set $\max = 6$, then we see that $$ P(\max = 6) = 1 - \left( \frac{5}{6} \right)^3 = 1 - \frac{125}{216} = \frac{91}{216} = 42.1\%. $$