Probability of A

If the unknown letter is an $A$ then there are $\frac{50!}{18!*9!*10!*13!}$ ways of having $18 A$ s, $9 B$ s, $10 C$ s and $13 D$ s. However, we could also have $17 A$ s, $10 B$ s, $10 C$ s and $13 D$ s or $17 A$ s, $9 B$ s, $11 C$ s and $13 D$ s or $17 A$ s, $9 B$ s, $10 C$ s and $14 D$ s. As a result, the overall probability of having $18 A$ s =

$\frac{50!}{18!*9!*10!*13!}$ $\div$ $($ $\frac{50!}{18!*9!*10!*13!}$ + $\frac{50!}{17!*10!*10!*13!}$ + $\frac{50!}{17!*9!*11!*13!}$ + $\frac{50!}{17!*9!*10!*14!}$ $)$ $=$

$\frac{50!}{18!*9!*10!*13!}$ $\div$ $\frac{8812*50!}{18!*10!*11!*14!}$

$=$ $\frac{10*11*14}{8812}$ $= \boxed{0.17}$ ( $2$ decimal places ).