First let's count the number of $2$ in $21!$. This can be done by counting the factor of $2$ in all the even numbers less than $21$. The answer is $18$. This means that $21!$ can be factorized into the following (we do not need to determine $k_3$, $k_5$, etc): $$21=2^{18}3^{k_3}5^{k_5}\cdots$$

Accordingly, it has this number of divisors: $$(18+1)(k_3+1)(k_5+1)\cdots$$

Odd divisors cannot have any factor of $2$. Hence, the count is $$(k_3+1)(k_5+1)\cdots$$

Therefore, the answer is $$\frac{(k_3+1)(k_5+1)\cdots}{(18+1)(k_3+1)(k_5+1)\cdots}=\boxed{\frac{1}{19}}$$