C&P

Now, since we can draw two of the same numbers and order doesn't matter, we don't do $9\choose 4$ to find the total amount of possibilites, but instead just do

$9 \cdot 9\cdot 9 \cdot 9 = 9^4 = 6561$

Now, we can use complementary counting to find the number of cases that DO NOT work and then subtract it from 1. 

This is because it's MUCH easier to calculate odd numbers rather than even numbers. 

Our first case is that all 4 numbers are odd. If all 4 numbers are odd, the product is odd, therefore meaning it cannot possibly be divisble by 4. 

Since out of the 9 numbers, 5 numbers are odd, we have a $5/9$ chance each generation that the number is odd. 

Since we need 4 odd numbers in a row, we simply have $({5 \over 9})^4 = {625 \over 6561}$

Our next case is where we get 3 odd numbers in a row and then a 2 or a 6. If we roll a 4 or 8, the product will automatically be divisble by 4. 

Thus, we have ${2 \times 5^3 \times {4 \choose 3} \over 6561} = {1000 \over 6561}$

Subtracting the sum from 1, we get the final answer of $1 - ({625 \over 6561} + {1000 \over 6561} ) = \color{brown}\boxed{4936 \over 6561}$

I would like verification of the final answer, but I believe this is the correct answer. 

Thanks! :)