There are \(6^4 = 1296\) ways to roll the dice.
Use stars and bars to count the number of ways to roll a 12. Each roll must be at least 1, so we subtract \(4 \times 1 = 4\) to get 8 stars and \(4 - 1 = 3\) bars.
This makes for \({8 + 3 \choose 3} = {11 \choose 3} = 165\) ways to roll a 12. But, we can't have cases like 6, 1, 1, 0; 6, 2, 0, 0; 7, 1, 0, 0; and 8, 0, 0, 0.
There are \({4! \over 2!} = 12 \) ways for each of the first 3, and there are \({4! \over 3!} = 4 \) ways for the last one, which sums to 40.
So, there are 165 - 40 = 125 ways to roll a 12, which makes for a probability of \(\color{brown}\boxed{125 \over 1296}\)