Yes, the events are independent, and every roll has a chance of $\frac{1}{6}$ to be a $6$. But that does not mean that getting a $6$ in $12$ rolls is also $\frac{1}{6}$.
Basically you are saying that it doesn't matter whether I roll the die $1$ time or a billion times, the probability of getting a $6$ is always $\frac{1}{6}$. But if you think about it, not getting a $6$ in $1$ roll is quite plausible, but not getting a $6$ in a billion rolls is outrageously unlikely. So, this should tell you that there is something wrong in your thinking about this.
OK, but what exactly is going wrong? Well, not getting a $6$ in 12 rolls means more events need to take place than just not getting a $6$ in 1 roll, and the more events need to take place, the less likely it becomes for all those events to take place, even if the probability for each of those events is still the same. So that is possibly where your confusion lies: you didn't distinguish between the probability of all of those events to take place and the probability of each of those events to take place.
Finally, having hopefully cleared up your confusion, let's explain the formula. Well, as I said, in order to not get any $6$ in $12$ rolls, $12$ events need to happen: $12$ times you need to not roll a $6$, and each of those individual events has a probability of $\frac{5}{6}$. So that means that there is a $(\frac{5}{6})^{12}$ probabiity for the overall even of not getting a $6$ at all to take place. Meaning that the chance of getting at least one $6$ is $1-(\frac{5}{6})^{12}$.
You don't always need a rigorous method, but since $144$ is a comfortable number, and there are only $4$ dice, I think it is possible to create a rigorous method without using a computer.
Notice that $5$ cannot be part of any solution - it is not divisible by $144$. We continue by separating this problem into cases: where one case has the first number be $1$, then another with $2$, $3$ and so on.
Case 1:
Since the first digit is $1$, then we have to factor $144$ in $3$ numbers. If the next digit is $2$ or $3$, then observe that the most extreme case (two $6$'s) only gets $36$, less than $72$ or $48$. However, when the next digit is $4$, we find the only other possibility $(1,4,6,6)$. Trying $6$ gives the same combination.
Case 2:
When the first digit is $2$, we need to factor $72$ in $3$ numbers. The next digit cannot be $1$, but when it is $2$ or $3$, we get the solutions $(2,2,6,6)$, and $(2,3,4,6)$. When we try $4$ or $6$, again we get the same combinations.
Case 3:
Now, with the first digit being $3$, we need to factor $48$ in $3$ numbers. $1$ doesn't work, $2$ gives the same solution $(2,3,4,6)$, but $3$ gives a new solution - $(3,3,4,4)$. We can check that $4$ and $6$ give the same solutions.
When the first digit is $4$ or $6$, there are no new solutions.
This shows that $(1,6,4,4)$; $(2,2,6,6)$; $(2,3,4,6)$; and $(3,3,4,4)$ are the only solutions, and we can use your method to calculate the total number of rearrangements.
Best Answer
You are rolling four dice, and you want to know the chance of at least one three. Because it is at least one there are a few ways this could happen- namely there could be one three, two threes, three threes or four threes. Now, we can work all of these out using something called the binomial coefficients (technically speaking the number of heads is a binomial random variable) - the formula is the probability of a three $(1/6)$ raised to the power of how many, times by the probability of not a three (5/6) times by how many aren't threes, and then multiplied by the number of ways to select the dice (again, binomial coefficients). So the chance of one three is $4*(1/6)(5/6)^3=500/1296$. The chance of two threes is $6*(1/6)^2*(5/6)^2=150/1296$. The chance of three threes is $4*(1/6)^3*(5/6)=20/1296$ and the chance of four threes is $1*(1/6)^4*(5/6)^0=1/1296$.
If we add these all together, we get $(500+150+20+1)/1296=671/1296$ which is the the same as $1-(5/6)^4$ as expected. This is the formal way of working it out, but the short cut is the complement. The way I target these questions is to work out which has the least leg work. You are asked for the chance of at least one which includes the options 1,2,3,4, where as the complement is just 0. If you were asked for the chance of at least two, this is 2,3,4 and the complement is 0 or 1 (which is easier to work out). It is only when you're asked for 3 or more, that it's easier the formal way.
In general, if $X$ is the number of something with maximum value $Y$ (number of heads in Y coin tosses, threes on a dice roll, etc) and you want to calculate the probability that $X>n$ (written $P(X>n)$) it easier to use the complement if n