A die is rolled 4 times, what is the probability of getting at least 2 sixes? (order matters) I know the complementary probability would be getting 0 or 1 sixes
so... 
P(getting at least 2 sixes) = 1 - P(getting 0 or 1 sixes)
If the order matters though, I suppose the calculation is different?
 A: HINT
The idea is the same. for no sixes, you have $6^4$ total permutations, of them $5^4$ are without sixes, so you get $(5/6)^4$.
can you compute probability to get exactly 1 six?
A: Not sure what you mean by "order" in this situation.  In general there are far more possibilities with permutations (where order is involved).
I programmed a simulation where the computer completes your scenario 10 million times.  Here is the probability I got:
13.17651% chance of getting at least 2 sixes out of 4 rolls.
This is obviously not an exact probability, but its range of error is likely within 0.01% or so.
A: It doesn't make sense to say order matters for a question like this. Perhaps what you mean is that it's important to make sure we're not making any unwarranted assumptions about order. For example, we wouldn't want to naively assume that asking the probability that at least two rolls are sixes is equivalent to asking the probability that the first two rolls are sixes, which is $\frac1{36}$, or something like that.
You have a good start. The probability of getting no sixes is the probability that you won't get a six on any roll, so that's $(\frac56)^4$. To find the probability of getting exactly one six, you multiply the number of positions where you can get that one six (a six on the first, second, third, or fourth roll) by the probability of such a scenario happening. To get exactly one six, and have it happen on the second roll, that roll has to be a six, and three rolls have to be anything other than six. So the probability would be $(\frac56)^3\cdot\frac16$. Multiply that by 4, and you get $4\cdot\frac16\cdot(\frac56)^3$. Add these together, and you get $(\frac56)^4+4\cdot\frac16\cdot(\frac56)^3$. Subtract this from one as you said, and you have your answer. 
