A fair 6 sided dice is rolled 4 times. What is the probability that at least 3 of the numbers will be either 1 or 6?

I would use a binomial probability: $$ \begin{align*} P(\text{at least 3 are 1 or 6}) &= P(\text{exactly 3 are 1 or 6}) + P(\text{exactly 4 are 1 or 6}) \\ &= {}_4C_3 \left(\dfrac{2}{6}\right)^3\left(\dfrac{4}{6}\right)^1 + {}_4C_4 \left(\dfrac{2}{6}\right)^4\left(\dfrac{4}{6}\right)^0 \\ &= 4 \left(\dfrac{1}{3}\right)^3\left(\dfrac{2}{3}\right) + \left(\dfrac{1}{3}\right)^4 \\ &= \dfrac{8+1}{81} \\ &= \dfrac{1}{9} \\ \end{align*} $$

Your sample space of $1296$ can be replaced by a far smaller and more convenient sample space.

But as an exercise we carry out the count of how many of the ordered quadruples satisfy the requirement of at least three $1$ and/or $6$.

First we count the number of cases where every entry is one of $1$ or $6$. There are plenty of choices other than $1111$ and $6666$, like $6111$. The first toss can take on anyone of $2$ values, and for each such value, there are $2$ possibilities for the second toss, and so on for a total of $2^4$.

Next we count the number of cases where we have exactly three $1$ and/or $6$.

Where the oddball throw occurs can be chosen in $\binom{4}{1}$ ways. For each of these ways, the number on the oddball throw can be chosen in $4$ ways. And now the remaining three slots can be filled with $1$ or $6$ in $2^3$ ways, for a total of $128$.

Thus $16+128$ elements of our sample space of $1296$ are "favourable." Now we can write down the probability.