The following is another approach to the "at least one $6$" problem, grossly less efficient than the approach through first finding the probability of the complement. But it introduces ideas that will be very important later, so it is worth thinking about.
We will be happy if we get at least one $6$. What is the probability that we will be happy? Well, we will be happy if we get exactly $1$ $6$. We will also be happy if we get exactly $2$ $6$'s. Also if we get exactly $3$ $6$'s. And also with exactly $4$ $6$'s.
These events are (pairwise) disjoint. For example, it is impossible to get exactly $2$ $6$'s and (simultaneously) also exactly $4$ $6$'s.
Now I will introduce a bit of notation that looks initially mysterious, and that is not really needed here, but will become important to you later. Let $X$ be the number of $6$'s that we get. Then, for example, $P(X=2)$ means the probability that we get exactly $2$ $6$'s.
We will be happy if $X=1$, also if $X=2$, and so on up to if $X=4$. Since these events are pairwise disjoint, the probability we will be happy is $P(X=1)+P(X=2)+P(X=3)+P(X=4).$
So "all" we need to do is to calculate the probabilities of these four events, and add up!
Let us, for instance, find $P(X=2)$. Think about about our $4$ tosses. Write Y (for yes) if on a toss we get a $6$, and N is we don't. So for example NYYN means we got a non-$6$, then a $6$, then a $6$, then a non-$6$. Note that on this sequence of tosses, the total number of $6$ is $2$.
The probability of NYYN is is $(5/6)(1/6)(1/6)(5/6)$. There are several other patterns in which we end up with exactly $2$ $6$'s. For example, there is YYNN, YNYN, some others. The probability of YYNN is $(1/6)(1/6)(5/6)(5/6)$, which is exactly the same as the probability of NYYN. In fact the probability of any single pattern with exactly $2$ $6$'s is $(1/6)^2(5/6)^2$.
So to find $P(X=2)$, we find the number of patterns with exactly $2$ $6$'s, and multiply that by $(1/6)^2(5/6)^2$. The number of patterns is the number of ways of choosing where the two Y will go.
There are $\dbinom{4}{2}$ such ways (I assume you have already met this). So $P(X=2)=\binom{4}{2}\left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^2.$
We can do a similar analysis for exactly $1$ $6$, exactly $3$, exactly $4$. The probability we will be happy is therefore $\binom{4}{1}\left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^3+ \binom{4}{2}\left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^2+ \binom{4}{3}\left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^1+ \binom{4}{4}\left(\frac{1}{6}\right)^4 \left(\frac{5}{6}\right)^0.$
Now calculate! Really, I almost mean it. Getting the right answer, which you already know from the "quick" method, is a useful test of accuracy and attention to detail.
Of course the above was an inefficient way of calculating the probability. But the same set of ideas shows, for example, that if we toss the die $20$ times, the probability of exactly $4$ $6$'s is $\dbinom{20}{4}(1/6)^4(5/6)^{16}$. The general idea has many applications, and not only to gambling!