I will illustrate my question with an example. Suppose we're trying to solve this problem:
Bob possesses a coin which has a probability of $\dfrac 14$ of turning up $T$ and a probability of $\dfrac 34$ of turning up $H$, and the flips are independent of each other. Bob performs the experiment of the flipping the coin until the first head shows up, and then stops and records the number of flips he did. What is the probability that it took Bob $3$ flips to get the first head?
This is the way I've been taught to do this sort of problem: The sample space is $\{H, TH, TTH, TTTH, ... \}$. We want to find $$P[(\text{first flip, T})\cap (\text{second flip, T})\cap(\text{third flip, H}) ]$$
$$= P[(\text{first flip, T})]\cdot P[(\text{second flip, T})] \cdot P[(\text{third flip, H})]$$
$$= \dfrac 14 \cdot \dfrac 14 \cdot \dfrac34$$
My question is, is the probability (for example) $P[(\text{second flip, T})]$ really equal to $\dfrac 14$? What if the experiment ends on the first flip? Then I would think $P[(\text{second flip, T})]=0$ or undefined. So instead of calculating
$$P[(\text{first flip, T})\cap (\text{second flip, T})\cap(\text{third flip, H}) ]$$
shouldn't we be calculating something like
$$P[((\text{first flip, T})\cap(\text{experiment lasts at least 2 flips }))$$
$$\cap ((\text{second flip, T})\cap(\text{experiment lasts at least 3 flips }))$$
$$ \cap((\text{third flip, H})\cap (\text{experiment lasts at least 3 flips}))]$$
?
Conceptually, I think it is easier and clearer to view this as a tree.
The point is that after a head, and after the third tail in a row, it does not matter what happens next. Your calculation is not going to be affected.
If you look at the nodes, there are 4 of relevance (as you noted), with probabilities of 3/4, 3/16, 3/64, 1/64. Obviously, these add up to 1, because one of these must occur. Areas that are further down the tree are not going to matter, because the probability at each node is the sum of the probability of all the nodes further down the tree from there.
Aside: This is too easy to solve with Markov chains, but Markov chains would be the right way to solve the harder versions of it...