Why is probability sometimes calculated using ordered pairs of outcomes rather than unordered pairs? For example, if we are tossing two coins, where each coin falls on either heads ($H$) or tails ($T$), we have the following possible outcomes: $\{H, H \}$, $\{H, T \}$, $\{T, T \}$.
However, when solving some exercises, I noticed that this is not the way to go when looking for the number of possible outcomes in probability. Rather, we conceive of the outcomes of coin tosses as ordered pairs. In this case, we have $(H,H)$, $(H,T)$, $(T,H)$, $(T,T)$.
I don't understand why the first approach is wrong and the rationale of why we need two ordered pairs $(H,T)$  and $(T,H)$ rather than just $\{H,T\}$. Therefore, I don't really understand what probability is about either. Please, help me understand.
 A: It helps to look at the probability spaces that each generates:
When working with unordered pairs:
$$\begin{array}{c|c}\text{Outcome} & \text{Probability} \\ \hline \{H,H\} & 0.25 \\ \{H,T\} & 0.5 \\ \{T,T\} & 0.25\end{array}$$
Note how different outcomes have different probabilities. On the other hand, working with ordered pairs:
$$\begin{array}{c|c}\text{Outcome} & \text{Probability} \\ \hline (H,H) & 0.25 \\ (H,T) & 0.25 \\ (T,H) & 0.25 \\ (T,T) & 0.25\end{array}$$
Sample spaces where every outcome is equally probable is called an Equiprobable space. In general, given the choice between two probability spaces, Equiprobable spaces are often (but not always) easier to work with, even though they have some redundant information. But, both probability spaces are valid ways to record the results flipping two fair coins, and in some contexts, you may prefer the probability space of unordered pairs rather than the Equiprobable space.
An example of when you must use a non-equiprobable space is when you are conducting Bernoulli trials. You have a single sample space with two outcomes that are not equiprobable. Now, using unordered pairs tends to be as common (or possibly even more common) than ordered pairs. If the probability of flipping heads on an unfair coin is $0<p<1$, then over the course of ten flips, you wind up with the probability space:
$$\begin{array}{c|c}\text{Outcome} & \text{Probability} \\ \hline \text{10 heads} & \dbinom{10}{0}p^{10}(1-p)^0 \\ \text{9 heads, 1 tail} & \dbinom{10}{1}p^9(1-p)^1 \\ \text{8 heads, 2 tails} & \dbinom{10}{2}p^8(1-p)^2 \\ \vdots & \vdots \\ k\text{ heads, }10-k\text{ tails} & \dbinom{10}{k}p^k(1-p)^{10-k}\end{array}$$
