I've watched several videos explaining qubits but I can't yet understand why they are typically represented as a pair of probabilities.
The videos explain it's more accurate to understand them as vectors in spheres. But then, if a qubit "value" can be modelled as a vector in a sphere, wouldn't it be much more accurate to represent it with a latitude and longitude [on that sphere] instead?
Or... maybe neither of these abstractions model the real thing well?
As others have said, yes you can describe things using latitude and longitude. This is basically the Bloch vector.
The problem is that while this works great for just one qubit, once you move to trying to describe more than one qubit, it's missing something. Think, for example, about $n$ qubits. There are $2^n$ classical values that $n$ bits can take. For example, if $n=3$, there are 8 classical values that 3 bits can take: 000, 001, 010, 011, 100, 101, 110, 111. This means that $n$ qubits can be in a superposition of those $2^n$ different states, and we need $2^n$ different parameters to describe that (OK, you might argue you can reduce that by 2 since there's an irrelevant global phase and a normalisation). If you just use a description of $n$ Bloch vectors, you've got at most 3 parameters for each (latitude, longitude, length), and hence at most $3n$ parameters (there are also some constraints between the lengths that I'm neglecting for simplicity). As $n$ increases, there's a lot that you're missing from this description - entanglement! Without all those extra parameters, quantum loses all its interest in terms of computational ability.
We actually do use latitude and longitude to describe qubits, because one very common way of writing the state is of a qubit is
$$ \begin{align} \cos{(\theta/2)}|0\rangle + e^{i \phi}\sin{(\theta/2)} |1\rangle \end{align} $$
where $\theta$ is the angle formed by the Bloch vector and the $z$-axis (so technically it is the complement of the latitude, which is measured up from the equator instead of down from the North pole), and $\phi$ is the longitude with respect to the $x$-axis. This is the most general form for a pure quantum state, which you can think of as having only quantum uncertainty (outcome is undefined before measurement), but no classical uncertainty (outcome is well-defined but unknown to the observer). The "pair of probabilities" referred to in those videos is likely the probability that one obtains either $|0\rangle$ or $|1\rangle$ upon measurement - knowing these only determines the vector's projection onto the $z$-axis (i.e. its latitude), but does not determine its longitude, so those probabilities alone are not sufficient to specify a pure quantum state of the above form.
