I can understand the intuition behind a two dimensional bloch circle, as it represents the probability distribution of a certain state vector. However, I fail to grasp what the third dimension adds to the visualization when it comes to probabilities. How does it help? What is it supposed to add to the model that the Bloch circle didn't?
Since length is always 1, then what's the point of adding a z axis?
A qubit is a two-level quantum system and hence it can be written as:
$$ |\psi \rangle = \alpha |0\rangle + \beta|1\rangle $$
where $|0 \rangle$ and $|1\rangle$ are the computational basis and they defined as $$ |0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \ \ \ \ |1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$
and $\alpha, \beta \in \mathbb{C}$. So we have that $|\psi \rangle \in \mathbb{C}^2 $ and you can think of $\mathbb{C}^2$ as $\mathbb{R}^4$. However, the overall phase in the quantum state doesn't matter, that is, $|\psi\rangle $ is equivalent to $e^{i\phi} |\psi \rangle$. Thus, you can factor this overall phase out of $|\psi\rangle$ and have only 3 parameters left. Essentially now you are considering the space $\mathbb{R}^3$. Along with the normalization requirement, that a quantum state must have unit norm, this forces it to be a unit-sphere, which we call the Bloch-Sphere.
You're not trying to visualise probabilities. You're trying to visualise the much richer structure afforded by probability amplitudes. For a single qubit, the probability amplitudes require 3 real parameters, hence three dimensional space.
The Bloch sphere represents state with the following probability amplitudes: $$|\psi\rangle = \cos(\theta/2)\ |0\rangle +e^{i\phi} \sin(\theta/2)\ |1 \rangle, $$ where $\theta$ and $\phi$ are the angles on the Bloch sphere. Each point on the sphere represents a different state. As mentioned by other answers these $\cos(\theta/2) $ and $e^{i\phi} \sin(\theta/2) $ are not probabilities of the state being in the state $0$ or $1$, rather they represents the probability amplitudes. To get a probability form a probability amplitude, we square the probability amplitudes. That is, $\cos^2(\theta/2) $ and $\sin^2(\theta/2) $ are the probabilities, and you see they add up to $1$.
You also notice that the angle $\phi$ does not show up in the expressions for probabilities. Does this mean that we can just ignore $\phi$ all together? No. If we could then indeed we wouldn't need a 2-dimensional sphere to represents the states.
The reason we cannot ignore the angle $\phi$ from our considerations is because of the superposition principle of quantum mechanics. Suppose we have, $$|\psi_1\rangle = \frac{1}{\sqrt{2}}( |0\rangle + |1 \rangle), \ \ \ \ \ \ \ \ \ |\psi_2\rangle = \frac{1}{\sqrt{2}}( |0\rangle +e^{i\phi} \ |1 \rangle.$$ Then the superposition principle tells us that we can make a third state out of their superposition, $$|\psi_3\rangle = N(|\psi_1\rangle + |\psi_2\rangle) = \sqrt{\frac{2}{3+\cos(\phi)}}\left(|0\rangle + \frac{1+e^{i\phi}}{2} |1\rangle \right) ,$$ where $N$ is some normalization factor written down explicitly in the second equality. Anyway, the point here is now $\phi$ is very important. For example the probability of the state $|\psi_3 \rangle$ being in the $|1\rangle$ state is, $$P(1) = \frac{1+\cos(\phi)}{3+\cos(\phi)}$$ which very much depends on $\phi$. In fact this probability ranges from $0.5$ when $\phi = 0$ to $0$ when $\phi = \pi$.
