What does it mean that an electron's position is described by probability? I just read an answer to a Phys.SE question about why electrons don't collide with protons where an answering user said:

In quantum mechanics, an electron doesn't have a definite position or
  momentum. It has a wave function from which the probability of finding
  it at a particular position or momentum can be calculated. An electron
  bound to a proton will probably be very near the proton.

Is an electron's path inherently non-deterministic -- i.e., is it unlike billiards in that if you know the starting velocity and mass of one ball you can calculate the rest of its path entirely. Or is it that, it is just impossible for observers to figure out this data, and thus we have to rely on probability calculations?
If it is not simply a case of being unable for us to observe, then I will be very confused. Does that mean "randomness" exists? That would seem extremely... illogical? But obviously I'm missing some pieces of understanding.
 A: Quantum mechanics is deterministic, in the sense that the evolution of the probability distribution (i.e wave function) can be obtained rigorously by solving the Schrodinger Equation. 
However, quantum mechanics is not "as deterministic as you would wish", in the sense that there is Heisenberg uncertainty principle. The uncertainty principle arises from the fundamental property that quantum observables do not always commute with each other, i.e. generally $\hat A\hat B\neq \hat B\hat A$, and as a result, if a quantum state has a definite value for $A$, it cannot at the same time have a definite value for $B$, and vise versa. This is in sharp contrast with the way we perceive the world works -- for example we say we know the state of a billiard ball only if we know both its position $x$ and its velocity $v$. Well quantum mechanics says this is simply impossible for quantum states (which means for all states in real life). 
But things are not that bad. The uncertainty principle says that the uncertainties $\Delta A$ and $\Delta B$ satisfy $\Delta A\Delta B\sim \hbar$. You see that while it is impossible to be absolutely certain about both ($\Delta A$ and $\Delta B$ cannot both be zero), but you are still pretty certain, in the sense that the "average uncertainty" $\hbar$ is really really a small number.
