If you do a (simple) calculation of a quantummechanical rotator in a spherical potential — i.e. calculate where an electron would be in the vicinity of an nucleus — you will find that the eigenfunction $\left | \Psi_1 \right \rangle$ corresponding to the lowest energy $E_1$ is dependent only on the distance of the electron from the nucleus ($\left | \Psi_1 \right \rangle = f(r)$ — use spherical coordinates because it makes your job a million times easier!) — which gives us a sphere like the s-orbitals are often drawn.
If you carry on, you realise that $\left | \Psi_1 \right \rangle$ is more or less a decaying exponential function, so its value is highest for $r = 0$. The probability of an electron being at a specific point is typically given as $\left \langle \Psi_1 \middle | \Psi_1 \right \rangle$ and evaluates as $\Psi^2_1$ in this case. Since the function has only one phase, the highest $\Psi_1^2$ value is — in the nucleus.
So the nucleus already carries the highest single probability of having the electron right inside it. It’s not even ‘special’ for the electron to be in the nucleus.
However, check Dave’s answer for the physical implications, which I didn’t even know about!