What are the odds of finding an electron a long way from an atom? As in metres away? Pretty self-explanatory, heard that there is a chance that you might find an electron on the other side of the universe, just wanted to know.
 A: Rigorously speaking, the probability to find the electron at a distance exactly equal to $r$ from the nucleus is $0$. On the other hand, we can define the probability to find the electron in a volume $d \mathbf{r}$ as 
$$P(d \mathbf r) =| \Psi(\mathbf r)|^2 d \mathbf r = |\Psi(r,\theta,\phi)|^2 r^2 \sin \theta \ dr \ d\theta \ d \phi$$
where I have introduced spherical coordinates.
If $\Psi(r,\theta,\phi)$ is in the form
$$\Psi(r,\theta,\phi)=R(r)\Phi(\theta, \phi)$$
we can define a radial probability density
$\rho(r) $ such that
$$P(r, r+dr) = \rho(r) dr= 4 \pi \ |R(r)|^2 r^2 dr$$
The previous formula is good if $dr$ is small compared to $r$. If it isn't (and we will write $\Delta r$ in that case), we will have to take the formula
$$P(r,r+\Delta r) = 4 \pi \int_{r}^{r+\Delta r} |R(r')|^2 r'^2 dr'$$
If we take the hydrogen fundamental state for a practical example (you can find a picture of the radial probability density here), we get
$$P(r,r+dr) = \frac {4} {a_0^3}\exp\left(-\frac{2r}{a_0}\right) r^2 dr$$
Let's take $r=1$m and $dr = 1 $cm: we get
$$P \simeq 2.7 \cdot 10^8 \cdot \exp(-3.8 \cdot 10^{10})$$
For all practical purposes, this number is equal to $0$ (just try and calculate it to see that).
If we take more reasonable values, such as $r=1$ nm and $dr=0.1$ nm, we get
$$P \simeq 2.7 \cdot 10^3 \cdot \exp(-38) \simeq 8.5 \cdot 10^{-14}$$
So you can see why the claim that "there is a chance to find the electron on the other side of the universe" is ridiculous: even at a "small" (on astrophysical scale) distance, $1$ m, the probability density is for any practical purpose equal to $0$.
Update (to answer some of the comments)
We could of course calculate the probability that the electron is found between $r$ and $\infty$: that would be
$$P(r,\infty)=4 \pi \int_r^\infty |R(r')|^2 r'^2 dr'$$
If we take again the hydrogen atom fundamental state, we get
$$P(r,\infty)=\frac {4} {a_0^3} \int_r^\infty \exp\left(-\frac{2r'}{a_0}\right) r'^2 dr'$$
the integral can be (easily?) solved to get
$$P(r,\infty) = \frac 1 {a_0^2}\exp \left(-\frac{2 r}{a_0}\right) \left(2 r^2+2 a_0 r+a_0^2\right) \simeq \exp \left(-\frac{2 r}{a_0}\right) \cdot  \frac{2 r^2}{a_0^2} $$ 
where I have used $r \gg a_0$. We obtain, for $r=1$m
$$P \simeq 3.8 \cdot 10^{20} \cdot \exp(-3.8 \cdot 10^{10})$$
This number is always dominated by the exponential term, so, in fact, the probability is always practically $0$.
We could ask ourselves what would change if we considered a different element, or a molecule. The answer is that not much would change. The point is that, as it is clear from the previous discussion, the villain here is the exponential term. This term will always appear for bound states in a potential well because the wave function must be normalized, that is to say
$$\int_{\mathbb{R}^3} |\Psi(\mathbf r)|^2 d \mathbf r = 1$$
So terms will appear that will kill the wave function for $r \to \infty$, and those terms will usually be exponential (I remember a very good discussion of the general problem of a 1D potential well in the Cohen-Tannoudji book...).
A free electron is better described as a wave packet, so the discussion would be different. But exponential terms will always come out to ensure normalization, so the general idea will not change.
