Is the Probability of Selecting 3 Random and Colinear Points nil? Recently, the mathematics YouTube channel released a video titled "Triangles have a Magic Highway - Numberphile". In the video, at 6:40, the expert being videoed says that the probability of any three random points lying on a single line is zero. 
I intuitively understand this in a real number plane. Given the infinite points in $R^2$, the probability of selecting three co-linear points is $\frac{3}{\infty}\approx 0$.
I'd like to see a rigorous mathematical proof, if there is such a thing. Further, is there any abstraction in which the probability changes? I.e. in $R^3$, there are seemingly more points to select from ((x,y,z), as opposed to only the set (x,y)). Are there more points to select from, giving a quantifiable $or$ somehow significant difference in the probability, or am I thinking of two, both non-quantifiable/insignificantly different infinities?
 A: Preliminaries
Firstly there is no such thing as choosing a completely random point in the plane. Why so? If we assume that there is ever such a thing, then notice that for it to be completely random, every point in the plane must be equally likely to be chosen. How likely? It cannot be absolutely zero likelihood, otherwise no point can ever be chosen. Therefore every point must be chosen with some nonzero likelihood $c > 0$. But no matter how small $c$ is, there is a natural number $n$ such that $\frac{1}{n} < c$, because $c$ is positive. (This is the archimedean property of the reals.) Now take any $n$ points in the plane. (We have infinitely many to choose from!) Choosing any of them precludes choosing any other one, so the total likelihood of choosing one of these $n$ points is $nc > 1$. Oh dear, we have a problem; a more than 100% likelihood of some event! Contradiction.
Distribution
Every mention of "random" must come with a distribution. Say we pick a point in the plane with distribution being some bivariate normal joint distribution. Note that the density function of a continuous distribution such as this is not at all the probability of choosing the point. In fact for any continuous distribution the probability of choosing any particular point is always zero. What you can say instead is that the probability of choosing a point in some region is the integral of the density function over that region, which is essentially the volume under the function. Of course, the integral of the density function over the whole plane should be $1$.
Probability of picking points from a continuous distribution
You can see that for a continuous distribution, regardless of the number of dimensions (at least one dimension), the probability of picking any specific point is zero, by definition of the density function. And all real numbers that are zero are absolutely equal. So there is no sense in which it is less likely to choose a point from a line as compared to choosing it from the whole plane containing the line, as long as in both cases we are talking about continuous distributions.
Similarly, when you look at any continuous distribution of triples of points in the plane, the probability that their are collinear is zero. The totally rigorous proof will need some analysis and basic measure theory, but here is a reasonably good approximate explanation. This is related to asking "What is the volume of a spherical surface?" Well zero of course, because we can cover all the points on the surface using cubes with arbitrarily small total volume. This is a layman version of the mathematical statement that the spherical surface has zero Jordan measure in 3-dimensional Euclidean space. If something has zero volume in some space, then it must also have zero probability in any continuous distribution over that space. (*)
In your case we have 2 dimensions per point so we have a 6-dimensional Euclidean space and we basically want the 6-dimensional "volume" of a 5-dimensional surface in the space, which is zero. In layman terms it is roughly because collinear triples of points have 5 degrees of freedom, 2 for the first point, 2 for the next point, and then 1 for the third point because it must be on the line given by the other two. By the earlier reasoning, zero volume implies zero probability.
Non-continuous distributions
Of course, one can have non-continuous distributions, which just means that some points can be chosen with nonzero probability, but there can only be countably many such points (**). Here is an example. Choose $(0,n)$ with probability $2^{-n}$ for every positive integer $n$. Total probability adds up to $1$, so it is a valid distribution from which we can draw a random point in the plane. Now you may think that this cannot be called "random point in the plane", since some points can never be chosen. So what? It is one of the many valid random distributions of points in the plane. This is why it is crucial to specify the distribution, otherwise nobody knows what kind of "random" you are talking about.
Proofs
For the mathematically inclined...
(*) Here is the intuition behind why zero volume implies zero probability for any continuous probability distribution. All the details depend on the exact definitions for volume and probability, but intuitively they all boil down to not allowing the ratio of probability to volume to be unbounded as a region is shrunk (but always with positive volume). Simply put, continuous distributions have the probability density smoothly distributed over the space and never have any point with infinite density. With this in mind, if we have a region $A$ with zero volume but positive probability $m$, we can take any sequence of regions $A_1 \supseteq A_2 \supseteq \cdots$ with volume going to zero and all of which contain $A$, but then every region in the sequence has probability at least $m$, and so $\frac{Prob(A_n)}{Vol(A_n)}$ increases without bound as $n$ increases, contradicting our notion of continuous distributions.
(**) Here is the proof that you can only have countably many points with positive probability. Let $S_n$ be the collection of points with probability at least $\frac{1}{n}$, for any positive integer $n$. Note that every point with positive probability will be in $S_n$ for some positive integer $n$. But for any positive integer $n$, $S_n$ cannot have more than $n$ points, otherwise their total probability would be more than $1$, and hence $S_n$ must be a finite collection. Since we have countably many finite collections, their union certainly is countable.
