Probability
Neither Math Professors nor information transmission are infallible.
What you are ascribing to a Math Professor makes little sense.
Your friend is right, both in the answer (the probability is $0$) and in his argument (dimension).
Specifically, a triple of points in a plane forms a 6-dimensional space, and the subspace of collinear points is 5-dimensional, so its measure (probability) will be $0$ for any "reasonable" definition of probability.
"The number of points" in both 6d space and 5d subspace does not enter the picture (the cardinality of both is continuum and is irrelevant).
Strictly speaking, it is not easy to define a probability measure on the whole plane. The measure cannot be both invariant under euclidean transformations and have the total probability of $1$.
Thus the usual solution is to restrict the points to a bounded subset.
Thus, let the points be selected uniformly and independently from the unit square $[0;1]\times[0;1]$ with the usual Lebesgue measure (aka area) as the probability.
Then the 3 points $A=(x_1,y_1), B=(x_2,y_2), C=(x_3,y_3)$ can be represented as a single point selected uniformly from $[0;1]^6$.
The subset of collinear points is described by a single equation
$$
\det \begin{bmatrix}
x_1-x_2 & x_1-x_3 \\
y_1-y_2 & y_1-y_3
\end{bmatrix} = 0
$$
which defines a co-dimension 1 subset and thus has probability $0$.
Comparing infinities
Your are asking, in bold, whether two infinities can be compared or not.
The answer is definitely YES.
It depends, however, on what you are trying to accomplish.
E.g., both integers and reals are infinite sets, but integers are countable while reals are not. Thus one can say that there are more reals than integers. (Moreover, this idea of cardinality gives us a very simple argument why irrationals and transcendentals exist: because the "alternatives" (rational and algebraic numbers) are countable and reals are not).
However, you are talking about probabilities which means counting only
for finite sets. For infinite sets you have to use measure theory.
Here you would say that when selecting a point at random from the segment $[0;1]$, the probability that you will end up with a number between, say, $\frac{1}{3}$ and $\frac{1}{2}$ is the length of the segment $[\frac{1}{3};\frac{1}{2}]$ which is $\frac{1}{6}$.
Here we just compared two (uncountable) infinities: the set of points in $[0;1]$ and the set of points in $[\frac{1}{3};\frac{1}{2}]$ (both continuums).
Best Answer
There is no such thing as a uniform distribution on the plane. Without specifying how the points are chosen, the question is not properly stated. However, if the points are chosen independently from some continuous distribution (absolutely continuous with respect to Lebesgue measure), the probability of the third point lying exactly on the line through the first two is $0$.