Let $\Gamma$ be a non-algebraic analytic curve. Let $z_1,\dots, z_n$ be points of $\Gamma$. Let $X_{n,d}$ be the set of algebraic curves of degree $d$ that contain $z_1,\dots, z_n$. Now consider a random point $z_{n+1}$ on $\Gamma$ and the analogous space $X_{n+1,d}$. I claim $\dim X_{n+1,d} <\dim X_{n,d}$ almost surely.
Why is this? Let $Y_{n,d}$ be the set consisting of pairs of a curve in $X_{n,d}$ and a point $z \in \Gamma$ contained in that curve. Then $Y_{n,d}$ maps to $X_{n,d}$ and the fibers, being the intersection of two analytic curves, are discrete, and, lying on a Jordan curve, are bounded, hence are finite. So $\dim Y_{n,d} \leq \dim X_{n,d}$. Now $Y_{n,d}$ also maps to $\Gamma$, and $X_{n+1,d}$ is the fiber of $Y_{n,d}$ over $z_{n+1}$. If an analytic space maps to a curve then the fiber over a random point of the curve has dimension one less with high probability, so $\dim Z_{n+1,d} = \dim Y_{n,d}-1 \leq \dim X_{n,d}-1$.
By induction, it follows that $\dim X_{n,d} \leq \frac{d+2}{2}-1 - n$ since $\frac{d+2}{2}-1 $ is the dimension of the space of algebraic curves of degree $d$.
Thus if $n \geq \frac{d+2}{2}$ then with high probability if $z_1,\dots, z_n$ are random points then $X_{n,d}$ is empty.
On the other hand, if $\Gamma$ is a non-algebraic curve of degree $d$ then for any points $z_1,\dots, z_n$, $X_{n,d}$ is nonempty.
We can calculate $X_{n,d}$ using linear algebra and the coordinates of $z_1,\dots,z_n$ so this gives an empirical method to determine with high confidence whether $\Gamma$ is an algebraic curve of degree at most $d$. Of course the calculations are not completely precise in practice so we will more accurately get a method to determine if $\Gamma$ is close to an algebraic curve of degree at most $d$.
Then of course we can heuristically assume that if $\Gamma$ is not an algebraic curve of degree at most $d$ for some fixed $d$ then it probably isn't an algebraic curve at all, and if $\Gamma$ is close to an algebraic curve within the limits of the precision of our calculations then it probably is an algebraic curve.