As an interesting exercise, I was wondering whether we could reformulate classical mechanics in such a way that we could use the same mathematical paradigm we use in quantum mechanics. I'll expose it here since I want to know whether this procedure is of any interest, or if it is correct.
For simplicity, I chose a classical particle with mass $m$ subject to a simple harmonic potential. Then, I assigned it a probability density function $P(x)$. Therefore, my goal now is to calculate said probability density so that I can know the probability of my particle being in an interval $x\in[x_0,x_1]$.
I begin by assuming that the probability distribution is inversely proportional to velocity. This seems quite natural to me, since the faster the particle goes, the less likely it will be for it to be in a given position when we measure it. Therefore, I have:
$$P(v) = \frac{N}{v}, \ \ N \in \mathbb{R}$$
where, of course, the probability density must fullfil the normalization condition:
$$\int_{-L}^LPdx = 1$$
Naturally, the classical particle is only allowed to move in the interval $x \in [-L,L]$, which means it has a mechanical energy:
$$E=\frac{1}{2}kL^2$$
where $k$ is the spring constant, or whatever other constant the harmonic potential might have depending on the system.
What I need to do now is to find an expression of $P$ in terms of $x$. This is easily done by using the conservation of energy:
$$E=K+U\Longrightarrow v = \sqrt{\frac{2E-kx^2}{m}}$$
or more concisely, dividing by the particle's mass:
$$v = \sqrt{2\varepsilon-\omega^2x^2}$$
where $\varepsilon=E/m$. Now, introducing the expression I found for $v(x)$ in my probability density, I finally have it as a function of $x$.
$$P(x)=\frac{N}{\sqrt{2\varepsilon-\omega^2x^2}}$$
All that is left to do is to calculate the normalization constant, $N$. Imposing the probability density fullfils the normalization condition:
$$\int_{-L}^L\frac{N}{\sqrt{2\varepsilon-\omega^2x^2}}dx = 1$$
$$\frac{N}{\sqrt{2\varepsilon}}\int_{-L}^L\frac{dx}{\sqrt{1-\omega^2 x^2/2\varepsilon}}=1$$
This is the integral of an arcsine, which yields the result:
$$N=\frac{\omega}{2\arcsin(\frac{\omega}{\sqrt{2\varepsilon}}L)}$$
Recalling that $\varepsilon=\frac{E}{m}=\frac{1}{2}\frac{k}{m}L^2 = \frac{1}{2}\omega^2L^2$, the arcsine vanishes and yields $\pi/2$. Therefore, I end up with:
$$P(x) = \frac{\omega}{2\frac{\pi}{2}}\frac{1}{\sqrt{2\varepsilon-\omega^2x^2}}$$
Again, the term relative to the particle's velocity can be simplified, and I finally get:
$$\boxed{P(x)=\frac{1}{\pi\sqrt{L^2-x^2}}}$$
