(This is duplicate of the same question on MathOverflow where it was not approved.)
The concept of geometric probability assumes that for each (uncountable) set $A$ corresponding to a random event the proper measure $\mathrm{mes}(A)$ is defined, so that the probability of the event can be computed as follows: $$ \mathrm{P}(A)=\frac{\mathrm{mes}(A)}{\mathrm{mes}(\Omega)}, $$ where $\Omega$ is the sample space.
However, choosing the measure is not always unambiguous. Let, for example, $$ \Omega=\{(\rho,\varphi)\in\mathbb{R}^2\ |\ 0\le\rho\le 1,\ 0\le\varphi\le\pi\},\\ A=\{(\rho,\varphi)\in\Omega\ |\ 0\le\rho\le 0.5,\ 0\le\varphi\le\pi\}. $$ On the one hand $\Omega$ and $A$ may be considered as rectangles in $\mathbb{R}^2$, so $\mathrm{mes}(\Omega)=\pi$, $\mathrm{mes}(A)=\pi/2$. On the other hand, we may think of $\Omega$ and $A$ as of semicircles with differing radiuses. The latter case gives $\mathrm{mes}(\Omega)=\pi/2$, $\mathrm{mes}(A)=\pi/8$. Each case results to different probabilities: $\mathrm{P}(A)=0.5$ vs. $\mathrm{P}(A)=0.25$.
My question is what is the reasonable way to distinguish such situations. Is it possible to formally recognize which measure should be selected?
P.S. Actually, the "random midpoint method" for resolving Bertrand paradox takes two random values $(\rho,\varphi)$, where $\rho$ is the distance from the center of the unit disk to the chord, $\varphi$ is orientation of the radius-vector $\vec{\rho}$. We have $\Omega=[0,1]\times[0,2\pi)$, $A=[0,0.5]\times[0,2\pi)$. I cannot figure out why should I (formally) prefer polar coordinates rather than Cartesian in this case.
You italicized the word formally which means that the answer is no.
Which measure is appropriate depends on your setting.
In your example, if $P(A)=1/2$, then you are dealing with Cartesian coordinates (rectangles) and if $P(A)=1/4$, then your coordinates are polar and you are dealing with semi-disks.
PS. Bertrand paradox explains it succinctly:
IOW, you need to start with geometry (which implicitly assumes the uniform measure) of your process rather than coordinates.
Specifically Re: your PS question:
The midpoint is selected uniformly from the the disk. I.e., your $\Omega$ is the disk of radius 1 and $A$ is the circle of radius 1/2. Which coordinates you use to represent them is up to you.