Boltzmann's definition of entropy is $\sigma = \log \Omega$, where $\Omega$ is the number of microstates consistent with a given macrostate. If I understand correctly, this means that it only makes sense to speak of the entropy of a physical system with respect to a definition of the "macrostate" of the system. For example, we can consider a system consisting of $N$ spins that can be either up or down. Suppose that the actual system is in the "alternating" microstate $\uparrow \downarrow \uparrow \downarrow \cdots \uparrow \downarrow$. We could stipulate that the macrostate of the system is the total spin excess $s$, and calculate the entropy of the actual physical system using the fact that $s = 0$. Or, instead, we could stipulate that the macrostate of the system is the spin excess $s_1$ of the first $N/2$ particles together with the spin excess $s_2$ of the second $N/2$ particles. In this case, we would calculate a different entropy for the same actual physical system, because this time we would use the extra information $s_1 = s_2 = 0$.
What I'm confused about is that the way I always hear the 2nd law stated is something like "The entropy of any closed system tends to increase," with no mention of the definition of macrostate with respect to which we are calculating entropy. I am wondering what the more precise statement of the 2nd law is, with explicit reference to how macrostates are defined.
My first thought was that perhaps the 2nd law says something along the lines of "With respect to any fixed definition of the macrostate of a closed system, the entropy of the system tends to increase." But this can't quite be right, because we can rig the macrostates so that entropy decreases just by adding in "impossible" microstates. For example, if we consider the system $\uparrow \downarrow \uparrow \downarrow \cdots \uparrow \downarrow$ described above, we could define macrostates as follows: For each integer $s$, we define a macrostate $S_s$, which consists of all microstates of $N$ spins with spin excess $s$, along with a whole bunch of quantum harmonic oscillator microstates. If we choose the number of QHO microstates that we include in $S_s$ carefully, we can ensure that entropy w.r.t. this definition of "macrostate" decreases whenever entropy w.r.t. the straightforward spin-excess definition increases, and vice versa.
So this makes me think that maybe we need some notion of the "possible" microstates of a system, and then the 2nd law should say something like "With respect to any fixed partition of the possible microstates of a closed system into macrostates, the entropy of the system tends to increase." But I'm struggling to even imagine what "possible" would mean here. Maybe this is all completely wrong?
Any help would be appreciated. Thanks!