You are unable to explain how reducing the space occupied by the initial configuration you proposed could change the temperature - which you say is 0 in all those cases - or the macrostate - which you call absolute zero macrostate in all those cases. The entropy would be the same whenever the particules are more concentrated than in your example - even though they would have lower probability of ocurrence. Don't you agree?

> With enough cubes and boxes such a state has a very low probability of occuring.

Sure. The thing is that if you calculate an entropy using Gibbs formula from the distribution of microstates for a given macrostate the value that you obtain depends on how many cubes are used to define the macrostate. There is no entropy of the microstate - the entropy depends on how you decide to define the macrostate. If the lattice is fine enough, and the particles indistinguishable, in the limit the entropy is zero - the macrostate becomes the same as the microstate and one single microstate is possible.

It all depends on the description we make of the system. The "I have balls in a box and I know their positions" example was incomplete.

If for example I have N of those balls in equilibrium in an isolated box of volume V and I know the energy E I know the equilibrium macrostate. The energy won't change because it's isolated. The macrostate will not change. The (equi)probability distribution of the possible microstates corresponding to the macrostate won't change. Gibbs entropy which is calculated using that distribution won't change.

If by sheer luck all gas particles moved to the exact left side of the container the energy wouldn't change, the macrostate wouldn't change, the entropy wouldn't change - it doesn't matter how unlikely that is. It doesn't matter if you know the position of each particle. The entropy for the thermodynamical system described in the previous paragraph is independent of your knowledge of where the particles are or how unlikely you think their positions are.

I agree that you could use alternative ways of defining the macrostate where that happens! (And calculating Gibbs entropy will give different results. One can even get zero entropy using the microstate as macrostate - and one may actually want to do that if the microstate is known!)

> Even at equilibrium it can lower in entropy. The equilibrium is simply the highest entropy state.

I would understand that you said either:

"the equilibrium is the highest entropy state but there may be fluctuations that shift the system temporarily out equilibrium"

or

"the equilibrium is the highest entropy state and possibly a distribution of states around it."

Both concepts are used. When you say that the equilibrium is simply the highest entropy state but the equilibrium entropy can also be lower I'm not sure if it can be read as one of those or you're saying something else entirely.