Why is the change in entropy is different for different interpretations? Forgive me if the title is a bit vague. Read on and I'll clarify.

So, today in class, my teacher was explaining entropy. He gave various examples explaining how the degree of randomness changes. One example in particular stuck out to me. Here it goes:

Imagine there is a container with a hypothetical boundary in the middle. Neon gas is on the left of the boundary, and Argon on the right. As long as the boundary exists, the gases can't cross it to intermix.
Now the boundary is removed, and the gases mix. This increases the randomness in the system and thus entropy increases. We get $\Delta S \gt 0$. Thus the process is spontaneous.



My Question:

How can we definitively say that entropy has increased (or that randomness has increased)?



My Reasoning:

So my teacher says that entropy increases. These following cases show that change in randomness depends on your interpretation.



Interpretation 1:

Before mixing (boundary exists):

If we take some volume of gas, say $1 \text{ cm}^3$, at a random position in the container, we might get:


Pure $\ce{Ne}$
Pure $\ce{Ar}$
A mixture of the two gases


The point is that we aren't sure what composition we get, because the distribution of gases isn't homogeneous.

After Mixing (boundary removed):

The gases diffuse and intermix to form a homogeneous mixture.

Now if we take $1 \text{ cm} ^3$ from a random position, we can be $100\%$ sure of the composition of the gases, because the mixture is homogeneous.

This means there is more uniformity now. Or, we can say that there is less randomness. This tells us that $\Delta S \lt 0$.



Interpretation 2:

Using a new approach now.

Before Mixing (boundary exists):

Take any single atom from either side of the boundary at random. My chances of correctly predicting what atom I get (either $\ce{Ne}$ or $\ce{Ar}$) are $50\%.$

After mixing (boundary removed):

I again pick any atom at random from a random point in the container. My chances of correctly predicting what atom I get is again $50\%.$

This tells us that since I still have the same chance of correctly predicting, the randomness of the system has not changed.

This gives us $\Delta S = 0$.



Same physical process, but we get 3 different values of degree of randomness ($\Delta S$) depending on our interpretation. We know that diffusion does take place, so the process is spontaneous, and hence $\Delta S \gt 0$.

My Questions:


How to accurately know whether the degree of randomness increases or not?
How is the exact value of $\Delta S$ measured in actual experiments?




Note:


Assume the gases to be ideal.
Assume the container to be sufficiently large.
From the perspective of Statistical Mechanics:

Multiplicities do depend on the external parameters that specify the energy level of the system.

For ideal gas of $N$ indistinguishable molecules, the multiplicity is given by $$\Omega= \frac{1}{N!}\frac{V^N}{h^{3N}}\,\frac{2\pi^{3N/2}}{(3N/2- 1)!}(\sqrt{2mE})^{3N-1}$$

So, indeed the multiplicity increases when the position space increases.

When the external parameter like volume of the container is increased, the energy levels come closer and then the particles can have more microstates for the same $E$.

Prior to the volume increase, the microstates of the system are $\Omega_{N,E,V_i}\;_;$ when the volume is increased, not will be there the microstates $\Omega_{N,E,V_i}$ still be available after the removal of the constraint, but also, there be addition of microstates. In general, $$\Omega_{V_f}\geq \Omega_{V_i}\;.$$

So, as the volume increases, the multiplicity increases and so does the entropy.

As requested by OP:

From the perspective of Thermodynamics:

Suppose the two dissimilar indistinguishable gases $A$ and $B$ having the same number of particles, energy, volume are partitioned from each other.

Now, when the partition is removed, the energy of the gas $A$ confined to a localised area gets dispersed over a greater volume. 

Thus the entropy of gas $A$ increases and that is given by \begin{align}\Delta S_A &amp;= Nk\,\ln \frac{V_f}{V_i}\\ &amp;= Nk\,\ln \frac{2V}{V}\\ &amp;= Nk\,\ln 2\;.\end{align}

Same happens for gas $B$ also.

Therefore, the change in entropy of the system is given by \begin{align}\Delta S_\textrm{system}&amp;= \Delta S_A + \Delta S_B\\ &amp;= 2Nk\ln 2 \gt 0\end{align}

Therefore, entropy increases when the partition is removed.