fluid dynamics - Excluded Volume Calculation in Van der Waals Equation

When accounting for the excluded volume for in the Van der Waals equation, it is assumed that the molecules are hard spheres and are of diameter. If we consider a cube of volume V, then we can say that the side of this cube is of length $V^{1/3}$. Consider the diameter of the molecules to be $\sigma$. Suppose that the number of molecules in this box to be $N$. If we anchor $N-1$ molecules at their positions and look at the excluded volume from the perspective of the $N^{th}$! molecule, we see that the center of this molecule can approach the walls of the cube only upto a distance of $\sigma/2$ and can approach the anchored molecules upto a distance of $\sigma$ from their centers as shown:.

Then the excluded volume for this molecule should be $V_{ex}=(V^{1/3}-\sigma)^{3}-(N-1)(\frac{4}{3}\pi\sigma^{3})$. This follows even if we consider any other molecule and anchor the rest. But, according to wikipedia, we would be overcounting. I don't see how. The correct expression should be $V_{ex}=(V^{1/3}-\sigma)^{3}-(N/2)(\frac{4}{3}\pi\sigma^{3})$. Can anyone please explain?

Answer

As mentioned in the wikipedia page $4 \times \frac{4 \pi r^3}{3}$ is the excluded volume per particle, so you have to sum over all the particles and divide by the number of particles. While summing up you divide by 2, because a pair of particles only contribute once to the excluded volume.