In the situation of two boxes separated by a membrane, there is a very strong difference with the mathematical Brownian motion :
$$\text{The box is finite.}$$
This has a very strong consequcne : the mean displacement cannot be zero anymore.
To understand why, consider for instance a particle starting very close to the border of the box. It can not go beyond the border and therefore, the displacement can only be directed toward the opposite direction. After a long time, the position's probability distribution of any particle becomes uniform in the box, whatever the starting position was.
When there are now a huge number of particles, after an equilibration time, they will fill all the available volume with a uniform density, whatever was their initial distribution.
EDIT: uniform distribution in a finite domain
To see that the proability density of a Brownian motion tends to the uniform distribution, a simple way is to consider a segment $[0,L]$. The differential equation governing the pdf of the particle's position $p(x,t)$ is the heat equation, also called diffusion equation
$$\frac{\partial p}{\partial t}(x,t)=D\frac{\partial^2p}{\partial x^2}(x,t).$$
We write $p(x,t)$ as a Fourier series in space
$$p(x,t)=\sum_{k\in\mathbb Z}a_k(t)\mathrm e^{\mathrm i2\pi kx/L}.$$
A term $a_k$ corresponds to oscillations of $p$ that have a space periodicity
$L/|k|$, so the larger $|k|$ the more $p$ oscillates rapidly with respect to position.
The diffusion equation splits into a series of ordinary differential equation, one for each $k$
$$\frac{\mathrm da_k}{\mathrm dt}(t)=-\frac{4\pi^2Dk^2}{L^2}a_k(t).$$
The solutions are $a_k(t)=a_k(0)\exp\left(-k^2\,4\pi^2Dt/L^2\right)$. We observe that only $a_0(t)$ is not decreasing, such that $p(x,t)\to a_0(0)$
as time $t\to\infty$. This proves that the distribution becomes uniform. We also learn that the larger is $|k|$, the faster the oscillations go to zero.
Spectral analysis of higher level shows that this statement remains true for any domain in any finite dimension, because the Laplacian $(-\Delta)$ has only positive eigenvalues and the uniform function $x\mapsto 1$ is always
an eigenfunction, with eigenvalue $0$.
