If \( n_l(t) \) is the number of particles in the left half after \( t \) moves, the change in \( n_l(t) \) in the time interval \( \Delta t \) is $$ \begin{equation*} \Delta n=\left(\frac{N-n_l(t)}{N}-\frac{n_l(t)}{N}\right)\Delta t, \end{equation*} $$ and assuming that \( n_l \) and \( t \) are continuous variables we arrive at $$ \begin{equation*} \frac{dn_l(t)}{dt}=1-\frac{2n_l(t)}{N}, \end{equation*} $$ whose solution is $$ \begin{equation*} n_l(t)=\frac{N}{2}\left(1+e^{-2t/N}\right), \end{equation*} $$ with the initial condition \( n_l(t=0)=N \). Note that we have assumed \( n \) to be a continuous variable. Obviously, particles are discrete objects.