If two people start at 0 and walk with random direction in one dimension (with constant step-size), the probability that their distance from each other increases is 0.5 after a step, so the probability that the standard deviation of their positions increases is 0.5 -- $P_2(\Delta\sigma>0)=0.5$.
With three or more (N) people, my intuition is that this probability is larger for larger N, but I can't find the relationship analytically. I could simulate it to get approximations of $P_N(\Delta\sigma>0)$, but I'm interested in how straightforward the analytical solution might be.
Edit 1: At the first step from 0, since $\sigma$ is 0, $P_N(\Delta\sigma>0)$ = $1-(\frac{1}{2})^{N-1}$. For later steps I suppose $P$ is dependent on the current $\sigma$.
Edit 2: I initially wanted to solve this for step sizes drawn from a uniform distribution [-1,1], but that may make things overly complicated to start. I'll focus on only step sizes of -1 or +1.
Some work:
In a step, each $i^{th}$ person walks a distance $\delta_i$ (-1 or 1) from $x_{i,0}$ to $x_{i,1}$.
$$ x_{i,1} = x_{i,0} + \delta_i $$ $$ \bar{x}_1 = \bar{x}_0 + \bar{\delta} $$ $$\sigma_1^2 - \sigma_0^2 = \frac{1}{N-1} (\sum_i |x_{i,1} - \bar{x}_{i,1}|^2 - |x_{i,0} - \bar{x}_{i,0}|^2)$$ $$ = \sum_i (\bar{\delta} - \delta_i)^2 + 2(\bar{x}_0-x_0)(\bar{\delta}-\delta_i)$$ $$P(\sigma_1^2 - \sigma_0^2 > 0) = ?$$