Most probable free path length The distribution of scattering time is
$nv\sigma e^{-nv\sigma t}$
where $\sigma$ is the collision-cross section, $n$ is the number density of molecules and $v$ is the average relative velocity. This distribution is strictly decreasing with $t$ so that it has maximum at $t=0$. Therefore the most probable free path should be $0$. Is this correct? Or have I misunderstood something?
 A: For that distribution, the most probable single value is zero.
But the average value is $n\nu\sigma$, which is non-zero.
The underlying physical cause is that there's an equal probability of scattering in every cm travelled.  If the particle doesn't scatter in the first cm, it'll get a chance in the second, then the third, etc.  But if it does scatter in the 1st, it's no longer available to have it's first scatter in the 2nd or 3rd; it's already scattered.  So the most probably place to scatter is the place with the most flux of unscattered particles, right at the start.
A: It’s easier to think about distance than about time.  The distance between collision events is indeed exponentially distributed, $PDF\propto \exp (-n\sigma x)$, but the corresponding travel time depends on relative speed, which obeys the Maxwell-Boltzmann distribution.  (It’s even possible that the scattering cross-section depends on relative velocity.)  The mode of $\exp (-n\sigma x)$ is indeed at x=0, but the mean is more relevant than the mode for most purposes, e.g., transport phenomena.  
