What does this equation mean?

Gavin Crooks sharpened this two years later into a statement about entire probability distributions rather than a single average. If $P_{\mathrm F}(W)$ is the distribution of work values measured over many repetitions of a forward protocol, and $P_{\mathrm R}(W)$ is the distribution measured over the time-reversed protocol run from its own correctly defined reverse-equilibrium starting ensemble, the two distributions are related by $\frac{P_{\mathrm F}(+W)}{P_{\mathrm R}(-W)} = \exp\!\left(\frac{W - \Delta F}{k_{\mathrm{B}} T}\right)$. [ 6 ] . Where W exactly equals $\Delta$ F , the ratio is one and the forward distribution crosses the reflected reverse distribution; away from that point, the ratio grows exponentially in how far W departs from $\Delta$ F , which is the precise, quantitative sense in which a trajectory doing much more work than the equilibrium minimum is exponentially more typical of the forward process than of its reverse. This is the arrow of time rendered as an odds ratio rather than an inequality: not “entropy increases,” but “this particular trajectory is $e^{(W-\Delta F)/k_{\mathrm B}T}$ times more likely to have been produced running forward than running backward.”