Understanding $\varepsilon$-advantage in cryptographically secure deterministic RNG
Took a while to grasp it, but with the help of a friend we came to a conclusion (and it turns out it was rather easy).

Recall that a function $f_n : \{0,1\}^{n} \mapsto \{0,1\}^{m(n)}$ is said to be one-way if for a positive polynomial $poly(\cdot)$, every randomized algorithm running in probabilistic polynomial time inverts $f_n$ with a negligible probability, i.e.  $\Pr[A\:inverting\: f_n] &lt; \frac{1}{poly(n)}$, with $n$ size of the input. This polynomial comes from the fact that for every possible computational path, the algorithm will follow it with a probability $2^{-n}$, which is bounded by $\mathcal O(\frac{1}{n^c})$ for every constant $c$. Clearly, we want the adversary's advantage to be as asymptotically small as possible.