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In the game-based definition, we say that $G: \{ 0, 1 \}^n \rightarrow \{ 0, 1 \}^{\ell(n)}$ is a pseudorandom generator if For all ppt distinguishers $D$, there exists a negligible function $\nu$ such that: $$Pr[D( r) = 1] - Pr[D(G(s)) = 1 ] \leq \nu(n) $$ Where $r \gets \{ 0, 1 \}^{\ell(n)}$ and $s \gets \{ 0, 1 \}^n$ are chosen uniformly at random. Now, $Range(G)\subset \{ 0, 1 \}^{\ell(n)}$. So there's a possibility that $r \in Range(G)$ even if it's picked uniformly at random. Are we assuming that grabbing a "bad" $r$ is unlikely, or is the game implicitly saying the two cases are: $r \in Range(G)$ and $r \notin Range(G)$?

I could see if $\ell(n) = 2n$ then grabbing a bad $r$ would be unlikely, but the stretch only needs to be at least $1$, so if $\ell(n)= n+1$ a uniform at random $r$ would be in the range with probability $2^n/2^{n+1} =1/2$. In this case, always saying it's from $G$ seems like it would win the game $3/4$ of the time.

eternalmothra
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