Let's define $n$-th generating fraction of a finite group $G$ as $gf_n(G) = \frac{\{(a_1, ... , a_n) \in G^n| \langle \{a_1, ... , a_n\} \rangle = G\}}{|G|^n}$. The trivial properties of this object include it being monotonously non-decreasing in respect to $n$, and that $gf_n(G) \neq 0$ iff $G$ in $n$-generated.
This object looks very similar to the $n$-th commuting fraction of a finite group $cf_n(G) = \frac{\{(a_1, ... , a_n) \in G^n| [a_1, ... , a_n] = e\}}{|G|^n}$, where $[a_1, ... , a_n] = [[a_1, ... , a_{n-1}], a_n]$ is iterated commutator. For them, there is a following theorem:
If $n \geq 2$ and $cf_n(G) > 1 - \frac{3}{2^{n + 2}}$, then $G$ is nilpotent of class $n$.
That left me wondering, do there exist some similar facts for generating fractions. I managed to prove the following one:
If $n \geq 2$ and $gf_n(G) > 1 - (\frac{3}{4})^n$, then $G$ is $(n - 1)$-generated
Proof:
If $G$ happened to be cyclic, then there is already nothing to prove. Thus, we can assume without the loss of generality, that $G$ is non-cyclic. Now, let's calculate the probability that among $n$ independent uniformly chosen random elements none of them lies in the Frattini subgroup. It is $1 - (1 - \frac{1}{[G: \Phi(G)]})^n \leq 1 - (\frac{3}{4})^n$ because $[G: \Phi(G)] \geq 4$ for finite non-cyclic $G$. Thus, $gf_n(G) > 1 - (\frac{3}{4})^n$ implies, that there is an $n$-element generating set with an element lying in $\Phi(G)$. That means, that if we remove this element, the remaining $(n-1)$-element set will still be generating. Thus, $G$ is $(n-1)$-generated Q.E.D.
However, my question is about another possible statement (which, if true, seems to be in more direct relationship with the similar statement for communion fraction):
Is it true, that $\exists k \in \mathbb{R}_+$, $c \in (0, 1)$, such that for any finite group $G$ if $gf_n(G) < kc^n$, then $G$ is not $n$-generated?
Let $P_i$ be the set of the first $i$ primes, let $m=\prod_{p\in P_i} p$ and let $G=(C_m)^n$. Then $gf_n(G)<\prod_{p\in P_i} (1-\frac{1}{p})\to 0$ as $i\to \infty$.
So, for every $n$ and every $\epsilon>0$, there exists a finite group $G$ with $0<gf_n(G)<\epsilon$. This implies that the answer to your question is no.