Let $G$ be a finitely generated amenable group.


Is there a positive integer $n$ such that $n$ random elements of $G$ generate it with positive probability?


Being more formal, note that $G^n$ is amenable, so it carries a translation invariant finitely additive probability measure $\mu$. I am asking whether there is a positive integer $n$ such that:


$$\mu (\{(g\_1, \dots, g\_n) \in G^n \ : \ \langle g\_1, \dots g\_n \rangle = G\}) > 0$$


Alternatively, we can take a Folner sequence $\{F\_k\}\_{k \in \mathbb{N}}$ in $G^n$ and ask whether 


$$ \liminf\_{k \to \infty} \frac{|\{ \text{Generating n-tuples} \}\cap F\_k|}{|F\_k|} > 0$$



Are finitely generated amenable groups positively finitely generated?
The answer is already no for $\mathbb{Z}$, assuming the question is whether this holds for every meaure. Let $n\in\mathbb{N}$, and let
$$
S \,=\, \{(a\_1,\ldots,a\_n)\in\mathbb{Z}^n \mid \gcd(a\_1,\ldots,a\_n)=1\}.
$$
I will demonstrate a Følner sequence in $\mathbb{Z}^n$ that does not intersect $S$, which leads to a translation-invariant, finitely additive probability measure $\mu$ on $\mathbb{Z}^n$ for which $\mu(S) = 0$.


The idea is that there are large cubes in $\mathbb{Z}^n$ that do not intersect $S$. Specifically, define a **$\boldsymbol{k}$-cube** in $\mathbb{Z}^n$ to be any product of the form
$$
A\_1 \times \cdots \times A\_n
$$
where each $A\_i$ is a set of $k$ consecutive integers. I claim that for each $k\in\mathbb{N}$ there exists a $k$-cube $C\_k$ in $\mathbb{Z}^n$ so that $C\_k\cap S = \emptyset$. Then $\{C\_k\}$ is a Følner sequence in $\mathbb{Z}^n$, and
$$
\liminf\_{k\to\infty} \frac{|C\_k\cap S|}{|C\_k|} \,=\, 0.
$$


  
To prove that the $k$-cubes $C\_k$ exist, let $k\in\mathbb{N}$, let $\mathcal{I} = \{1,\ldots,k\}^n$, and choose a family
$$
\{p\_{(i\_1,\ldots,i\_n)}\}\_{(i\_1,\ldots,i\_n)\in\mathcal{I}}
$$
of $k^n$ distinct primes. By the Chinese remainder theorem, there exists a tuple $(c\_1,\ldots,c\_n)$ of integers so that
$$
c\_j \equiv -i\_j\pmod{p\_{(i\_1,\ldots,i\_n)}} 
$$
for each $(i\_1,\ldots,i\_n) \in\mathcal{I}$ and each $j\in\{1,\ldots,n\}$. Then
$$
C\_k \,=\, (c\_1,\ldots,c\_n) + \mathcal{I}
$$
is a $k$-cube in $\mathbb{Z}^n$ that does not intersect $S$, since
$$
p\_{(i\_1,\ldots,i\_n)}\;\bigl|\; \gcd(c\_1+i\_1,\ldots,c\_n+i\_n)
$$
for each $(i\_1,\ldots,i\_n) \in \mathcal{I}$.


**Edit:** This technique can also be used to show that there exists a Følner sequence $\{F\_k\}$ in $\mathbb{Z}$ itself so that for all $n\in\mathbb{N}$
$$
\liminf\_{k\to\infty} \frac{|\{\text{generating $n$-tuples}\}\cap F\_k^n|}{|F\_k^n|} = 0.
$$
Specifically, we can let each $F\_k$ be a disjoint union
$$
F\_k \,=\, A\_{k,1} \uplus A\_{k,2} \uplus \cdots \uplus A\_{k,k}
$$
where each $A\_{j,k}$ is a set of $k$ consecutive integers, and these sets are chosen to have the property that $\gcd(a\_1,\ldots,a\_k)\ne 1$ for all $(a\_1,\ldots,a\_k)\in A\_{k,1}\times\cdots\times A\_{k,k}$. If we fix $n$, then the probability that a randomly chosen element of $F\_k^n$ has two coordinates from the same $A\_{j,k}$ goes to $0$ as $k\to\infty$, and thus a randomly chosen element of $F\_k^n$ has a high probability of not generating $\mathbb{Z}^n$.


