Probability of generation of ${\mathbb Z}^2$

According to Proposition 1 in the paper

G. Maze, Gérard, J. Rosenthal, U. Wagner: Natural density of rectangular unimodular integer matrices, Linear Algebra Appl. 434, No. 5 (2011), 1319-1324, ZBL1211.15044,

the probability that $n$ random vectors generate $\mathbb{Z}^{n-1}$ is $$p_n = \prod_{j=2}^n \zeta(j)^{-1}.$$

For $n=2$ this gives $p_2=\zeta(2)^{-1}=6/\pi^2$, whereas for $n=3$ we obtain $$p_3= \zeta(2)^{-1} \zeta(3)^{-1} \simeq 0.505739038$$

Let me convert my comments to an answer. Let $u_n$ be the probability that a triple in $([0,n-1]^2)^3$ generates $\mathbb{Z}^2$, and let $v_n$ be the probability that a triple in $((\mathbb{Z}/n)^2)^3$ generates $(\mathbb{Z}/n)^2$. Certainly $v_n\geq u_n$, and I think that $v_n$ should be asymptotic to $u_n$. Using the Chinese Remainder Theorem and the structure of $(\mathbb{Z}/p^k)^2$ we see that $v_n$ is the product of $v_p$ for all primes dividing $n$. A little linear algebra gives $v_p=(1-p^{-2})(1-p^{-3})$. Thus, the expected density is $$ v_\infty = \prod_p (1-p^{-2})^{-1}(1-p^{-3})^{-1} = (\zeta(2)\zeta(3))^{-1} \simeq 0.5057390381 $$ agreeing with the answer that Francesco Polizzi just entered while I was typing this.