Probability of generation of ${\mathbb Z}^2$ What is the probability that three pairs $(a,b) $ , $(c,d) $ and $(e,f) $ of integers generate $\mathbb Z^2$? As usual the probability is the limit as $n\to \infty$ of the same probability for the $n\times n$ square. It is well known that for $\mathbb Z $ the probability of two numbers to generate is $6/\pi^2$.
 A: 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 $$p_3= \zeta(2)^{-1} \zeta(3)^{-1} \simeq 0.505739038$$
A: 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.
