Probability of all combinations of k numbers among n being coprime

Question

A simple argument shows that if we choose $n$ positive integers at random, the probability of their greatest common divisor being 1 is $1/ \zeta (n)$ (in the sense that if we choose the numbers among $\lbrace 1, 2, 3, \dots , N \rbrace$ uniformly and independently, the above probability tends to some number $p(n)$ as $N \rightarrow \infty$).

In the above sense, what is the probability $p(n,k)$ that among n integers $x_1, x_2, \dots, x_n$ chosen at random we have $\gcd (x_{i_1}, x_{i_2}, \dots, x_{i_k})=1$ for all possible combinations of the $x_{i_j}$ (with $s \neq t$ if for $u \neq v$, $x_{i_u}=x_s$ and $x_{i_v}=x_t$) ?

Answer 1

For any prime $q$, define $$ \ell(q,n,k) = q^{-n} \sum_{j=0}^{n-k} \binom nj (q-1)^j, $$ so that $\ell(q,n,k)$ is the probability, if a biased coin that comes up heads only $1/q$ of the time is tossed $n$ times, that at least $k$ heads are obtained. Equivalently, $\ell(q,n,k)$ is the probability, if $n$ numbers are chosen uniformly and independently from the set $\{0,1,\dots,q-1\}$, that at least $k$ of the numbers equal $0$.

Then the same argument giving the $1/\zeta(n)$ result shows that the probability that the greatest common divisor of every $k$-subset of $n$ "randomly chosen" integers is $1$ equals $$ \prod_q \big( 1 - \ell(q,n,k) \big), $$ where the product is over all primes $q$. This product is convergent as long as $k\ge2$; it diverges to $0$ if $k=1$ (appropriately).

For example, if three integers are chosen at random, the probability that they are pairwise coprime is $$ \prod_q \big( 1 - \ell(q,3,2) \big) = \prod_q \bigg( 1-\frac1q \bigg)^2 \bigg( 1+\frac2q \bigg) \approx 0.286747. $$ (A million trials with random integers between $1$ and $10^{60}$ yielded $286912$ pairwise coprime triples, so this limiting probability seems accurate.) I don't believe this product has a closed form in terms of the zeta function.