First of all, note that there is no canonical notion of equidistribution on a countable set like the integers.

When asking for the probability that $k$ 'randomly chosen' integers are coprime, it is more-or-less intuitively clear what is meant by 'randomly chosen'. This is basically because the density of integers divisible by a given prime $p$ is (up to 'differences from rounding') the same in any interval $\{1, \dots, n\}$, namely $1/p$.

In your question this is not the case: for example in $S_4 \cap \{1, \dots, 10\}$, 5 of 5 numbers (= 50 percent) are even, while in $S_4 \cap \{1, \dots, 10^6\}$, 1070 of 1273 integers are even (which is about 84 percent) and in $S_4 \cap \{1, \dots, 10^{30}\}$, already 445064 of 462692 are even, which is about 96 percent.

So in order to make your question well-defined you need to be explicit about what you mean by 'randomly chosen'.

First of all, note that there is no canonical notion of equidistribution on a countable set like the integers.

When asking for the probability that $k$ 'randomly chosen' integers are coprime, it is more-or-less intuitively clear what is meant by 'randomly chosen'. This is basically because the density of integers divisible by a given prime $p$ is (up to 'differences from rounding') the same in any interval $\{1, \dots, n\}$, namely $1/p$.

In your question this is not the case: for example in $S_4 \cap \{1, \dots, 10\}$, 5 of 10 numbers (= 50 percent) are even, while in $S_4 \cap \{1, \dots, 10^6\}$, 1070 of 1273 integers are even (which is about 84 percent) and in $S_4 \cap \{1, \dots, 10^{30}\}$, already 445064 of 462692 are even, which is about 96 percent.

So in order to make your question well-defined you need to be explicit about what you mean by 'randomly chosen'.