Imagine a set $S$ of $10^{12}$ consecutive integers $n, n + 1, n + 2, n + 3, \ldots, n + 10^{12}-1$, where the exact identity of $n$ will be partially determined randomly as described below, and otherwise left unspecified.
First we decide whether $n$ is divisible by $2$, with probability $1/2$ for each possibility. That determines which members of $S$ are even and which are odd. Among those that are even, we toss another coin to decide which are divisible by $2^2$, then similarly for $2^3$, etc. It is not quite certain that any will be divisible by $2^{40}$, since that's a bit bigger than $10^{12}$, and there's a probability a bit less than $1/2$ that at least one will be divisible by $2^{41}$, etc.
Then we do likewise with $3$, then with $5$, and so on through the list of all primes.
There is a certain probability that a number will not be thus identified as divisible by any prime. This will be, in effect, a prime number but not any particular prime number, just as $n$ is not any particular number.
Can anyone say anything of interest about the probability distribution of the number of such mystery primes, or the probability distributions of their locations? Will they be the same as if we let the value of $n$ be an actual number and then found limits of these probabilities as $n\to\infty$?
Later edit: Henry's answer called to my attention what I should have thought of before: the expected number of primes by which $n$ is divisible would be $1/2 + 1/3 + 1/5 + 1/7 + 1/11 + \cdots$, so $n$ would be infinite. And none of the numbers in this interval of length $10^{12}$ would be prime.
This is a different kind of infinite number from those I've thought about before. But it's a perfectly tractable one: one could examine its properties.
This seems rather unlike the infinite integers of nonstandard analysis, because with each of those one can say that there is some larger infinite integer that is prime. It's not so clear how that would make sense with something like this. But yet it's quite similar to the infinite integers of nonstandard analysis in some ways: every third one is divisible by 3, every ninth one by $3^2$, etc.
Another comment: Actually, in one sense they're not so different from the infinite integers of nonstandard analysis. Almost all of them have infinitely many distinct prime factors. One difference is that with those infinite integers, each has a largest prime factor, which is infinitely large. And although one can say that for each infinite integer $n$, there is a prime $p$ greater than $n$, in almost every case its difference from $n$ will not be finite.