In factoring consecutive numbers, how soon do we expect to see the smallest prime not yet seen? For $A\subseteq\{1,2,3,\ldots\}$ one can take $\Pr(A)$ to mean $\displaystyle \lim_{n\,\to\,\infty} \frac{|A\cap\{1,\ldots,n\}|} n,$ if one doesn't insist on probability being countably additive.
Suppose we have a long sequence of positive integers $m, m+1, m+2,\ldots, m+k$ and all of them are${}\ll\sqrt N.$ Among prime numbers${}<\sqrt N,$ suppose there are many that are not among the factors of any numbers in this long sequence and many that are. ("Many" must mean at least one, but typically many more.)
Can anything of interest be said about the probability that among the prime factors of $m+k+1,$ the smallest prime number not yet occurring in the factorizations of numbers in $[m,m+k]$ will appear?
 A: All primes up to $k+1$ divide one of $m,m+1,\dots,m+k$, so the primes that might divide $m+k+1$ are the primes greater than $k+1$. For each such prime $p$, the probability that it divides none of $m,\dots,m+k$ is presumably $1-(k+1)/p$, while the probability that it divides $m+k+1$ is presumably $1/p$. (I say "presumably" because I don't know how $m$ and $k$ are chosen; if $k$ is fixed then these probabilities are accurate.) In particular, the probability that $p$ is the smallest prime not dividing $m(m+1)\cdots(m+k)$ and that it does divide $m+k+1$ is
$$
\bigg( \prod_{k+1<q<p} \frac{k+1}q \bigg) \frac1p,
$$
where the product is over primes $q$ in the given range (each factor is the probability that $q$ does divide $m(m+1)\cdots(m+k)$).
Therefore the probability that the smallest prime not dividing $m(m+1)\cdots(m+k)$ does divide $m+k+1$ should be
$$
\sum_{p>k+1} \bigg( \prod_{k+1<q<p} \frac{k+1}q \bigg) \frac1p.
$$
Numerical calculations of this expression up to $k=8000$ suggest that this probability decays like, maybe, $1/(\sqrt k(\log k)^2)$? Very hard to guess exactly; I'm sure the asymptotics could be worked out with enough pain.
