TITLE: If you picked a number between $1$ to $10^n$ with no zero digits,find the probability that the digits' product is less than the number's square root?
QUESTION [7 upvotes]: What is the probability if you picked a number with no zero digits between $1$ to $10^n$ that the digits' product is less than the number's square root?
You can't pick a number like $371790$ because it has a zero.
An example for it being bigger than the square root is $\;2\times2\times2\times2\times9\times9>\sqrt{222299}.$
An example for it being smaller than the square root is ;$2\times2\times2\times2\times9\times2<\sqrt{222292}.$
If there is a limit of the probability as n goes to infinity, what is it?

REPLY [2 votes]: Let's consider a slight variation. Let $P_n$ the probabily that a natural number choosen randomly among the numbers with exactly $n$ decimal digits (that is, in $[10^{n-1},10^n-1]$ none of which zero has that property. That is (taking decimal logarithms) we are interested in the event
$$ \frac{\log r}{n}  < \frac{1}{2n} \log s\tag1$$
where
$$r = d_0 \times d_1 \times \cdots d_{n-1}$$
$$s = d_0 + d_1 10 + d_2 10^2 + \cdots d_{n-1} 10^{n-1}$$
Then we can assume $d_i$ are iid , uniform in $[1,9]$. Let $c_i = \log(d_i)$. Let $\mu_c =E[c_i] = \frac{1}{9}\sum_{k=1}^9 \log (k)=0.61775$. Also, because $0\le c_i <1$ the variance is bounded: $\sigma_c^2 <1/4$ (actually $\sigma_c^2 =0.0867\cdots$).
Then, for large $n$, the LHS of $(1)$ tends to a gaussian $N( \mu_c, \sigma^2_c/n)$
For the RHS:
$$\frac{1}{2n} \log(s)=\frac{1}{2n}(n + x_n) \tag2$$
where $x_n = \frac{d_{n-1}}{10}+\frac{d_{n-2}}{100}+\cdots$. Hence $0<x_n<1$
and $ \frac12 -\frac{1}{2n}<\frac{1}{2n} \log(s)<\frac12$
Then for large $n$ the LHS will concentrate around $\mu_c=0.61775$ and the RHS will be below $0.5$, hence $P_n \to 0$.

Some numerical results: