Average product of the digits in a car's licence plate

Question

A car's licence plate is on the form

$$\underbrace{A...A}_{k\,\text{letters}} \quad \underbrace{19\ldots9}_{n\,\text{digits}}$$

What is the average product of the digits in a car number with $n$ digits? Note: the digits in a licence plate never starts with a leading $0$.

I was thinking that the product of every number containing a $0$ is $0$. And there are $9 \cdot 10^{n-1}$ digits in total, where $9^n$ does not contain a single zero. However, I am unsure if the average product of these numbers is simply $(4.5)^{n}$. Leading to what I suspect is the final answer

$$ \frac{(4.5)^n}{9 (10^{n-1} - 9^{n-1})} $$

Is this correct or is there something faulty with my logic?

Answer 1

The average product of $n$ nonzero digits is not $4.5^n$. For $n=2$ the average product is $25$, while $4.5^2=20.25$ Multiplying big numbers together makes the product much larger, so there is an upward bias.

Answer 2

Hint: if $X_1, X_2, \ldots X_n$ are independent, $\mathbb E[X_1 X_2 \ldots X_n] = \mathbb E[X_1] \mathbb E[X_2] \ldots \mathbb E[X_n]$. The average of $0,1,\ldots,9$ is $4.5$, but for the first digit the average of $1,2,\ldots, 9$ is $5$.

But I don't know in what jurisdiction the numbers on a license plate never start with $0$.