Does almost every whole number integer contain any of the digits zero through nine?

Question

For example, how many whole numbers contain an eight? Well, for whole numbers less than ten, it's just eight itself, so that's 10% and for whole numbers less than 100, there are 8, 18, 28, 38, 48, 58, 68, 78, the 80s, and 98, so that's 19 out of 100, which is 19%; the percentage went up, and for whole numbers less than 1,000, we have those whole numbers again, but with a 1 through a 9 in front, counting all of the 800s. So, that's 19*9=171+100=271. 271/1,000=27.1%, so it's going up. We can do the similar thing for whole numbers less than 10,000: 9*271=2,439+1,000=3,439. 3,439/10,000=34.39%. It's going up every time. Because of this happening, is it possible that all whole numbers contain any of the digits zero through nine (just one out of the ten)?

Answer 1

The number of integers smaller than $10^n$ and containing at least one $8$ is equal to $$10^n - 9^n$$

This is because the number of integers without eights is $9^8$.

This means that the numbers with at least one eight represent $$1 - \left(\frac9{10}\right)^n$$ of all numbers, a number limiting to $1$ as $n$ becomes large.

Answer 2

Yes, as you look at larger and larger $N$, the fraction of numbers in the range $1,\ldots,N$ containing all digits in their decimal expansion approaches $1$. For example, if $N = 10^M$ then we are choosing $M$ digits at random. The probability that you don't contain all digits is at most $10(9/10)^M$, and this tends to zero as $M$ tends to infinity.