As some may know, Champernowne's constant is one of the only known constants proven to be normal. The number is constructed by concatenating whole numbers as you count up and appending them behind a decimal point. In base 10, the constant is 0.12345678910111213...
My main question here is that in order to be "simply normal", each digit must be equally likely to appear throughout the entire number. This, however, does not seem to hold for 0 since it's skipped in the counting. For instance, since we don't start counting at 0, the first time it appears is in the back of 10. The first time 00 appears, is in the back of 100. I used some Python code to write out the first 68888897 digits of the constant and then calculated the probabilities of each digit occurring.
Here is the output:
The number 0 occurs 5888896 times giving a probability of 0.085483964128501
The number 1 occurs 7000001 times giving a probability of 0.10161290577783529
The number 2 occurs 7000000 times giving a probability of 0.10161289126170796
The number 3 occurs 7000000 times giving a probability of 0.10161289126170796
The number 4 occurs 7000000 times giving a probability of 0.10161289126170796
The number 5 occurs 7000000 times giving a probability of 0.10161289126170796
The number 6 occurs 7000000 times giving a probability of 0.10161289126170796
The number 7 occurs 7000000 times giving a probability of 0.10161289126170796
The number 8 occurs 7000000 times giving a probability of 0.10161289126170796
The number 9 occurs 7000000 times giving a probability of 0.10161289126170796
As you can see, with this counting style, 0 seems underrepresented. That being said, I know the number has been proven to be normal so my question is this: if we were to take the constant and remove 1, 11, 111, 1111, etc. from the counting, would the number that remains still be normal? If not, why is the lack of zeros not an issue in the original constant? If it is, then how do the digits still get represented equally as the limit tends to infinity?