- 23 Jun '15 16:03Assuming you don't use leading zeroes, the chances that a given integer has a '9' in it is:

1 - (8/9)(9/10)^n

Where 'n' is the length of the integer in digits.

I guess you might take the limit of that as 'n' approaches infinity? But that would make it very close to "all of them" I think - 23 Jun '15 22:15

Yes.*Originally posted by forkedknight***Assuming you don't use leading zeroes, the chances that a given integer has a '9' in it is:**

1 - (8/9)(9/10)^n

Where 'n' is the length of the integer in digits.

I guess you might take the limit of that as 'n' approaches infinity? But that would make it very close to "all of them" I think

100% of the integers have at least one occurrence of the digit "9".

- 29 Jun '15 00:51

If N = 1*Originally posted by forkedknight***Assuming you don't use leading zeroes, the chances that a given integer has a '9' in it is:**

1 - (8/9)(9/10)^n

Where 'n' is the length of the integer in digits.

I guess you might take the limit of that as 'n' approaches infinity? But that would make it very close to "all of them" I think

1 - (8/9)(9/10) = 1 - (8/10) = 1/5

One fifth of all the one digit integers is not 9. Fail formula. - 05 Aug '15 02:26

Cantor cant count.*Originally posted by FabianFnas***But - and this is not a joke:**

Among all integers there are exactly as many numbers with a nine in them as they are numbers without a nine in them. No approximately but exactly! Not as n reach infinity, but when n *is* infinity!

This is provable. Cantor showed us how. - 05 Aug '15 18:19

Well the the answer should be 50%, and I don't think that's true...*Originally posted by FabianFnas***But - and this is not a joke:**

Among all integers there are exactly as many numbers with a nine in them as they are numbers without a nine in them. No approximately but exactly! Not as n reach infinity, but when n *is* infinity!

This is provable. Cantor showed us how.