Interesting mathematical proof that that 100% of all numbers contain at least one 3 and despite there being an infinite number of numbers that don’t…

How can this be? The solution is so surprising, it is difficult, if not impossible to believe that 100% of integers contain the digit three at least once. The simple fact that the number 8, for example, has exactly zero threes in it seems to dispute this.

Consider this: what percentage of the first ten numbers contains at least one three? That’s easy- ten percent; three and only three. What percentage of the first one hundred number contains at least one three? A slightly inflated nineteen percent. What percentage of the thousand numbers contains at least one three? Twenty-seven point one (27.1) percent.

The percentage of numbers with threes in them rises can be expressed as 1 – (.9)^n, where n is the number of digits. It reaches 99% at about the point where n has 42 digits.

The ratio of “threed” to “three-less” numbers at infinity would be 1 – (.9)^(Infinity), or 1.

It is interesting to note that there are *also* an infinite number of integers which do not contain the digit three. The simple progression “1, 11, 111, … ” illustrates this fact.

The Grey Labyrinth