Quinn Angstrom
nearly every number has a 7 in it
Quinn Angstrom
if you selected a random number out of all numbers that exist, the chances that you'd pick one without a 7 in it are infinitely small
Master Chieve
6
Niamh Vibes
this is why he resigned
Quinn Angstrom
only 10% of single-digit numbers have a 7
Quinn Angstrom
but 19% of two digit numbers have a 7
Quinn Angstrom
each extra digit adds more ways for there to be a 7, and therefore a higher percentage of 7-having numbers
Quinn Angstrom
and each digit has ten times as many numbers in it as the last digit
Quinn Angstrom
so as you expand outwards it's weighed more and more massively in favor of every number having a 7
Quinn Angstrom
that being said, picking a random number out of all numbers is literally impossible
Niamh Vibes
That’s fascinating
Master Chieve
It’s easy to pick a random number out of all numbers
Master Chieve
Watch this
Master Chieve
14
Master Chieve
Bam. I’m a genius
Quinn Angstrom
you can't even pick a random number out of two
Master Chieve
Maybe you can’t
Master Chieve
I’m built different
Master Chieve
(Shitposting aside this did give me a “huh!” Moment)
Rama
So what proportion of base 10 natural numbers has each digit in it
Rama
It'd be .199999 repeating for any of them, right?
Quinn Angstrom
.9999999 repeating
Quinn Angstrom
for each one
Rama
Hm
Quinn Angstrom
consider this
Quinn Angstrom
what are the odds that a trillion-digit number wouldn't have a 5 in it
Quinn Angstrom
the average one would have about a hundred billion fives in it
Rama
Important clarification: are we talking about "at least one five" or "the expected number of 5s in a number"
Quinn Angstrom
at least one five
Rama
So, from 1-10, .1 numbers have at least one five
Quinn Angstrom
yes
Rama
1-100, .19 numbers have at least one five
Rama
1-1000, wouldn't it be .199 and that pattern extends to infinity?
Quinn Angstrom
0-999 has three places where it can slot a five, and each one happens 10% of the time
Quinn Angstrom
the overlaps are minimal
Quinn Angstrom
so it's about like .28-.29
Quinn Angstrom
to account for all xx5, x5x, and 5xx numbers
Rama
Okay so the odds of no 5 is (0.9)^number of digits
Rama
And the odds of yes 5 is 1-that
Quinn Angstrom
that sounds right yeah
Quinn Angstrom
so by 100 digits, the odds of no 5 are about 0.000026
Rama
So the answer is lim(x->infinity) 1-(0.9^x)
Quinn Angstrom
in other words, 1
Rama
Which I don't have a calculator handy but yeah it's probably .999999