How can the probability that a number contains the digit 3 be 1? Based on this Numberphile video which claims almost all integers contain a $3$, I have a few questions on the reasoning behind recurring decimal numbers like $0.9999\ldots =1$
What they have shown is that $$\lim_{n \to + \infty} \frac{10^n-9^n}{10^n} = 1$$
this basically means that probability of eg. a $3$ occurring in a set of numbers like for $1-10, 1-100,$ increases as the upper bound gets large.
So you are more likely to see a $3$ when you take $1-100000$, than $1-10$ as the probability gets higher.
So what I would like to know is as '$n$' approaches $∞$ does probability of seeing a '$3$' equals $0.99999....$?
But since $0.9999... = 1$ wouldn't this not make sense, since there are infinitely many numbers that do not have a '$3$'?
All I need is for an explanation as to why this logic is wrong. Simpler answers are most appreciated.
Note 
I am not looking for the reason as to why 0.9999...=1.
 A: Your question seems to be "How is it possible for an event to have 100% probability if there are exceptions?"  The answer is, this is just one of many counter-intuitive situations when dealing with infinite sets.
With finite sets, if an event has any exceptions, its probability is necessarily < 100%.  However, with infinite sets, it's possible for for there to be some exceptions (even an infinite number of exceptions) and the event to still have 100% probability.  In other words, with infinite sets, "is true with probability 1" and "is always true" have different meanings!
See almost surely for more information.
A: You seem to think that the numbers $0.9999999\dots$ and $1$ are two different things. They are not. They are the exact same thing. The difference between $0.999\dots$ and $1$ is the same as the difference between "The third round rock rotating around the sun" and "the Earth". They are two different ways of representing the exact same thing.
What the video shows is that the limit of the probability of seeing a $3$ is equal to $1$. That does not mean that the probability itself is equal to $1$ for any single value of $n$.
What it does mean is that if $n$ is really big, then the probability is really close to $1$. It doesn't equal $1$, of course, since you can always pick a number with no threes in it.
A: The probability increases as the range increases, like you say; the probability that a 3 appears when we choose a random number between 1 and 100000 is much greater than when we pick a number between 1 and 100.  As we let the range increase, the probability increases; when we do not have an upper bound, the probability is $0.999\ldots = 1$.  So yes, the probability that, taking a completely random integer, it has a 3 as one of its digits is 1.
Ok now wait; there are lots of integers without a 3 (infinitely many), so how can this be?
The problem is the interpretation of "probability 1". We tend to think this means that taking a random number, it would be impossible for it not to contain a 3 (which is clearly not true).  But this interpretation only works when we are talking about the probability of an event from a finite sample space. When the number of possibilities are infinite (as in this situation, where there are infinitely many integers to choose from), this has a slightly different meaning. It means that the event happens "almost surely". So taking a random number, the probability it does not contain a 3 is zero, but is not impossible. It would just be like splitting an atom when you throw a dart at a dartboard.
A: Yep, there are infinitely many numbers without a $3$, but there are many more with a $3$.
Indeed, in $n$ digit numbers, $9^n$ of them have no $3$, while $10^n-9^n$ do. The ratio is
$$\left(\frac{10}9\right)^n-1$$ which tends to infinity exponentially, meaning that the numbers without a $3$ become more and more rare.

Also note that "the probability of seeing a $3$ equals $1$" doesn't mean that it is absolutely impossible to have no $3$.
A: The logic is both right and wrong, depending what kind of math you're using. But I'm a programmer, rather than a mathematician, so take what I say with a pinch of salt.
In programming at least, once you start working with different infinities, perhaps using libraries which allow you to handle the Aleph numbers of well ordered infinite sets, you also start playing with different kinds of zeroes, or rather, with different kinds of infinitesimals, which get all called "zero" in almost all branches of mathematics (algebra, calculus), in the same way as they treat all types of infinity as a single "inverse-zero", 1/0, the conceptual "upper bound" of the integers.
Once you start using math that is aware of relative infinities, then 0.999... is not one, it's just infinitesimally far from one. The fact that there are two ways of thinking about this, mathematically, is why there are so many arguments about it: as in all great arguments, both sides are right.
If you are using math which is aware of infinite sets, then 0.999... != 1, but 0.999... == 1 - (1/infinity).
If we randomly select an integer, the chance of there being few enough digits in it that we could express just its magnitude (even using shortcuts like powers of powers, terms like "googol" and "graham's number", or knuth's up-arrow notation) in a single lifetime is infinitesimal: there are only a finite number of magnitudes that we can express, and an infinite number of magnitudes.
Since "everyone knows" that dividing any integer by infinity gives zero, then there are apparently zero integers that we can express. But using math which is aware of different infinities, there aren't zero: just an infinitesimally small fraction of integers that we can express.
The chance of it being a number without a 3 in it, is infinitely larger than that. Because there is an infinite quantity of numbers with no 3s in, and only a finite list of numbers that could not be expressed by humans.
But it is still infinitesimally small, because for every number with no threes, there is an infinite number that does contain threes.
Picturing no-threes is hard. I find this easier to picture when considering the "countable" integers and the uncountable 'real' numbers between. It's clear to me that there are infinite integers; between each of those, there are infinite real numbers.
The chances of hitting an integer when panning through the numbers from 0.5 to 1.5, are 100%. The chances of randomly selecting an integer in a random pick form that range are infinitudinously small.
