Hypercube in infinite dimension I would like the following problem.
You’ve got a 10 x 10 x 10 cube made up of 1 x 1 x 1 smaller cubes. The outside of the larger cube is completely painted red. On how many of the smaller cubes is there any red paint? 
Normally the answer would be : $10^{3}-8^{3}$
If we generalize to dimension n, we would have : $10^{n}-8^{n}$
Now consider a that you have a ball inside the hypercube of n dimension. What is the probability that it would inside of the painted smaller cubes ? the answer would be $\frac{10^{n}-8^{n}}{10^{n}}$
If n tend to infinity, the probability is 1. How is that possible ? 
Can we go from a discrete probability on a finite cardinal to an infinite set like that ? is it mathematically accurate? How would you explain the result? 
 A: I don't know if my reasoning is correct, but consider this:
On a 3d cube, the cube 1 and 10 counting from every of three dimensions would have some color on it. So a hypercube of any dimension would have cube number 1 and 10 from every dimension that would have paint on it. When n is infinity, you have unit 1 and 10 from every one of those infinite dimensions that has paint on it.
Because inifinity is...infinite, you would have an infinite number of painted units (units numbered 1 and 10). But you would also have infinity of not painted units. Since in an infinity you can have one to one correspondence between one sequence and another sequence that goes to infinity, you would have a one to one correspondence of painted units to non-painted units. 
It doesn't matter that there is much less cubes numbered 1 or 10, than cubes from 2 to 9. Infinity is unmeasurable, and there is one to one correspondence on any sequence no matter how bigger the spaces between each number in one sequence than another are.
A: "How is that possible?" You are dealing with $n$ dimensions, with $n$ tending to infinity. Do not base yourself on intuitions from a 3D world.
Instead of cubes, I suggest you picture finite series of numbers between $1$ and $10$. You pick $1$ value for each term of the series. If one of the term of the series is either $1$ or $10$, you are on the edge of the cube. This gives a better intuition of the probability being close to $1$ for a large $n$.
You can go to infinity like that if certain conditions are verified, usually that the probability for infinite series is defined and exists. Here, you are going from a probability on a discrete set to a probability on an uncountable infinite set. The limit you are looking for can be defined as the integral of $f : [0,1] \rightarrow {0,1}$ where $f(x)$ equals $1$ iff there is no $0$ or $9$ in the (infinite) decimal notation of $x$. This is not considered a well-defined integral. 
Besides, $f$ is equal to $1$ on an uncountable infinite set, and to $0$ on another uncountable infinite set...
Unless someone did extend the definition of integral, or probability, for such cases, and did not tell me (which is possible, I don't know everything ever defined in mathematics), your approach is not mathematically valid.
It is up to you to do it for this case. But you will also have to prove plenty of properties in order for your result to mean something interesting.
