Math question: How does probability cope with this?

[Kuroneko]

I mean a random variable x with a uniform probability distribution P(x) where x is in the Cantor set (C). So given a random number from C, what's the probability that x is in some uncountable subset U of C?

The main problem with this is simply that it's not clear what it means for a distribution to be uniform in this case. We could generalize uniformity to be relative, i.e., a probability measure μ is uniform relative to measure ν iff μ = cν for some constant c. Our standard sense of uniformity would be relative to the Lebesgue measure (and indeed that's exactly what we're doing when we think of uniform probability as proportional to interval length), but of course the Cantor set has Lebesgue measure zero, so no probability measure over the Cantor set is going to be compatible with it in this sense.

As an example, let U be [0, .5] intersected with C (in other words, the first half of the Cantor set). Then it would naively seem that the chance of x being in U is 1/2, but I'm not sure how to show it.

Let's try it. Build the Cantor set C in the standard manner: let C_0 = [0,1] and C_{n+1} = {x/3,2/3+x/3: x in C_n}, thus removing the middle-third of each subinterval, and C be the intersection of all C_n. Let f:[0,1]→C be defined by Sum[ a_n/2^n ] ↦ Sum[ 2a_n/3^n ], where each a_n is in {0,1} and n>0. Finally, for a random variable X with uniform distribution over [0,1], let Y = f(X). Then Y takes values only in the Cantor set and has the additional property that for a fixed n, if takes any subinterval contained in any particular Cantor iterate C_n, the probability that Y is in that subinterval is directly proportional to its length.

This is something close to the sense you're using, and I suppose that it is reasonable to call uniform. The probability measure itself would be the μ(f^{-1}), where μ is the Lebesgue measure.

In other words, you cannot paint it - it has an infinite area and so you'd be there forever and never have enough paint.

You can't paint it only if you're required to have a uniform thickness of your coat of paint or otherwise have some unfortunate lower bound conditions on it.

1 doesn't mean 'certain to happen' and 0 doesn't mean 'certain not to happen'. They're infinitesimally close to being certain, so they're called 'almost certain'.

Indeed. That's why in analysis, 'almost everywhere' means 'everywhere but on a set of measure zero'. It's the exact same thing with probability, just with the additional condition that the measure is a probability measure.

[Apollonius]

So there actually isn't anything in the universe that we can say is certain not to happen? How can the proposition that the moon will one day turn into a purple Smurf be equally likely as those other things that you say have 0 probability yet still may occur?

[Apollonius]

Well, part of it is visualization, part of it is 'speaking the language', so to speak, and part of it is simply understanding the rules of symbol manipulation. We do it the way musicians hear notes or chord progressions inside their heads: years of practice.

I don't think understanding of math is entirely analogous to understanding of music. Sure, in both fields there are things that you learn by practicing often enough, and then there are things you don't really understand, but you accept that they are a certain way "just because", and you use them in the ways shown to you, because you know that they work.
But then, music is in a way easier, because what ever you do, there will be an immediate sensory feedback, either through your ears, or your fingers, or (if you're tone deaf) from your teacher's tortured (rarely delighted) facial expression. After some years of hearing what notes sound like when you actually play them, you will be able to recall their sound from memory by merely looking at a score, just like the word "elephant" creates the image of an elephant in your mind.
It's because of this that I believe that geometry is much more like music than other branches of math, because it's tangible.

For many years, I was unable to understand why (a+b)² = a²+2ab+b². It didn't make any sense to me. It just looks wrong. My teachers just said that's the way it is, you better learn it or fail the class. It wasn't until 10 years after I finished highschool, that I realised that you have to take that "squared" part literally. It's just a bunch of actual squares and rectangles, and you add them up, one by one. You can draw that on paper or in your head, and you see how it makes sense, because there is something you can actually look at, and it isn't symbols.
But how can you do the same with Int_c^c ρ(x)dx = 0? (Heck, I don't even have an idea what that means, so maybe it wasn't even a good example). How often do you feel like having to visualise shapes, movement and processes to help you do what you're doing?

[Kuroneko]

So there actually isn't anything in the universe that we can say is certain not to happen?

If you mean certain not to happen treating the laws of physics as certain, then sure. If, on the other hand, you mean certain not to happen in some more absolute sense, then no, nothing is ever absolutely certain.

How can the proposition that the moon will one day turn into a purple Smurf be equally likely as those other things that you say have 0 probability yet still may occur?

In terms of picking random numbers, an impossible outcome would have both probability zero and probability density zero.

For many years, I was unable to understand why (a+b)² = a²+2ab+b². It didn't make any sense to me. It just looks wrong. My teachers just said that's the way it is, you better learn it or fail the class. It wasn't until 10 years after I finished highschool, that I realised that you have to take that "squared" part literally.

You can--I've seen those algebra blocks, and they're nifty as teaching aids. However, you don't have to, since the distributive law is very intuitive: x(y+z) = xy+xz (don't believe it? just try with particular numbers). Then (a+b)² is just the the distributive law applied twice: (a+b)(a+b) = (a+b)a + (a+b)b = aa+ba + ab+bb = a²+2ab+b².

It's just a bunch of actual squares and rectangles, and you add them up, one by one. You can draw that on paper or in your head, and you see how it makes sense, because there is something you can actually look at, and it isn't symbols.

Well, one isn't going to get far in mathematics without a certain amount of skill at visualizing things, but speaking purely personally, I've always preferred symbols to geometrical thinking unless it is very inconvinient. Which is more appropriate depends on the problem, but a good example of a case where symbols are superior by far would be Surlethe's distributivity of cross-products problem. Geometrically, it requires a fairly complicated diagram and some explanation, but symbolically in components, it's just the distributive law. [1]

But how can you do the same with Int_c^c ρ(x)dx = 0? (Heck, I don't even have an idea what that means, so maybe it wasn't even a good example).

Imagine a graph of some function ρ(x). The symbol Int_a^b ρ(x)dx is the integral of ρ(x) with respect to x, which is the signed area of the region between ρ(x) and the x-axis (as in, negative portions of ρ(x) count as negative area) and x between a and b. This makes it intuitively obvious that the integral from c to c is zero, as you simply get a line segment of zero area in that case.

The fundamental theorem of calculus even has a very general and geometrically interesting form that relates integrals of a derivative over a region with the integrals over the boundary of that region. In the case of the region being an interval between a and b in the real line, the boundary is just two points {a,b}, and so we get Int_a^b[ dF ] = F(b)-F(a).

[1] For the curious, [A×(B+C)]^i = ε^i_jk A^j (B+C)^k = ε^i_jk A^jB^k + ε^i_jk A^jC^k = [A×B]^i + [A×C]^i. Here, superscripts represent vector components rather than exponents.