What is the probability that a randomly selected integer is even? Preface:
The point of this question is to find an effective means of defining the probability of a particular infinite set of events within another infinite set of events, using $2\Bbb{Z}\subset\Bbb{Z}$ as a simple (I hope) example. This is the first step towards answering more essential questions like "what is the likelihood that the solution to a differential equation can be obtained analytically?", which in turn form the basis for effective decision making (e.g. should I attempt an analytical solution?).
I would like to define the probability in such a way that the ordering of the set does not affect it (if this is at all possible). This is likely to require a different notion of 'size' from cardinality. In particular, while the cardinality of $\Bbb{Z}$ and $2\Bbb{Z}$ are the same, there is some sense in which the integers are 'larger' than the even numbers by virtue of being a superset.
The 'size' of a set, considered in this manner, must account for the number of distinct things in the set - i.e. replacement fails.
In the general case, the answer to the question "what is the probability that an element of $B$ is in $A$ given $A\subset B$ is trivially $0$ whenever $A$ is finite, and $1$ whenever $B\setminus A$ is finite, independent of the ordering of $A$ and $B$, because the probability of a finite number of events within an infinite number of events is infinitesimally small.
This approach fails when $B\setminus A$ and $A$ are both infinite.

That being said, I would argue that there is at least some probability $0<p<1$ such that $n\in\Bbb{Z}.P(n\in2\Bbb{Z})=p$. The reasoning is quite simple:
Q: Suppose that $n$ is a natural number. What is the probability that $n$ is an integer? A: $1$, because the natural numbers are a subset of the integers.
Generalization: if $A\subseteq B$, then the probability that a randomly selected element of $A$ is an element of $B$ is $1$ - if not, then it must be possible to select an element of $A$ which is not in $B$ (contradiction)$^1$
Q: Suppose that $n$ is a natural number. What is the probability that $n$ is imaginary?A: $0$, because no natural number is imaginary.
Generalization: if $A\cap B=\emptyset$, then the probability that a randomly selected element of $A$ is an element of $B$ is $0$ - if not, then the intersection of $A$ and $B$ must contain at least one element (contradiction)$^2$
From this, it seems reasonable to say that if $A$ is a nonempty subset of $B$ (such that $|A|=|B|$), then the probability that a randomly selected element of $B$ is in $A$ is greater than $0$ (because the intersection is nonempty) and less than $1$ (because the complement of $A$ is nonempty). In other words, if $A\subset B$, then the probability that $a\in A$ given $a\in B$ lies in the interval $(0,1)$.$^3$ As $2\Bbb{Z}\subset\Bbb{Z}$, the probability of a randomly selected integer being even should fall somewhere in the interval $(0,1)$.
Intuitively, I want to say that $p=0.5$, and I can provide a loose justification as to why this should be. However, I am hoping to find a more formal (and preferably consistent) means of addressing this and similar problems.

$^1$ The brief application of some nonstandard techniques potentially suggests that if the set $A\setminus B$ is nonempty but finite, then the [standard part of the] probability of a randomly selected element of $A$ being in $B$ is $1$. Conversely, the [st.p. of] probability that a randomly selected element of $A$ is in $B$ is less than $1$ iff $A\setminus B$ is not finite.
Caveat: The cardinalities of $A$ and $B$ must be interpreted as hyperintegers (the same cardinal may correspond to more than one hyperinteger) - hence $A$ and $B$ must be countable. This may or may not lead to a problem with the continuum hypothesis, but I haven't looked into it. For now, this is just a rough guess based on the properties of infinity discussed by Keisler & co.
$^2$ Again, a nonstandard interpretation would suggest that if $A\cap B$ is finite, then the probability of a randomly selected element of $A$ being an element of $B$ is effectively $0$. Conversely, the probability that a randomly selected element of $A$ is in $B$ is greater than $0$ iff $A\cap B$ is not finite.
$^3$ This only makes sense when neither $A$ nor $B\setminus A$ is finite.
A: As some of the comments point out, the usual way is to define a finite density and then take a limit. For example the probability of $k$ being even can be defined via:
$$p_n(k)=\frac{|\{2\mathbb{Z}\cap [n]\}|}{|\{\mathbb{Z}\cap [n]\}|}$$
with $p(k)=\lim_{n\rightarrow\infty}p_n(k)=1/2$. 
It’s important to understand that the resulting limit is NOT a probability because there is no uniform distribution on $\mathbb{Z}$. It’s just a density, which correlates with our intuition of probability. 
Unfortunately it’s not possible to make this density permutation invariant. For example, construct a bijection that looks something like $1,2,3,5,4,7,9,11,13,6,15,17,19,21,23,25,27,29,8,...$, eg where the even numbers get exponentially spaced apart. In this case the density will be 0 in the limit. 
