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.

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    $\begingroup$ I don't understand the analogy with complex numbers. As to the given question, the simplest approach is to let $p_N$ be the usual probability when you limit the integers to those of absolute value less than $N$. Then try to take the limit as $N\to \infty$. That works perfectly well here, and you get $\frac 12$. No need for any abstract apparatus. $\endgroup$ – lulu Jun 28 at 21:22
  • $\begingroup$ @lulu That would suggest that an alteration to the ordering of the integers would alter the probability that a randomly selection integer is even. Ex $\Bbb{Z}=1,-1,3,-3,5,-5,7\ldots,2,-2,4,-4,6,\ldots\implies P(n\in 2\Bbb{Z})=0$ $\endgroup$ – R. Burton Jun 28 at 21:25
  • $\begingroup$ See natural density $\endgroup$ – Jair Taylor Jun 28 at 21:25
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    $\begingroup$ Yes, of course. If you don't specify an ordering, I have no idea how you'd distinguish one infinite set of integers from another. Set theoretically, all infinite countable sets are the same. $\endgroup$ – lulu Jun 28 at 21:27
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    $\begingroup$ @copper.hat : It seems OP wants to somehow implement the idea of an uniform distribution to the integers, without running into the obvious obstacle. Thus the flight into NSA ideas. $\endgroup$ – LutzL Jun 28 at 23:27

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.

  • $\begingroup$ Would it be possible to take an 'average' of the density over the set of all possible permutations? Perhaps using calculus (seeing as how the set of such permutations is uncountable)? $\endgroup$ – R. Burton Jun 28 at 22:41
  • $\begingroup$ @R.Burton: Neat question. If I'm not mistaken, that would always give 1/2. Every permutation that gives you density $x$ in the limit, has a complementary permutation that gives you $1-x$. I'm a bit hazy on whether that's a bijection though. $\endgroup$ – Alex R. Jun 28 at 23:11

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