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 $$0 such that $$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.

• 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. – lulu Jun 28 at 21:22
• @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 $\Bbb{Z}=1,-1,3,-3,5,-5,7\ldots,2,-2,4,-4,6,\ldots\implies P(n\in 2\Bbb{Z})=0$ – R. Burton Jun 28 at 21:25
• – Jair Taylor Jun 28 at 21:25
• 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. – lulu Jun 28 at 21:27
• @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. – 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.
• @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. – Alex R. Jun 28 at 23:11