There are a number of sets in math that have infinite cardinality but still have a superset. For example, the set of even integers has infinite cardinality, but it is a subset of all integers. The cardinality of the set of integers is infinity, but it has a superset, real numbers. These are just a few examples.

So my question is if a random number was chosen from one of these supersets, what is the probability that it would be a part of one of its infinite subsets? There is an infinity of possibilities that are inside the subset, but also an infinity that is not.

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    $\begingroup$ Why should the cardinalities of any of the sets be relevant? Probabilities are based on measure, not cardinality. $\endgroup$ – Sandejo Dec 25 '20 at 2:32
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    $\begingroup$ "The cardinality of the set of integers is infinity, but it has a superset, natural numbers." -- I think you mean to set that the superset of $\Bbb Z$ is $\Bbb Q$, probably. Otherwise, $\Bbb N$ is a subset of $\Bbb Z$. $\endgroup$ – Eevee Trainer Dec 25 '20 at 2:49
  • $\begingroup$ It depends, how do you pick them? Because the order changes the probability (asuming equiprobability), you can reorder a sequence so that the sum of the even cardinalities and the sum of the odd cardinalities stays in whichever proportion you decide, you can rig it. $\endgroup$ – yugikaiba Dec 25 '20 at 4:07

The probability of an integer $x$ (where $|x| < n$ for a fixed $n$) being even, when uniformly randomly chosen, in the limit as $n \to \infty$, is equal to $1/2$.

The reason for the phrasing of this is that you need to be very careful about how you define this.

"Uniformly randomly" essentially ensures each integer is equally likely. Hopefully this choice is obvious.

The reason we first consider $x \in \Bbb Z$ such that $|x|<n$ is because it doesn't make much sense to deal with infinite sets at the outset. For instance, there are infinite many integers, infinitely many even numbers, and infinitely many odd numbers. (All "the same infinity" if you appeal to cardinality, namely $\aleph_0$.) How should you define the proportion which are either? Naively, you could claim $$\frac{\text{cardinality of even integers}}{\text{cardinality of all integers}} = \frac{\aleph_0}{\aleph_0} = 1$$ but then (aside from being somewhat ill-defined) that claims all integers are even (or odd), which doesn't match up with intuition. So we first want to work with things we know: finite sets.

So, define $$S_n := \left\{x \in \Bbb Z \; \middle| \; |x|<n\right\} \qquad E_n := \left\{ x \in S_n \; \middle| \; x \text{ is even } \right\}$$ Obviously, $$|S_n| = 2n-1 \qquad |E_n| = 2\left\lceil \frac n 2 \right\rceil -1$$ Then $$\text{probability of $x \in S_n$ being even} = \frac{|E_n|}{|S_n|} = \frac{2 \lceil n/2 \rceil - 1}{2n-1}$$ (obviously uniformly randomly chosen). To generalize these conclusions to all $x \in \Bbb Z$, we take the limit $n \to \infty$, since $S_n$ becomes $\Bbb Z$ in the limit $n \to \infty$, and $E_n$ likewise represents all of the even integers in the limit. The limit of the ratio can be verified to be $1/2$: $$\lim_{n \to \infty} \frac{|E_n|}{|S_n|} = \lim_{n \to \infty} \frac{2 \lceil n/2 \rceil - 1}{2n-1} = \frac 1 2$$ This is also easily seen graphically:

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If you want to look further into stuff like this, you should look into topics such as measure theory, where these notions are discussed in more generality. Questions like these can be rigorized and handled best in that context, particularly once discussions are taken to probability theory.

You can also generalize this sort of process to your other questions. I just focused on the probability an integer being even since it's simple and intuitive. It's basically the same idea every time.


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