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The probability that a positive natural number drawn at random is odd, is $\frac{1}{2}$.

Is there a way of calculating this fact using the cardinalities of the sets? Or is there some measure of cardinality, or of the number of elements within each set, which captures this fact?

If we say the sample space is $\mathbb{N}$ and the successful outcomes are $O=\{2n-1:n\in\mathbb{N}\}$ then we might (obviously incorrectly) infer from the cardinalities of the sets that:

$(\lvert\mathbb{N}\rvert=\lvert O\rvert)\implies P(n$ is odd$)=1$

It would seem to me that the closest we can come is to say the densities of $O$ and $\mathbb{N}\cap O$ in $\mathbb{N}$ are equal to each other and make our deduction by that means. Is that the normal approach within set theory?

Robert Frost
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    "The probability that a positive natural number drawn at random is odd, is $\frac{1}{2}$." This entirely depends on the distribution using which you pick the numbers – dimpol Jan 23 '17 at 14:15
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    When you say "a positive natural number drawn at random" you should always say from which distribution. Actually, there is no uniform distribution over natural numbers, because the probability of choosing either one of them would be arbitrarily small, i.e. 0. – ajr Jan 23 '17 at 14:16
  • With any such problem you need to say precisely what you mean. One standard thing is to cap the selection at some large $N$ and then let $N$ tend to $\infty$. For finite $N$ you can make your cardinality argument work. – lulu Jan 23 '17 at 14:26

5 Answers5

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No, there isn't any way to do it using the cardinality of the set. The cardinality of the set of odd numbers, the cardinality of the set of prime numbers, and the cardinality of the set of natural numbers are all the same: $\aleph_0$. But obviously the probability of choosing an odd number, a prime number, or a number at all are not the same - respectively, they should be $\frac{1}{2}$, $0$, and $1$. Within the field of set theory, then, there really isn't a way to calculate the density.

You can define probability using limiting density: take $\lim_{n \to \infty}\frac{|X \cap n|}{n}$, where $X \cap n$ is the set of members of $X$ less than $n$. This limit will give the correct probabilities for everything I noted above. However, I wouldn't call this a set-theoretic approach - this is more number-theoretic in character, because it depends heavily on the order of the natural numbers.

Reese Johnston
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My answer is yes.

To do this requires the concept of density of a set of integers. There are a variety of densities.

Look them up.

A Google search for "density of sequence of integers" is a good start.

marty cohen
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This is a good idea, but there's a better way to think about probability on infinite sets than cardinality. Although in the particular case of sets of size $\aleph_0=|\mathbb{N}|$ density is used, it's important to note that density does not give a probability. It gives a limit of probabilities. In general, the correct generalization involves something called measure which you can read about here.

The reason we try to use other concepts for the particular case of $\mathbb{N}$ is that using measure only lets you tell the difference between finite and co-finite sets on $\mathbb{N}$. Things like "choose a number at random, is it even" wind up undefined. For sets larger than $\mathbb{N}$ such as $\mathbb{R}$ this stops being a problem for some complicated reasons, and in general measure is the preferred generalization. It lets you say "given two sets, $A\subseteq B$, what's the probability that we choose a point uniformly at random from $B$ and it falls within $A$?" And gives the satisfyingly familiar answer of $\frac{\mu(A)}{\mu(B)}$ where $\mu(S)$ is the measure of $S$.

In fact, probability is "properly" defined using measure!

Stella Biderman
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Something slightly deviates, but might still work on your case.

(Here natural numbers refers to positive integer).

Fix the sample space to be no greater than $n$ (denoted $P_n$), it is easy to show that

$\frac{1}{2}\le P_n(odd)\le \frac{n+1}{2n}$.

Interpretation in terms of infimum.

It can be easily shown $\inf\{P_n(odd)\} =\frac{1}{2}$.

However, this does not mean random select a number in $\mathbb{N}$ and the probability of oddity is 0.5.

Interpretation in terms of a limit.

Basic limit property shows $\lim\limits_{n\rightarrow \infty} P_n(odd)=\frac{1}{2}$.

However, the interpretation should be careful:

for all $\epsilon>0$, there exists $N\in \mathbb{N}$ such that, if $n\ge N$, then for the sample space $\Omega: \{i, i\in \mathbb{B}\ and\ i\le n\}$, and random variable $X: \Omega\rightarrow \Omega$ by $X(x)=x$, $|P_n(odd)-\frac{1}{2}|\le \epsilon$.

Robert Frost
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laziestperson1
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No. But one can use the concept of numerosities to alculate such probabilities.

First I should correct you. While it is true that the numerosity of odd numbers is half the numerosity of integers, it is not true for positive odd numbers that their numerosity is half of positive integers.

The positive integers have numerosity $\omega -\frac{1}{2}$, the positive odd numbers have numerosity $\frac{\omega }{2}$. Thus, their ratio is infinitesimally close to $2$ but not exactly $2$.

Anixx
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