Why does "the probability of a random natural number being prime" make no sense?

I read an essay about prime numbers. In it the author suggests that a natural first question to ask is,

What is the probability of a random natural number being prime?

but proceeds to dismiss it as "not making sense". I wonder what's wrong with the question. Is its meaninglessness related to the fact that there is an infinite number of naturals?

Solution 1:

There's no uniform random distribution on the natural numbers, so the problem is with the phrase "random natural number". One cannot, for example, "pick a natural number at random", a fact which can be illustrated by the following thought experiment: How many digits should a natural number picked at random have?

On the other hand, we can view the question asymptotically: If $\pi(x)$ is the number of prime numbers less than $x$, we can study the ratio $$\frac{\pi(x)}{x}$$ as $x$ grows large. By the prime number theorem, $$\lim_{x \to \infty} \frac{\pi(x)}{x} \ln(x) = 1,$$ so since $\ln(x)$ is unbounded, the percentage of numbers less than $x$ that are prime becomes arbitrarily small as $x$ becomes arbitrarily large. In the asymptotic sense, then, the "probability of a natural number being prime" is zero, because $$\lim_{x \to \infty} \frac{\pi(x)}{x} = 0.$$

Edit: Let me elaborate a bit on why there isn't a uniform random distribution on $\newcommand{\N}{\mathbb{N}}\N$ (or on any countably infinite set, for that matter). To talk about probability, we must have a measure, i.e., a function $\mu$ sending "nice" subsets of $\N$ to $[0, +\infty]$. (What constitutes a "nice" subset is a technical detail that doesn't really matter here; actually, as with topological spaces and open sets, which sets are measurable is part of the data of the measure space.)

The main axiom for measures is that they're countably additive, that is, additive over countably infinite, pairwise disjoint collections of sets. So, if $A_0, A_1, A_2, A_3, \ldots$ is a collection of measurable subsets such that $A_i \cap A_j = \emptyset$ whenever $i \neq j$, then $$\mu\left( \bigcup_{i \in \N} A_i \right) = \sum_{i \in \N} \mu(A_i).$$ If we try to define a uniform measure on $\N$, let $A_i = \{i\}$. Since the measure is uniform, we must have $\mu(\{i\}) = \mu(\{j\})$ for all $i, j \in \N$. So by countable additivity, $$\mu(\N) = \mu\left( \bigcup_{i \in \N} \{i\} \right) = \sum_{i \in \N} \mu(\{i\}) = \sum_{i \in \N} \mu(\{0\}).$$ If $\mu(\{0\}) = 0$, then $\mu(\N) = 0$, so $\mu$ assigns zero to every set. If $\mu(\{0\}) > 0$, then $\mu(\N) = +\infty$. But a probability measure is, by definition, a measure such that the whole space has measure 1. Therefore, there is no uniform probability measure on $\N$.

On the other hand, we can often define uniform probability measures on uncountably infinite sets with no problem. For example, the Lebesgue measure on $[0, 1]$ is defined so that $\mu([a, b]) = b - a$ for any $0 \leq a < b \leq 1$. The reason this works is that measures don't have to be uncountably additive, just countably additive.

Solution 2:

The main issue with the question is that the density of primes decreases over increasing upper-limit interval subsets of $\mathbb N$. In fact, taking the density of the primes over the entire set of natural numbers evaluates to zero, therefore the probability would technically be zero. It does makes sense to say, for a given $n\in \mathbb N$, what is the probability of a number $q\in [1,n]$ being prime? And then the question has an exact, nonzero answer as the ratio of primes less than or equal to $n$ ($\pi(n)$) compared to $n$: $\pi(n)\over n$.