Question: 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?

Answer: 
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 $x$](https://en.wikipedia.org/wiki/Prime-counting_function), we can study the ratio
$$\frac{\pi(x)}{x}$$
as $x$ grows large. By the [prime number theorem](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Measure_%28mathematics%29), 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](https://en.wikipedia.org/wiki/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.
There is a sense in which the probability of a random natural number being prime approximates to $\frac{1}{log N}$, as Paul and Daniel each get to, and subject to Daniel's elegant exposition of the measure issue.

It's worth bearing in mind that for most of the twentieth century, a lot of mathematicians (and in particular, statisticians) burdened themselves with a very narrow definition of probability: that it represented the frequency of an occurrence in a long-enough series of independent identical trials.

Using that definition, it isn't really meaningful to ask whether a **specific** random natural number is prime or not: it either is (probability=1) or it is not (probability=0); and until you check, you don't know whether you are the former state or the latter. Every time you check that specific number, you will always get the same answer. So the estimate of $\frac{1}{log N}$ isn't meaningful (and nor is any other estimate, though if you choose 0, you'd be right more often than not), using that definition of probability.

There is a broader, more useful definition of probability: one that is a superset of the above definition. And that is, that it is the quantification of our reasoned belief that a specific event will occur. Under this definition, $\frac{1}{log N}$ is a meaningful estimate, and may be useful.