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?
 A: 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 $n$: $\pi(n)\over n$.
A: As in the other answers, indeed, there is difficulty in defining probabilities on somewhat-sparse subsets of infinite sets.
Nevertheless, in practice, there is a version of this question that has some substance, namely, to ask the "probability" that a number chosen "near" a given large-ish integer $N$ is prime, giving an answer depending on $N$. The prime number theorem essentially says the "density" of primes "near" $N$ is $1/\log N$, so a heuristic notion of this probability is $1/\log N$.
To make this more than a heuristic is not so difficult, using the prime number theorem, but of course it can also be rendered nonsensical if one desires, depending on how skeptical one wishes to be about probabilities in such situations. 
A: 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.
A: 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.
A: The probability that a random number n is prime is either 0 or 1; it will take us some time to find out which. In that sense probabilities make little sense.
But let’s take some large n, then count the primes p with n - n^1/2 < p < n + n^1/2, and calculate the probability that a random p in that range is a prime. The exact probability will depend on n, but it will be close to 1 / ln n, and the larger n gets the closer it gets to 1 / ln n.
So if I first pick a large n, then the probability that a random p in a large enough circle around n is prime is about 1 / ln n, depending on n and the size of the circle.
(This doesn’t work with any fixed size of the circle, we know there are intervals of size say 2,000,000 containing no primes at all and can even write them down. I haven’t proven that a size of n^1/2 is big enough but my guess is that it is).
