One way of stating the Prime Number Theorem is that the "probability" that a large number near x is prime is 1/log(x). (Here, as always, all logs are natural.)
A heuristic derivation of the prime number theorem, Frank Morgan, via Andrew Gelman, with some embellishment by me: let P(x) be the "probability" that a large integer x is prime. Then how do P(x+1) and P(x) compare? Say x is prime, which it is with "probability" P(x); it "divides" x+1 (and all larger numbers) with probability 1/x. (Of course, this doesn't actually make sense; the idea is that we're modeling the numbers divisible by x as a random set with density 1/x, which should have the same large-scale properties.) Other than this, x and x+1 are equally "likely" to be prime. So the function P satisfies
(*) P(x+1) = P(x) [P(x) (1 - 1/x)] + (1 - P(x)) P(x)
since x+1 is "prime" with probability P(x) (1-1/x) if x is "prime", and P(x) otherwise. Divide through by P(x) to get
P(x+1)/P(x) = P(x) (1-1/x) + (1-P(x))
P(x+1)/P(x) = 1 - P(x)/x.
Now, P(x+1) can be approximated by P(x) + P'(x), so we have
1 + P'(x)/P(x) = 1 - P(x)/x
P'(x)/P(x) = -P(x)/x
which has the general solution P(x) = 1/(C + log x).
This is a bit of a lie. For one thing, the difference equation (*) actually seems to have solutions that differ from those of the differential equation by a constant factor, which seems to depend on the initial conditions. (This amounts to changing the base.) For another thing, the assumption that x+1 might be divisible by x is, um, stupid if we're actually talking about prime numbers. (It's probably possible to rephrase this heuristic derivation as a rigorous result about random sets, though.) Still, it gets the prime number theorem to within a constant factor, which isn't bad for such a simple argument.