On the Wikipedia page of Goldbach's conjecture, a [heuristic justification][1] is given, which did not completely satisfy me.  It roughly goes as follows:

> * randomly define a subset integers in accordance with the *prime number
> theorem*
> * Let $K_n$ be the random variable, counting the *number of ways* the
> natural number $2n$, can be written as
> a sum of two members of this set.
> 
> Then $E[K_n]\rightarrow \infty$ .

The problem is that, although the mean goes to infinity, it still might be true that the probability that all evens can be written in *at least one way* is zero. 

So I thought of a different heuristic, and I am curious about whether anything is known about it:

> Let $\mathcal P$ be the collection of
> all subsets of odd numbers whose
> density agrees with the prime number
> theorem, and let $\mathcal G$ be the
> collection of subsets for which
> Goldbach's property holds (i.e. every
> even number can be written in at least
> one way with two members of the set). 
> Let $\mu$ be the uniform product
> measure of the space $\{0,1\}^{\mathbb
> N}$.  Then the quantity  $$
> \frac{\mu(\mathcal P \cap \mathcal
> G)}{\mu(\mathcal P)} $$ 
>  is (significantly) greater than zero. 
> 
> Edit: As pointed out in the comments,
> $\mu(\mathcal P) = 0$, so this
> quantity is  meaningless as it is, but
> I think it can be formalized in some
> way.

 

I do not know if this is easy or almost as difficult as the original problem.  But it would be a very convincing heuristic for me in that, it would tell me *how much of Goldbach's conjecture is already explained by the prime number theorem*.

I would appreciate answers, or references to any known results, or reasons if this kind of heuristic is not relevant, if that is the case.

  [1]: http://en.wikipedia.org/wiki/Goldbach%27s_conjecture#Heuristic_justification