On the Wikipedia page of Goldbach's conjecture, a heuristic justification 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 waysthe 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 $K_n>0$ for all $n$ 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.

`$\{1,\ldots,n\}$`

with the right density, $\mathcal{G}_n$ those for which Goldbach's property holds for all integer less than or equal to $n$, and estimate the probability of $P\in\mathcal{G}_n$ for $P$ drawn uniformly in $\mathcal{P}_n$. The answer of Charles should apply to this setting. $\endgroup$ – Benoît Kloeckner Jul 12 '10 at 20:39