TITLE: Probably true, but provably unprovable
QUESTION [35 upvotes]: I'm wondering if anyone has found, or can find, a sequence of statements $P(n)$ ($n \in \mathbb{N}$) such that:

Heuristic arguments using probability theory suggest that all the statements $P(n)$ are true.
One can prove in some widely accepted axiom system $X$, preferably by making the heuristic arguments rigorous, that "$P(n)$ holds for infinitely $n$".
One can prove in some widely accepted axiom system $Y$ that "$X$ cannot prove $\forall n P(n)$".

The goal here would be to find statements that are true 'just because they are probably true', not because we can put our finger on a reason why any individual one must be true.  
An example of such a sequence of statements might be 'if $p_n$ is the $n$th prime, there are infinitely many repunit primes in base $p_n$'.   However while this example meets condition 1 we seem far from having enough understanding of mathematics to prove 2, and 3 seems hopeless.  I think we need a less charismatic sequence of statements that are more carefully crafted to the task at hand.

REPLY [4 votes]: I will take the question as reworded in the comments: 1) asks for a plausible probabilistic argument for $\forall nP(n)$ using any framework you like, 2) asks for a proof in some widely accepted axiom system $X$ of "$P(n)$ for infinitely many $n$", and 3) asks for a proof in some widely accepted axiom system that $X$ cannot prove $\forall nP(n)$. 
Let us call a function $f$ from the positive integers to the positive integers Collatz-like if it is piecewise linear, the pieces being finite in number and depending only on the congruence class of the argument modulo some $m$. The ur-Collatz function is given by $C(n)=n/2$ if $n\equiv0\bmod2$, $C(n)=(3n+1)/2$ if $n\equiv1\bmod2$. Another example is given by $D(n)=n/2$ if $n\equiv0\bmod2$, $D(n)=(5n+1)/2$ if $n\equiv1\bmod2$. 
We make ourselves interested in the question, given a Collatz-like function $f$, is there a positive integer $s$ such that the sequence $s,f(s),f(f(s)),f(f(f(s))),\dots$ goes to infinity. 
Heuristically, $C(n)$ is, on average, much like $3n/4$, hence, decreasing, while $D(n)$ is much like $5n/2$, hence, increasing. Henceforward, we restrict our attention to those functions that are (heuristically, on average) decreasing. We enumerate them, and let $P(n)$ be the statement that there is no input to the $n$th such function such that the iterates of the function go to infinity. I find $\forall nP(n)$ plausible, on heuristic grounds. 
Now, infinitely many of our functions $f$ satisfy the condition $\forall mf(m)\le m$. The heuristic argument becomes a trivial proof that for all these infinitely many functions, $P(n)$. 
Finally, John Conway, Unpredictable Iterations, in Proc. 1972 Number Theory Conference, University of Colorado, Boulder, CO. 1972 , pp 49-52 (MR 52 #13717), proved that in the set of Collatz-like functions there are those for which $P(n)$ is algorithmically undecidable. So, we can't prove $\forall nP(n)$.