TITLE: When has the Borel-Cantelli heuristic been wrong?
QUESTION [52 upvotes]: The Borel-Cantelli lemma is very frequently used to give a heuristic for whether or not certain statements in number theory are true. 
For example, it gives some evidence that there are finitely many Fermat Primes, that is primes of the form $F_n=2^{2^{n}}+1$. Since the asymptotic density of the primes around $x$ is $\frac{1}{\log x}$, we expect that $$\text{Pr}\left(F_n\text{ is a prime number}\right)\approx \frac{1}{2^n}.$$ As the series $\sum_{n=1}^\infty \frac{1}{2^n}$ converges, Borel-Cantelli suggests that there will be finitely many Fermat Primes.
More examples:

Infinitude of the Mersenne Primes, since $\sum_{n=1}^\infty \frac{1}{n}$ diverges.
Heuristic justification for the ABC conjecture on Tao's blog.
Two different heuristics that there are only finitely many elliptic curves of rank greater than $21$. The first is due to Granville, and the second is recent work of of Garton, Park, Poonen, Voight and Wood.

In the case of Mersenne primes and Fermat primes, some major assumptions are made about independence, but even then, Borel-Cantelli can only ever be a heuristic since a "probability $0$ event" could still occur. 

My question is: How reliable is this heuristic? Are there any known cases/past conjectures where the Borel-Cantelli heuristic has been incorrect?

Edit: As Terence Tao mentions, the Borel-Cantelli heuristic fails for the $n=3$ case of  Fermat's last theorem due to algebraic structure. Examples like this are exactly what I am looking for.

REPLY [14 votes]: In your question you mention a heuristic that there are finitely many Fermat primes. However, an order argument shows that $2^{2^n}+1$ can only be divisible by primes of the form $k2^{n+1}+ 1$. This changes the estimate to infinitely many Fermat primes, and even predicts a constant fraction of Fermat numbers being prime(!)
Maybe an even more sophisticated heuristic brings the estimate back to finite. Either way, at least one of the heuristics must be wrong.
EDIT: But wait! The same order argument that shows that only primes of the form $k2^{n+1}+1$ can be factors, also shows that those primes are intrinsically more likely to be factors. A prime $p_k=k2^{n+1}+1$ divides $F_n=2^{2^n}+1$ if and only if $2$ has order $2^{n+1}$ modulo $p_k$. There are $2^n$ residue classes modulo $p_k$ with this order. So, the probability that $2$ is one of these residue classes, and hence the probability that $p_k$ is a factor of $F_n$, is heuristically $\sim 1/k$.
I believe this again predicts finitely many Fermat primes. At this point, I would not be surprised to see a yet more sophisticated argument predicting the opposite.