TITLE: Heuristically false conjectures
QUESTION [43 upvotes]: I was very surprised when I first encountered the Mertens conjecture. Define 
$$ M(n) = \sum_{k=1}^n \mu(k) $$
The Mertens conjecture was that $|M(n)| < \sqrt{n}$ for $n>1$, in contrast to the Riemann Hypothesis, which is equivalent to $M(n) = O(n^{\frac12 + \epsilon})$ .  
The reason I found this conjecture surprising is that it fails heuristically if you assume the Mobius function is randomly $\pm1$ or $0$. The analogue fails with probability $1$ for a random $-1,0,1$ sequence where the nonzero terms have positive density. The law of the iterated logarithm suggests that counterexamples are large but occur with probability 1. So, it doesn't seem surprising that it's false, and that the first counterexamples are uncomfortably large. 
There are many heuristics you can use to conjecture that the digits of $\pi$, the distribution of primes, zeros of $\zeta$ etc. seem random. I believe random matrix theory in physics started when people asked whether the properties of particular high-dimensional matrices were special or just what you would expect of random matrices. Sometimes the right random model isn't obvious, and it's not clear to me when to say that an heuristic is reasonable. 
On the other hand, if you conjecture that all naturally arising transcendentals have simple continued fractions which appear random, then you would be wrong, since $e = [2;1,2,1,1,4,1,1,6,...,1,1,2n,...]$, and a few numbers algebraically related to $e$ have similar simple continued fraction expansions. 
What other plausible conjectures or proven results can be framed as heuristically false according to a reasonable probability model?

REPLY [39 votes]: This is quite elementary, but surprised me when I first saw it, and I still think it's remarkable. 
The number of pairs of integers $(x, y)$ such that $x^2 + y^2 \leq n$ is asymptotically $\pi n$, since they are the lattice points inside a circle of radius $\sqrt{n}$. Therefore the average number of ways of writing a positive integer as a sum of two squares is $\pi$. Or $\pi/8$ if we regard solutions as the same when they differ only in signs or the order of the terms.  
One would therefore expect a positive proportion of the natural numbers to have a representation as a sum of two squares. Not a $\pi/8$-fraction, since some integers have several representations, but some slightly smaller positive density, since identities like $4^2 + 7^2 = 1^2 + 8^2$ look pretty much like random coincidences.
But actually almost no numbers are sums of two squares. Whenever the prime factorization of $n$ contains some prime $p\equiv 3$ (mod 4) to an odd power, $n$ cannot be a sum of two squares, as is easily seen by considering the equation modulo powers of $p$. And by Dirichlet's theorem, almost all numbers have some such prime to power 1 in their factorization.