Average rank of elliptic curves, excluding those of low rank It's conjectured that, asymptotically, half of elliptic curves have rank 0, half have rank 1, and elliptic curves of rank $\geq 2$ have density 0. But what if we disregard elliptic curves of rank 0 or 1: Are there any conjectures about the average rank of elliptic curves of rank $\geq 2$?
More generally, for any integer $n \geq 2$, what value should one expect for the average rank of elliptic curves of rank $\geq n$?
 A: These types of questions are pretty speculative.  One should be aware firstly that there is no reason for there to be roughly equal numbers of rank 2 curves and rank 3 curves.
Mark Watkins has a paper where he comes up with a conjecture, using random matrix theory, that the number of rank 2 elliptic curves with conductor up to $X$ is asymptotically $c X^{19/24} (\log X)^{3/8}$.  The paper is: Mark Watkins, Some heuristics about elliptic curves.
Experiment. Math. 17 (2008), no. 1, 105–125. At the end of section 4 of the paper, he remarks that possibly there are around $X^{\frac{21-r}{24}}$ elliptic curves of rank $r$, for each $r \geq 2$, compared to around $X^{5/6}$ total elliptic curves.
A: Regarding the work of Watkins as mentioned by Matt Young, he has a different paper where the question of rank distribution in quadratic twist families is considered. There the RMT prediction for rank 2 is well shown by the Rubinstein data to demonstrate about $D^{3/4}$ twists of rank 2 (or more) in the even parity subclass, while Watkins suggests that 3/4 is too high for rank 3. He does not posit anything exactly, but the data for the congruent number curve, divided into 2 natural classes (Section 3.3) gives best-fit exponents of 0.44 and 0.55, so about $D^{1/2}$ is probably a best current guess for rank 3, though in initial regions of data collection the logarithmic factors can be difficult. 
http://archive.numdam.org/article/JTNB_2008__20_3_829_0.pdf‎
The natural linear extrapolation for these types of problems might be semi-valid at least initially, so $D^{1/4}$ for rank 4 and then some power of logarithm for rank 5 are as knowledgeable as guesses as any. But the 2-torsion plays a role here, and it is not clear whether it affects the exponent on the $D$-power. Watkins has a recent preprint (joint with 5 others) where data for the congruent number curve is given, finding "lots" of rank 6 examples, but "few" of rank 7 (and none of rank 8). Granville has a heuristic (see Section 4 loc. cit.) that might suggest ranks 5 and 7 are the correct cutoffs in the generic and 2-torsion cases.
http://magma.maths.usyd.edu.au/~watkins/papers/RANK7.pdf
But really no one has any factual idea, and a number of caveats can be listed, concerning specially parametrised (sparse) families. Indeed, the rank 28 example of Elkies starts from a rank 17 special family and he gains 11 from searching on specializations, and the same was approximately true for the NSA curve, they had rank 24 starting from I think a rank 13 family of Mestre or Nagao. So again 11 more than the family rank. However, generically it seems one should not expect more than 10 "small" points (polynomial height) on an elliptic curve except in such special families, and one actually reachieves the bound of 21 when appending 11 as above.
Edit: Elkies says that the NSA searched in a rank 11 family, so they beat the family rank by 13 in fact, see page 5 of his arxiv.org/pdf/0709.2908v1
A: A very simple random matrix heuristic says, based on the function field model, that the rank of a random elliptic curve is the dimension of the invariant subspace of a random element of $O(n)$ for large $n$.
We can easily compute the dimension of the space of matrices that preserve exactly $k$-dimensional subspace. We can write each such matrix as a nondegenerate invariant subspace of dimension $k$ plus an orthogonal matrix with no fixed vectors of dimension $n-k$. The dimension of the space of $k$-dimensional subspaces is $k(n-k)$, and nondegenerate ones are generic. The dimension of the space of matrices in $O(n-k)$ that with no fixed points is $(n-k)(n-k-1)/2$, because that's a dense set of one of the connected components. So the total number is $k(n-k)+(n-k)(n-k-1)/2 = (n-k)(n+k-1)/2$.
This dimension takes the same value at $k=0$ and $k=1$ but takes rapidly decreasing values after that. So this heuristic suggests that most elliptic curves of rank at least $n$ have rank exactly $n$.
