[throwaway080383]

It's not accurate. Looking at the arxiv link (https://arxiv.org/abs/1702.02325), the idea is you fix some elliptic curve E, and you look at its "twists" E^d, where the parameter d is a nonzero integer.

If you only look at twists where |d| < N, you can ask "What proportion of these are rank zero, rank one, rank two, ...?" The Theorem in the paper is that as you let N go off to infinity, these proportions tend to 1/2, 1/2, 0, 0, 0, ... respectively.

Here's the thing: this does not correspond to a measure on the set of rational elliptic curves, and indeed, there is no reasonable way to define a uniform probability measure on a countable set. Consequently, statements like "half of all elliptic curves..." are kind of misleading and meaningless.

[throwaway080383]

To back up my last claim, let me prove that 100% of positive integers are even in the same spirit:

Fix any odd positive integer x, and consider its "twists" x{d}, which for positive d I define to be x*2^d. Every integer is of the form x{d} for a unique choice of x and d. Now, for fixed N, if I consider all "twists" for which d<N, the proportion which are even is (N-1)/N. Thus, as N tends to infinity, the proportion tends to 1.

[rocqua]

How does a twist x{d} being even equate to the integer d being even?

[throwaway080383]

The twist x{d} is even as soon as d>0.

[rocqua]

Yes, but that doesn't mean the integer d is even.

[throwaway080383]

I never say anything about d being even. Rather, as soon as d > 0, x{d} is even, so "almost all" x{d} are even. As you vary x and d, you cover all integers exactly once, so from this perspective, "almost all" integers are even.

My point is that this is very similar to how elliptic curves are being counted here.