Is there a non-enumerative proof that, in average, less than 50% of tiles in domino tiling of 2-by-n rectangle are vertical?


It is a nice exercise with rational generating functions (or equivalently, linear recurrence relations) to show that for a random domino tiling of a $2\times n$ rectangle, with $n$ large, we can expect about $\frac{1}{\sqrt{5}}\approx 44.7\%$ of the tiles to be vertical. In the spirit of [Non-enumerative proof that there are many derangements?](https://mathoverflow.net/questions/86118/non-enumerative-proof-that-there-are-many-derangements), I wonder if there is an easy way to see, without computing this fraction, that this average must be some constant < 50%.


**EDIT**: Maybe to sharpen the request, is there any heuristic which works here and also carries over to the case of a $k\times n$ rectangle (with say $k$ fixed and $n$ large).



Non-enumerative proof that, in average, less than 50% of tiles in domino tiling of 2-by-n rectangle are vertical?
Here is at least a heuristic argument for why we should expect fewer vertical dominoes than horizontal dominoes. As we increase the length of the strip (to the right, say), let us think about how the rightmost four squares are covered. There are three cases: (1) two horizontal dominoes; (2) two vertical dominoes; (3) the last two squares are covered by a vertical domino and the preceding two squares are covered by horizontal dominoes (that "stick out" to the left and cover two additional squares).


Imagine I'm trying to build one copy of every tiling of the $2\times n$ strip and I'm trying to gauge whether I'll need to buy more horizontal dominoes from Harry or more vertical dominoes from Victoria. For every tiling of a $2\times (n-2)$ strip I'll need two horizontal dominoes for case (1) and two vertical dominoes for case (2); this is a tie. But for case (3), I need two horizontal dominoes and only one vertical domino for each tiling of a $2\times (n-3)$ strip. So I'm going to need to buy more dominoes from Harry than from Victoria.


