TITLE: Non-enumerative proof that, in average, less than 50% of tiles in domino tiling of 2-by-n rectangle are vertical?
QUESTION [20 upvotes]: 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?, 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).

REPLY [5 votes]: Essentially also a reformulation:
Tilings of $n\times 2$ with vertical and horizontal dominos are in bijection with Zeckendorf expansions of integers in $[0,F_{n+1}-1]$ (perhaps up to a small shift of $n$), expanded to the correct length by adding leading zeroes: From left to right write $01$ for two parallel horizontal tiles and write $0$ for a vertical tile. The Zeckendorf expansion of a random integer in $[0,F_{n+1}-1]$
involves $F_n$ with asymptotic probability $1/\omega^2$ (for $\omega=(1+\sqrt{5})/2$ the golden number). The inequality $2/\omega^2>1/\omega$ implies now the result (and, more precisely, the probability for a vertical tile is thus indeed $\omega/(2+\omega)=1/\sqrt{5}$).