TITLE: Number of linear orderings of a set to have balanced frequencies of triple orders
QUESTION [6 upvotes]: Let $S$ be a set of $n$ elements and let $Q = (s_1, s_2, \ldots, s_n)$ be a linear ordering of $S$.  We write $s_i <_Q s_j$ when $s_i$ appears before $s_j$ in $Q$.
I want to construct a set (or possibly a multi-set) of orderings $\mathcal{Q} = \{Q_1, \ldots, Q_k\}$ such that for every $a,b,c \in S$, each of the $6$ possible ordering of $a,b$ and $c$ occurs with the same frequency in $\mathcal{Q}$.  Said differently, I want, for every $a,b,c \in S$, 
$$\Pr_{Q \in \mathcal{Q}}[a <_Q b <_Q c] = 1/6$$
Now, the set of all $n!$ orderings satisfies this property.
Can we make $\mathcal{Q}$ smaller?  What is the minimum number of orderings to get this property?  In particular, can $|\mathcal{Q}|$ be polynomial in $n$?
For $n =3$ and $n = 4$, six orderings suffice (see this stackexchange post).  It is tempting to conjecture that $6$ is always enough...but that'd be a bit surprising.

REPLY [5 votes]: With a colleague we are currently working on a similar problem: we want all 6 orderings to appear, though we do not care whether they happen with the same frequency. The lower bound transfers to your problem, however:
Let us assume that you have a "good" set of permutations $P_1,...,P_k$, and think of the positions of the value $1$ in all these permutations. For every value different from 1, store in a binary vector whether it appears before (noted 0) or after (noted 1) in the ordering. Because all 3-orders must appear, all n-1 vectors your built must be different. Thus, $k\geq \log_2(n-1)$. This technique is used, for instance, in this paper: http://arxiv.org/abs/1502.06888
To be sure that all 3-orders appear, we currently need $\leq 2.85\cdot \log_2(n)+O(1)$ permutations, and we are trying to close the gap. The construction wouldn't work for equal frequencies, however.