12 players play 4 games. In each game they play in groups of four, and, within each four, they play as pairs. So, in 4 games each player must have 4 different partners, and should play with as many of the 12 as possible (in their groups of four). Ideally, by the end, they would each have played in the same group of 4 as everyone else. Is there a method for this? I tried looking at Steiner systems but I got a bit lost.
The answer to the linked question works for lots of games, but I can't work out how to apply it to just 4. Tournament bracket for a 4-player game and 13 players in total
This is not a complete answer, but some comments. Suppose we have such a design in which everyone plays with everyone else. Each person plays in $4$ games each with $3$ other people, so she plays with one lucky person twice and everyone else once. Therefore the $12$ people consists of $6$ pairs each of whom play with each other twice and everyone else once.
Here is a solution without the "resolvable" aspect, i.e., the following is a set of $12$ blocks of size $4$ in a set of size $12$ such that every pair occurs together in a block, but the blocks cannot be organized into 4 lots of mutually disjoint blocks.
Consider the set $S = \{0, 1, 4, 6\} \subset \mathbf Z / 12\mathbf Z$. Observe that every nonzero element of $\mathbf Z / 12\mathbf Z$ is represented uniquely as a difference $x - y$ with $x, y \in S$, except $6$ which is represented twice (as $6-0$ and $0-6$). Therefore the set of $12$ blocks $\mathcal B = \{z + S : z \in \mathbf Z / 12 \mathbf Z\}$ have the property that every pair $x, y$ appears in exactly one block, except if $x-y = 6$ in which case they appear in two blocks.