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My searching has come up with, at best, this question:
What's the fairest turn sequence for n players?
which as far as I can tell does not answer my question.

I have a game with 6 players - let's call them A, B, C, D, E and F. My goal is to create a sequence of turns such that over a long enough period, each player would play after each one of the other players the same number of times. In other words, for player A the subsequences BA, CA, DA, EA and FA should appear an equal number of times.

It is also important that each player gets an equal number of turns, and that they never have to wait too long between turns. I suspect both of these will be required anyway for a good solution, but I might as well mention them.

Double turns (e.g. ABCCB) are not allowed.

The "first player advantage" topic (discussed in the other question I linked to) is of a lesser concern, but if it's possible to integrate it into a solution it wouldn't hurt.

How do mathematicians approach this type of problem?

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Consider a directed graph on the vertices $A,B,C,D,E$ with a an arc from every vertex to every other vertex. The in-degree and the out-degree of every vertex is $4$. Therefore, there is an Euler cycle that traverses every one of the $20$ edges exactly once. If we set the turns according to this cycle, then all the criteria are fulfilled. That is, if $A$ has the first turn, and $AB$ is the first edge in the cycle, then $B$ gets the second turn. If the second edge is $BD$, then $D$ gets the third turn, and so on. When we get all the way through the cycle, we jus repeat it, as many times as desired.

Each player follows each other player once in $20$ turns, and so plays $4$ times in every $20$ turns.

There is a well-known algorithm for constructing an Euler cycle in a directed graph.

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  • $\begingroup$ OP asked about $6$ players, but the same approach works. $\endgroup$ – Ross Millikan Oct 5 '19 at 23:16
  • $\begingroup$ @RossMillikan I don't know where I got $5$ from; of course it works for any number. $\endgroup$ – saulspatz Oct 6 '19 at 0:00

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