How do you seed a "non-binary" tournament?

Question

In a typical tournament, seeding is arranged to provide proportionate advantage to competitors based on their perceived relative ability. (Good teams get to play bad teams, bad teams have to play good teams, mediocre teams face mediocre teams.) This philosophy results in the familiar tournament seeding (for 16 competitors) shown below:

(1 vs. 16) (8 vs. 9) (4 vs. 13) (5 vs. 12) (2 vs. 15) (7 vs. 10) (3 vs. 14) (6 vs. 11)

But what if there are more than 2 competitors per match, without teams, with only one winner per match per round - say, a game where there are several people sitting around the table but only one person who advances? (Chinese checkers could be done this way.)

For example, how would we seed a tournament with 9 competitors, where every match had three competitors and one winner, and we're still trying to provide proportionate advantage/disadvantage to each player? Is it as simple as

(1 vs. 9 vs. 6) (2 vs. 8 vs. 5) (3 vs. 7 vs. 4)?

Expanding this to 27 competitors would yield:

[(1/27/18) (9/19/10) (6/22/13)]

[(2/26/17) (8/20/11) (5/23/14)]

[(3/25/16) (4/24/15) (7/21/12)]

Or does that not provide sufficient advantage/disadvantage? In this scenario, 1 has a greater advantage than 2 (that is, 1's average opponent in each round is weaker than 2's average opponent), and 2's is greater than 3, up to 9. But 10-18 each have the same average difficulty of opponent, as do 19-27. In other words, 1-9 are well-differentiated, but the 18th seed is no worse off than the 10th seed, and the 27th seed is no worse off than the 19th, at least in round one.

Is there a way to maintain fairness, while achieving complete differentiation of seeding advantage - with every seed being in a more advantageous (not just equally advantageous) situation than every seed below it, and vice versa - the way we can with a "binary" tournament?

Can we extrapolate this to 4 competitors per match, 5 per match, n per match?

Answer 1

My suggestion would be to ensure that the seed numbers in each round add up to a constant (assuming the highest seeds go through to the next round). So for example with $9$ seeds and contests of $3$, you could use numbers adding up to $15$:

(   1   5   9   )
(   2   6   7   )
(   3   4   8   )

and with $27$ players and contests of $3$, you could use numbers adding up to $42$:

(   1   14  27  )
(   5   12  25  )
(   9   10  23  )

(   2   18  22  )
(   6   16  20  )
(   7   11  24  )

(   3   13  26  )
(   4   17  21  )
(   8   15  19  )

This will meet your desire to see every seed being in a more advantageous situation than every seed below it, since the sum of its opponents' seed numbers in any round will be equal to the standard sum for that round minus its own seed number