When generalizing the Whist problem to schedules with byes, then I think the one thing that it is important to plan for, is that all players should play the same number of games, otherwise the schedule will be perceived as unfair by the players who have fewest games.

Definition:

A whist schedule with byes WSB(4t+b) is a schedule for 4t+b players and 4t+b rounds of play. Each round consists of t tables of whist play, and b players (1<=b<=3) who receive a bye. The schedule should have the following properties:

- Each player plays in exactly 4t rounds and has exactly b bye rounds.

- A player's 4t partners are all different.

- Each player opposes every other player at least once and at most twice.

- Pairs of players who never partner should play in opposition twice.

If all of these conditions are met then the schedule will be have optimal social balance, in the sense that the number of times players sit together at a table, either as partners or opponents, is at least twice and at most three times. Other properties that may be desirable are:

- Players should never have byes in consecutive rounds.

- Pairs of players who both have a bye in the same round, should never have byes together again.

When b=1 then the situation reduces to the classic whist schedule for 4t+1 players that you have already seen. If b=2 or 3, then things are more interesting, for example here are some cyclic solutions for 10 and 11 players:

Table 1 Table 2 Byes

( 1 5 v 7 10) ( 6 4 v 9 3) ( 2 8)

( 2 1 v 8 6) ( 7 5 v 10 4) ( 3 9)

( 3 2 v 9 7) ( 8 1 v 6 5) ( 4 10)

( 4 3 v 10 8) ( 9 2 v 7 1) ( 5 6)

( 5 4 v 6 9) (10 3 v 8 2) ( 1 7)

( 7 6 v 4 8) ( 1 9 v 5 3) (10 2)

( 8 7 v 5 9) ( 2 10 v 1 4) ( 6 3)

( 9 8 v 1 10) ( 3 6 v 2 5) ( 7 4)

(10 9 v 2 6) ( 4 7 v 3 1) ( 8 5)

( 6 10 v 3 7) ( 5 8 v 4 2) ( 9 1)

Table 1 Table 2 Byes

( 6 4 v 7 2) (11 1 v 9 5) ( 8 3 10)

( 7 5 v 8 3) ( 1 2 v 10 6) ( 9 4 11)

( 8 6 v 9 4) ( 2 3 v 11 7) (10 5 1)

( 9 7 v 10 5) ( 3 4 v 1 8) (11 6 2)

(10 8 v 11 6) ( 4 5 v 2 9) ( 1 7 3)

(11 9 v 1 7) ( 5 6 v 3 10) ( 2 8 4)

( 1 10 v 2 8) ( 6 7 v 4 11) ( 3 9 5)

( 2 11 v 3 9) ( 7 8 v 5 1) ( 4 10 6)

( 3 1 v 4 10) ( 8 9 v 6 2) ( 5 11 7)

( 4 2 v 5 11) ( 9 10 v 7 3) ( 6 1 8)

( 5 3 v 6 1) (10 11 v 8 4) ( 7 2 9)

I think many such schedules for b=2 and b=3 exist, although I have no general proof of this.