Round Robin Tournament Scheduling

Uno Tournament 6

schmidthaus · 8 · 1031

schmidthaus

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on: February 01, 2024, 10:30:03 PM
I have been hosting an annual Uno Tournament for 5 years now.  More and more people want to come each year and I'm struggling to find rotations for so many people.  I already have 27 responses and the party is over 3 weeks away.

Setup:
I have 6 tables, each table has a different color tablecloth (red, yellow, green, blue, black, white).  Each table has a different type of Uno game to play.  The tables can fit up to 6 people.  There are no partners, every person plays for themselves.  

Aims:
The primary aim is that everyone gets to play every kind of Uno (and therefore every table at least once).  The secondary aim is to have a good mix of players so people can meet many other people.  

I don't mind having people who have a bye each round (they get to refresh drinks and eat snacks!), plus the day of the party there are always last-minute cancelations that I can't control for.  


Ian Wakeling

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Reply #1 on: February 05, 2024, 02:49:51 PM
I am not aware of any resource that may provide useful schedules here.  Actually your scheduling problem is not easy because there is a trade off between playing each type of Uno once, and having a good mix of players.  As your priority is clearly the former, the mix of players must necessarily be compromised.  I can perhaps make a few schedules, but it sounds like you will not know the final number until the last minute, so I am not sure how much help I can be.

Here is an example.  Let's say you have 30 people and want 6 rounds of play with 6 tables of 5.  The following schedule might work:

        Table 1          Table 2          Table 3          Table 4          Table 5          Table 6
 R1 (10  6 18  3 12) ( 2 13 15 26 11) (14 23  7 22  5) ( 1 20  4 17 25) (21 16 19  8 27) (28 24 30  9 29)
 R2 (27  1 19 14  2) ( 7 10  8  9 28) (16 24  3 13 25) (30  6 22 11 15) (20 29 17 12  5) (23 21  4 26 18)
 R3 ( 9 15 22 26 16) (30 25 18 17 14) (19 29 10  4 11) (24 27 23 21 12) ( 6 13 28  1  7) ( 3  8  5 20  2)
 R4 (24 17  4  7  8) ( 3 22  1 21 29) (15 27 18 20 28) ( 5 26 13 19  9) (23 30 25 10  2) (11 16 14  6 12)
 R5 (11 28 25 21  5) (23 19 20  6 24) (26 12  8  1 30) ( 2  7 29 16 18) ( 4  9 14  3 15) (13 27 22 10 17)
 R6 (29 23 13 30 20) (27 12  4 16  5) ( 6  9 17 21  2) (28  3  8 14 10) (24 18 22 26 11) (15 25  1 19  7)

Above the players occur exactly once in each round/row and exactly once on each table.  No pair of players sit together at a table more than twice. Although it varies from player to player, each player will have some others whom they never meet, and a few players who they meet twice.  Does that seem useful?


schmidthaus

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Reply #2 on: February 11, 2024, 10:44:48 AM
It is, thank you!  You understood my predicament well.  As you surmised, there will have to be a compromise of mixing of players, and I'm good with that.  The party currently has 34 RSVPs, I will ask for recommitments a week out where I expect the number to drop, and then day of experience tells me that there will be a couple more drops.  

Because every table features a different version of Uno (Classic, Attack, Flip, Spin....you'd be surprised how many there are!) it is most important that everyone gets a chance to play every kind.  

What I chiefly want to avoid is what happened last year - I tried using a round-robin generator that is mainly used for baseball tournaments.  Each table was a "field" and I told it each field had 3 games each day, with no team playing twice in the same day (yielding 6 teams = 6 players at the table) and limited the number of weeks to the number of rounds just based on my experience of duration of the party.  

At a glance, this seemed to work but in execution, there was a problem.  It seems the generator had also assigned a home-field to teams.  We had a player that it turned out that 6 of 9 games he was at the red table and did not get to play all the different versions.  Some players had this happen to a lesser degree.  

So thank you for the above schedules that favor the table rotation first!  Thank you so much for your help!!


Ian Wakeling

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Reply #3 on: February 11, 2024, 12:18:19 PM
I have another idea.  A lot of the difficulty here is specifically because you have 6 tables, and this is just a difficult number to work with.  Are you able to find a 7th version of Uno?  If you can that's great, or if not still have 7 tables, but assign the last one as a bye, so it becomes the drinks and snacks table.  Now you can use the following:

(1  8 15 22 29 36) (2 10 18 26 34 42) (3 12 21 23 32 41) (4 14 17 27 30 40) (5  9 20 24 35 39) (6 11 16 28 33 38) (7 13 19 25 31 37)
(2  9 16 23 30 37) (3 11 19 27 35 36) (4 13 15 24 33 42) (5  8 18 28 31 41) (6 10 21 25 29 40) (7 12 17 22 34 39) (1 14 20 26 32 38)
(3 10 17 24 31 38) (4 12 20 28 29 37) (5 14 16 25 34 36) (6  9 19 22 32 42) (7 11 15 26 30 41) (1 13 18 23 35 40) (2  8 21 27 33 39)
(4 11 18 25 32 39) (5 13 21 22 30 38) (6  8 17 26 35 37) (7 10 20 23 33 36) (1 12 16 27 31 42) (2 14 19 24 29 41) (3  9 15 28 34 40)
(5 12 19 26 33 40) (6 14 15 23 31 39) (7  9 18 27 29 38) (1 11 21 24 34 37) (2 13 17 28 32 36) (3  8 20 25 30 42) (4 10 16 22 35 41)
(6 13 20 27 34 41) (7  8 16 24 32 40) (1 10 19 28 30 39) (2 12 15 25 35 38) (3 14 18 22 33 37) (4  9 21 26 31 36) (5 11 17 23 29 42)
(7 14 21 28 35 42) (1  9 17 25 33 41) (2 11 20 22 31 40) (3 13 16 26 29 39) (4  8 19 23 34 38) (5 10 15 27 32 37) (6 12 18 24 30 36)


This has best possible mix of players and it also has the property that it will work no matter how many players turn up (between 14 and 42).  Simply delete the unused player numbers from the schedule above.
« Last Edit: February 11, 2024, 12:19:51 PM by Ian Wakeling »


schmidthaus

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Reply #4 on: February 11, 2024, 04:33:31 PM
Actually yes, I have more than enough uno versions for a 7th table!  And it doesn't matter to me how many people per table as long as it is relatively close (like we wouldn't want 2 people at one table and another have 6).  

I hadn't thought about the 'bye' idea.  We've had 'bye' player numbers when someone no-shows at the last minute (grrr).  If a different solution isn't available, then the bye is a viable option as long as everyone has the same number of byes since they will be scored.  


schmidthaus

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Reply #5 on: February 11, 2024, 05:11:21 PM
Question about this: "Simply delete the unused player numbers from the schedule above."

In the past, I have a posterboard with a sign-in where the line number becomes their player number.  I've done both assigning in order, and letting them choose a random line (player) number.  Do I understand correctly that it would be best for this example to have them sign in sequentially, not randomly? 
« Last Edit: February 11, 2024, 10:26:58 PM by schmidthaus »


Ian Wakeling

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Reply #6 on: February 12, 2024, 03:33:46 AM
Yes, you are correct, I did intend that you signed people up sequentially.  There is something you may not have noticed with the schedule above, if you have 42 players, then each will play with 7 x 5 = 35 other players out of a possible 41 other players.  The 6 players whom a player does not meet are precisely the same players as the others within their own group of 7 - so if you are player 8 for example, you will not meet players 9 to 14; here the groups of 7 are the same as the 6 columns within any one table.  If people arrive in small groups because they have traveled together, and you assign sequentially then there is a good chance that the members of that group will not play each other - of course this might be what you want as the members of the group are more likely to play with people they don't already know.  If it is not what you want, then I would pick a minimum number of players that you are sure will turn up - say 28 - and then assign the first 28 at random and then go sequentially after that.


Ian Wakeling

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Reply #7 on: February 12, 2024, 03:44:04 AM
Regarding the question about the numbers at each table. This should be as balanced as possible as long the player numbers you don't use are the high numbers.  So if you used player numbers 1 to 30, then each round should have 5 tables of 4 and 2 tables of 5.