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51

Requests / Re: Uno Tournament 6

« Last post by Ian Wakeling 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.

52

Requests / Re: Uno Tournament 6

« Last post by Ian Wakeling 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.

53

Requests / Re: Uno Tournament 6

« Last post by schmidthaus 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? 

54

Requests / Re: Uno Tournament 6

« Last post by schmidthaus 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.  

55

Requests / Re: Uno Tournament 6

« Last post by Ian Wakeling 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.

56

Requests / Re: Uno Tournament 6

« Last post by schmidthaus 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!!

57

Requests / Re: King/Queen of the Beach

« Last post by Ian Wakeling on February 10, 2024, 03:50:29 AM »

I have answered your question here.

58

Requests / Re: Mix double 12 schedule

« Last post by Ian Wakeling on February 10, 2024, 03:41:57 AM »

The schedule below does balance the male-female partnerships, but at the expense of the imbalance of the opposition pairs.  Each player has one opposite sex player who they never oppose, and two other players who they oppose twice. For example, consider M1: the pair (M1 F3) never oppose, while (M1 M4) & (M1 F4) oppose twice.  Hope that helps.

[(M2 F4):(M5 F3)] [(M6 F2):(M3 F6)] [(M1 F5):(M4 F1)]
[(M5 F6):(M4 F5)] [(M3 F4):(M1 F2)] [(M2 F3):(M6 F1)]
[(M5 F1):(M6 F6)] [(M3 F2):(M4 F3)] [(M2 F5):(M1 F4)]
[(M2 F6):(M3 F5)] [(M1 F1):(M5 F2)] [(M4 F4):(M6 F3)]
[(M1 F6):(M6 F4)] [(M5 F5):(M3 F3)] [(M4 F2):(M2 F1)]
[(M5 F4):(M3 F1)] [(M4 F6):(M1 F3)] [(M2 F2):(M6 F5)]

59

Requests / Re: Mix double 12 schedule

« Last post by kitch3 on February 09, 2024, 03:34:16 PM »

Could you modify it so there is only 6 rounds and every male plays once with every female 

60

Requests / Re: 14 (possibly higher, up to 20) player, single court, 3,4 or 5 players per side?

« Last post by Ian Wakeling on February 09, 2024, 12:40:03 PM »

I am aware that you can have a balanced solution for 16 players and 30 games of 4 vs 4.


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

There might be something similar for 18 players and 51 games of 3 vs 3, and 20 players and 38 games of 5 vs 5.