Round Robin Tournament Scheduling

Round Robin Euchre for 17 Players

Jones · 6 · 5639

Jones

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on: January 23, 2011, 11:58:16 AM
Greetings,

My Friends and I have a monthly Euchre Party with typically 18 people.  However, once in a while someone can't make it so we're down to 17.  Are there any good scenarios for 17 players?  Maybe with 3 tables and 2 to 3 people sitting out per round.  It's an evening party, so more than 10 rounds gets a little late.  Everyone playing against each other is a plus but not a must.

Thanks in advance,
Jones


Ian Wakeling

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Reply #1 on: January 24, 2011, 10:43:20 AM
Hi,

Can you please confirm the numbers, as I am thinking with 17 people that there would be 4 tables of 4 with one person sitting out per round.  Perhaps it would help if you describe what normally happens when all 18 people make it to the party.

Ian.
« Last Edit: January 24, 2011, 10:45:54 AM by Ian »


Jones

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Reply #2 on: January 25, 2011, 04:32:59 PM
Sure.

Currently, when all 18 are there, we play 4 tables, 9 rounds, 2 players sit per round, each player sits out once.  

If we end up only having 17 people show up, we bump down to 16 players ( 1 person volunteers to sit out for the night) playing with 3 tables, for 12 rounds, 4 players sit out per round, Each player sits out 3 times. This tends to push the night later than we'd like.

Although, last time we had 17 we tried 4 tables with no byes for 8 rounds while the person sitting out for the night served drinks and subbed in as needed.  However, when we did this, a few rounds resulted in duplicate pairs.

If we went with 4 tables of 4 with one sitting out per round we'd end up playing 17 rounds which is more than the group would prefer.  However if this is the only way, than we can continue to drop to 16 as stated above.
« Last Edit: January 25, 2011, 04:39:17 PM by Jones »


Ian Wakeling

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Reply #3 on: January 26, 2011, 05:46:42 AM
As 17 is a prime number, then 17 rounds is indeed the only way to give everyone the same number of games.  However it is possible to come close to balancing the byes.  In the 10 round schedule below, all players, apart from player B, have exactly 7 games and 3 byes.  Player B, only has 2 byes and so gets one more game.  A player's partners are all different, and a player's opponents are all different.  The pairs GN and IJ never occur as partners or opponents, so the schedule is almost perfect.

  Table 1     Table 2     Table 3      Byes
(D M v N F) (K G v A L) (E Q v B O) (I H C P J)
(B H v I N) (E J v C D) (O P v A K) (M G L Q F)
(O M v B D) (C F v I H) (G L v P Q) (J E N K A)
(C G v B M) (Q N v A I) (F L v O J) (D K E P H)
(E K v J N) (A Q v M F) (P L v B C) (D I H G O)
(E B v A H) (C J v N O) (G P v I D) (K Q L F M)
(Q F v C K) (P A v J H) (O E v G I) (M B D N L)
(D A v C L) (G M v H K) (F P v E N) (O J I B Q)
(J F v G B) (L M v I E) (Q K v D H) (A N C P O)
(D K v I B) (L O v N H) (P J v Q M) (E G C A F)


There are some resources here that may be of interest if your numbers change.

Hope that helps.


Jones

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Reply #4 on: January 26, 2011, 08:42:22 PM
Thank you Ian.  Your knowledge is much appreciated.

I will be sure to present this to the group, however, I foresee it becoming an issue with player B getting an extra round.
I'm guessing we'll end up sticking with bumping down to 16 players with one person volunteering to sit out for the evening.  

As I stated earlier, we ran into an issue with duplicate pairs when we tried 16 players, 4 tables, 8 rounds, no byes.  Do you have any quick solutions to this?


Ian Wakeling

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Reply #5 on: January 30, 2011, 11:40:35 AM
It seems a little unfair to me that one out of the 17 ends up serving the drinks all night!   But here are my suggestions for 16 players.

The best solution I can offer is for 12 rounds:

(K H v F P) (B O v E N) (J D v L A) (G I v M C)
(L E v G M) (C P v F O) (K A v I B) (H J v N D)
(I F v H N) (D M v G P) (L B v J C) (E K v O A)
(J G v E O) (A N v H M) (I C v K D) (F L v P B)
(M N v P D) (F A v G C) (B E v K J) (L O v H I)
(N O v M A) (G B v H D) (C F v L K) (I P v E J)
(O P v N B) (H C v E A) (D G v I L) (J M v F K)
(P M v O C) (E D v F B) (A H v J I) (K N v G L)
(N L v F E) (P A v K G) (D C v O H) (J B v M I)
(O I v G F) (M B v L H) (A D v P E) (K C v N J)
(P J v H G) (N C v I E) (B A v M F) (L D v O K)
(M K v E H) (O D v J F) (C B v N G) (I A v P L)


above all the partners are different and all pairs of players oppose at least once and at most twice.

For 8 rounds you could use the following:

(J K v D E) (B H v L O) (M I v F A) (N C v P G)
(L J v P I) (N B v A C) (K H v M G) (D F v E O)
(A O v G J) (B M v D P) (C H v K F) (I L v N E)
(D H v P N) (E J v G C) (A B v I O) (F L v K M)
(B J v L K) (C O v M N) (D G v I F) (P H v A E)
(P B v F G) (H M v J I) (N A v K E) (O D v L C)
(A L v C F) (G N v J M) (E H v D B) (I K v O P)
(M O v E P) (D L v A G) (H I v K C) (N J v B F)


here the properties of the schedule are similar to the first, except that the pair BM never play in opposition.

Ian.
« Last Edit: January 30, 2011, 11:42:50 AM by Ian »