Round Robin Tournament Scheduling

### Variation on Social Golfer - Poker League

rye-train · 6 · 4915

#### rye-train

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• Posts: 3
on: August 17, 2009, 08:04:19 PM
We are looking to build a balanced (or at least as much as we can) round robin schedule for a poker league.   The one wrinkle that poker brings to the equation is you do not play head to head, but instead at a table where you are playing as 9 or 10 people.   The idea is if we balance the schedule and award points for how well you do each match then eventually the best player will be identified over a series of matches.

We are looking to have 20 people or so participate but can adjust slightly from there if it helps balance the schedule.  We want people to play in at least 10 matches and would like it to take 20 to 30 matches in total to complete (we are basically looking to play once a week over 9 months - with some off weeks for holidays).

The perfect numbers would be 20 people in total in the league, playing 10 matches each with each match consisting of 10 players (so in this scenario they would play the other 19 players on average 5 times over their 10 matches).

Is there a way to figure out the optimal schedule?

#### Ian Wakeling

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Reply #1 on: August 19, 2009, 08:06:26 AM
Can you confirm the format of play.  Is it the case that the 20 players meet at a venue and then split into two groups of 10, playing simultaneous matches.  Or is each match an independent event involving 10 players?

Thanks,

Ian.

#### rye-train

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• Posts: 3
Reply #2 on: August 19, 2009, 05:02:07 PM
Quote
Can you confirm the format of play.  Is it the case that the 20 players meet at a venue and then split into two groups of 10, playing simultaneous matches.  Or is each match an independent event involving 10 players?

Thanks,

Ian.

Sorry, I guess I should have made that clear each event would be 10 people with the other 10 not playing that night.
Jim

#### wbport

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• Posts: 129
Reply #3 on: August 20, 2009, 07:51:44 AM
Quote
Sorry, I guess I should have made that clear each event would be 10 people with the other 10 not playing that night.
Jim
Several ways to do that: 1-10 & 11-20, or odds one night and evens the next round, or 1-5 + 11-15 in one round and the rest the next round.  To do it randomly go here and, not only can you assign pairing numbers but determine who plays when in each pair of rounds:  first 10 in one round and last 10 in the next.

#### Ian Wakeling

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Reply #4 on: August 20, 2009, 11:59:57 AM
The independence of the games does make things a lot easier, for example here is a schedule for your preferred scenario:

` 12  4  1  5  9 10 20  3 17 19 13  5  2  6 10 11  1  4 18 20 14  6  3  7 11 12  2  5 19  1 15  7  4  8 12 13  3  6 20  2 16  8  5  9 13 14  4  7  1  3 17  9  6 10 14 15  5  8  2  4 18 10  7 11 15 16  6  9  3  5 19 11  8 12 16 17  7 10  4  6 20 12  9 13 17 18  8 11  5  7  1 13 10 14 18 19  9 12  6  8  2 14 11 15 19 20 10 13  7  9  3 15 12 16 20  1 11 14  8 10  4 16 13 17  1  2 12 15  9 11  5 17 14 18  2  3 13 16 10 12  6 18 15 19  3  4 14 17 11 13  7 19 16 20  4  5 15 18 12 14  8 20 17  1  5  6 16 19 13 15  9  1 18  2  6  7 17 20 14 16 10  2 19  3  7  8 18  1 15 17 11  3 20  4  8  9 19  2 16 18`

Above there are 20 players and each player plays in exactly 10 games.  There are 190 possible pairs of players, of which 150 occur together in exactly 5 games, while the remaining 40 pairs occur in exactly 4 games.

For certain combinations of the number of players and number of players per game there will be balanced schedules where all pairs of players occur together equally often.  The most suitable of these would be:

19 Players, 9 players per game, 9 games each, and all pairs meet in exactly 4 games:

`  2  5  6  7  8 10 12 17 18  3  6  7  8  9 11 13 18 19  4  7  8  9 10 12 14 19  1  5  8  9 10 11 13 15  1  2  6  9 10 11 12 14 16  2  3  7 10 11 12 13 15 17  3  4  8 11 12 13 14 16 18  4  5  9 12 13 14 15 17 19  5  6 10 13 14 15 16 18  1  6  7 11 14 15 16 17 19  2  7  8 12 15 16 17 18  1  3  8  9 13 16 17 18 19  2  4  9 10 14 17 18 19  1  3  5 10 11 15 18 19  1  2  4  6 11 12 16 19  1  2  3  5  7 12 13 17  1  2  3  4  6  8 13 14 18  2  3  4  5  7  9 14 15 19  3  4  5  6  8 10 15 16  1  4  5  6  7  9 11 16 17`

21 players, 9 players per game, 15 games each, and all pairs meet in exactly 6 games:

`  4  6  8 10 13 14 16 18 21  5  7  9 11 12 15 17 19 20  1  2  3 12 15 11 18 21 16  2  3  1  6  8  4 19 20 17  3  1  2  7  9  5 13 14 10  1  2  3 14 10 13 20 17 19  2  3  1  8  4  6 15 11 12  3  1  2  9  5  7 21 16 18  6  7  8  9 12 14 19 21  1  6  7  8  9 10 15 16 20  2  7  6  9  8 11 13 17 18  3  8  9  4  5 12 14 18 17  2  8  9  4  5 15 10 19 21  3  9  8  5  4 13 11 20 16  1  4  5  6  7 14 12 20 16  3  4  5  6  7 15 10 17 18  1  5  4  7  6 11 13 21 19  2 12 13 14 15 18 20  1  7  8 13 12 15 14 16 19  2  4  9 13 12 14 15 17 21  3  5  6 14 15 10 11 18 20  3  9  4 15 14 10 11 19 16  1  6  5 14 15 11 10 21 17  2  7  8 10 11 12 13 20 18  2  5  6 10 11 13 12 16 19  3  8  7 11 10 13 12 21 17  1  9  4 18 19 20 21  1  6  9 13 15 19 18 21 20  2  4  7 10 12 18 19 21 20  3  5  8 11 14 20 21 16 17  1  8  5 12 10 21 20 16 17  2  6  9 14 11 21 20 17 16  3  7  4 13 15 16 17 18 19  1  4  7 14 11 16 17 19 18  2  8  5 15 13 17 16 18 19  3  9  6 10 12`

Hope that helps.
« Last Edit: August 20, 2009, 03:45:47 PM by admin »

#### rye-train

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• Posts: 3
Reply #5 on: August 20, 2009, 02:44:23 PM
Thank you gentleman!

Out of curiosity, how does one go about coming up with the solution to these? Is it some sort of rotation algorithm?   or is it too complicated to explain?