What do you think of this format?

Running Chess Tournaments
1

This was a variation on standard quads, to try to equalize the average rating of a player’s opponents. In regular quads, each of the groups has one player rated higher than the rest, and one player rated below the other 3. To make these pairings, all of the players were assigned pairing numbers by rating, and then filled in on pre-numbered pairing sheets. That meant that just like quads, pairings did not depend on the results of earlier rounds. Also, just like quads, there was no TD discretion about pairings - they were pre-determined by rating.

Unlike quads, though, most of the players had opponents both above and below themselves in rating. That was only impossible for the very top and bottom entries.

Just as in quads, all of a player’s opponents were within a small range, no more than 3 places up or down on the overall list. Also, just as in quads, no player had the same color all 3 rounds.

Prizes were easy to calculate: 18/4 = 4.5, so there were 4 and a half equal prizes. In this case, it was $50 for each of 1st, 2nd, 3rd, and 4th, and $25 for 5th. Player #1(3.0) got $50 for 1st, #9(2.5) got the same amount, $50, for 2nd, and players #2, 5, 8, 11, 12, 15, and 17, (each 2.0) split the remaining $125 for $18 each.

Is this new? It will produce legal pairings for any number of entries, meaning no duplicate games and a balance of colors, as long as a few detailed instructions are followed. I can share or publish the system.
-fb

[code] Rate Rnd 1 Rnd 2 Rnd 3

  1. |2420 | B 4 | B 3 | W 2 |
    | | 1.0 | 2.0 | 3.0 |
  1. |2157 | W 3 | W 4 | B 1 |
    | | 1.0 | 2.0 | 2.0 |
  1. |2147 | B 2 | W 1 | B 5 |
    | | 0.0 | 0.0 | 0.5 |
  1. |2059 | W 1 | B 2 | W 6 |
    | | 0.0 | 0.0 | 1.0 |
  1. |2020 | B 6 | B 7 | W 3 |
    | | 0.5 | 1.5 | 2.0 |
  1. |2004 | W 5 | W 8 | B 4 |
    | | 0.5 | 0.5 | 0.5 |
  1. |1997 | B 8 | W 5 | B 9 |
    | | 1.0 | 1.0 | 1.0 |
  1. |1936 | W 7 | B 6 | W 10 |
    | | 0.0 | 1.0 | 2.0 |
  1. |1812 | B 10 | B 11 | W 7 |
    | | 0.5 | 1.5 | 2.5 |
  1. |1731 | W 9 | W 12 | B 8 |
    | | 0.5 | 0.5 | 0.5 |
  1. |1578 | B 12 | W 9 | B 13 |
    | | 1.0 | 1.0 | 2.0 |
  1. |1532 | W 11 | B 10 | W 14 |
    | | 0.0 | 1.0 | 2.0 |
  1. |1528 | B 14 | B 15 | W 11 |
    | | 0.0 | 0.5 | 0.5 |
  1. |1523 | W 13 | W 16 | B 12 |
    | | 1.0 | 1.5 | 1.5 |
  1. |1479 | B 18 | W 13 | B 17 |
    | | 1.0 | 1.5 | 2.0 |
  1. |1439 | W 17 | B 14 | W 18 |
    | | 0.0 | 0.5 | 1.5 |
  1. |1273 | B 16 | B 18 | W 15 |
    | | 1.0 | 1.5 | 2.0 |
  1. |1201 | W 15 | W 17 | B 16 |
    | | 0.0 | 0.5 | 0.5 |
    ----------------------------------[/code]
2

I’d be interested to see what rules you use to pair it.
Though you prize structure was lacking with your broad range of ratings.

With such a broad range of ratings and doing the modified quads, I’d more likely have done prizes for ratings ranges.
With this format it seems the 1200 player at the bottom could possibly go 3-0 w/o playing anyone over 1550.

Does the bonus points rating go by average rating of people you play, or entire tournament?

3

An interesting idea, a “continuous quad” where only the top and bottom players in the entire tournament are paired down or up every round, unlike a quad where half the players are paired down or up every round. I guess another advantage over quads is that it can accommodate a number of players that is not a multiple of four, more smoothly. And like a set of quads, it doesn’t produce an overall winner but rather competitive games.

How even are the pairings, except for the very top and very bottom players? Try this experiment: Using your algorithm, pair a tournament of 20 players rated 1500, 1510, 1520, …, 1690 . Report the average opponents’ rating for each player. Report if any player has three Whites or three Blacks.

Could it be extended to four rounds so that colors are equalized? Can colors be equalized then? Repeat the above experiment reporting average opponents’ ratings and whether any player doesn’t have two Whites and two Blacks.

How does it handle an odd number of players – how does it assign byes? Repeat the above experiment(s) with 21 players rated 1500, 1510, 1520, …, 1700.

4

That’s the point of quads, which put players in small groups of similar ratings. It gives any player, regardless of their general level, a reasonable chance to win a prize by playing well compared to their peers. This keeps that feature.

In fact, the only player who complained about the fairness of the pairings was the bottom 1200 player, since he was ‘playing up’ every game, and no-one else was. My answer was simple - his pairings were not any tougher than they would have been in regular quads, a common and popular format.

OK, here’s your example. BTW, I already used this for a tournament, and wouldn’t have unless I was sure that it would work without causing a serious color problem.

[code]
A B C D E F G H
1 . 1690 . 2, 3, 4 . 1670 . -20 . b4 w3 b2 . 1670 . -20
2 . 1680 . 1, 3, 4 . 1673 . - 7 . w3 w4 w1 . 1673 . - 7
3 . 1670 . 1, 2, 4 . 1677 . 7 . b2 b1 w5 . 1673 . 3
4 . 1660 . 1, 2, 3 . 1680 . 20 . w1 b2 b6 . 1670 . 10
5 . 1650 . 6, 7, 8 . 1630 . -20 . b6 w7 b3 . 1647 . -3
6 . 1640 . 5, 7, 8 . 1633 . - 7 . w5 b8 w4 . 1643 . 3
7 . 1630 . 5, 6, 8 . 1637 . 7 . b8 b5 w9 . 1627 . -3
8 . 1620 . 6, 7, 8 . 1640 . 20 . w7 w6 b10 . 1623 . 3
9 . 1610 . 10,11,12 . 1590 . -20 . b10 w11 b7 . 1607 . -3
10 . 1600 . 9,11,12 . 1593 . - 7 . w9 w12 w8 . 1603 . 3
11 . 1590 . 9,10,12 . 1597 . 7 . b12 b9 w13 . 1587 . -3
12 . 1580 . 9,10,11 . 1600 . 20 . w11 b10 b14 . 1583 . 3
13 . 1570 . 14,15,16 . 1550 . -20 . b14 w15 b11 . 1567 . -3
14 . 1560 . 13,15,16 . 1553 . - 7 . w13 b16 w12 . 1563 . 3
15 . 1550 . 13,14,16 . 1557 . 7 . b16 b13 w17 . 1547 . -3
16 . 1540 . 13,14,15 . 1560 . 20 . w15 w14 b18 . 1543 . 3
17 . 1530 . 18,19,20 . 1510 . -20 . b20 w19 b15 . 1520 . -10
18 . 1520 . 17,19,20 . 1513 . - 7 . w19 b20 w16 . 1517 . -3
19 . 1510 . 17,18,20 . 1517 . 7 . b18 b17 w20 . 1517 . 7
20 . 1500 . 17,18,19 . 1520 . 20 . w17 w18 b19 . 1520 . 20

A: player number
B: player rating
C: opponents in standard quads
D: avg. opponent rating in standard quads
E: difference between player’s own rating and avg. rating of
opponents in standard quads
F: crosstable for new system
G: avg. opponent rating in new system
H: difference between player’s own rating and avg. rating of
opponents in new system [/code]

As you can see, column H is usually smaller (in absolute value) than column E, and never larger.

I really don’t know. It would be difficult to figure out. I have been told by TD’s senior to me that even this 3-round ‘continuous quad’ would be impossible, but I worked out the algorithm.

The first step is to add a house player, if needed, to make an even number. But, not necessarily divisible by 4, just by 2, as my original experiment of 18 players shows.

As for requested byes, it doesn’t provide for them, but neither do Round-Robin pairings (of which quads are just one example.) Players are scheduled to play all 3 rounds, and if any do drop out, the only possibilities are to recruit a new house player within the appropriate rating range, or give the remaining person a full-point bye.

5

Slick! I like it. :smiley:

6

Do not forget that Quads are based on the basic Round Robin format for pairings. What this means is that the players [1 & 2] higher up get an extra White. Therefore, to make things more equitable you should either draw for position [like larger round robin tournaments], or list the players in reverse order of their rating. By listing in reverse order of rating you will give the lower rated players 2 Whites and 1 Black. This would be one other way to try to offset what may be a disadvantage in rating difference in the Quad.

-Larry S. Cohen

7

If in order of rating, 1 & 2 would have the same color (draw or flip) against 3 & 4 for the first round and swap both opponents and colors for the 2nd round. For the final pairings between 1 & 2 and 3 & 4, draw or flip for colors.

8

The only thing about this that is sort of out of place is you are advertising overall place prices and the doing modified class pairings.

I think the plus score format of prizes would be best for this sort of tournament. That way you don’t get into disputes over that sort of thing. There is another thread discussing that.

I’d be interested in your pairing algorithm.