Due to time constraints, Diplomacy tournaments usually have a small number of rounds. Often less than the number of boards played in each round. This can result in many players having similar records which leads to a high probability of ties. This is especially true in DIAS tournaments where centers are not scored, and all survivors receive an equal share in any draw. This article deals with the issue of breaking ties especially in DIAS tournaments.
The first tie breaker should always be head to head competition. This applies to any tournament. If the tournament scoring system declares two players to be tied when all of their games are considered then recalculate considering only the games where the players directly played against one another. After this calculation there is still a significant chance of a tie. The players could have played no games against one another, or played just one and both received the same result. So there should still be some backup system for breaking ties.
Before I go further, let me say a few words about the DIAS philosophy which will help to decide on a method to declare one record "better" than another when the scoring system says they are the same. The DIAS system places negotiation skill as the highest concern. The only question for scoring is survival or elimination. If a country exists it has a voice in international diplomacy. If it does not it does not. Imagine the game ends in a three way draw with supply centers divided 16-15-3. The country with three supply centers has demonstrated skillful diplomacy in preventing the larger countries from crushing him. Indeed, he has gotten them to voluntarily give him an equal share in the draw. Truly a diplomatic achievement.
This is the reason that DIAS tournaments avoid scoring supply centers. Of course, the assumption is that the game ends by solo victory or a voluntary draw. If the game were ended by an involuntary deadline then perhaps the two larger powers were well on their way to eliminating the small power and finishing in a perfect 17-17 draw. In this case the smaller power should not be scored very high. The DIAS format is more appropriate when games are allowed as much as possible to be ended only by the actions of the players. Scoring supply centers more heavily weights military skill, and is more appropriate when games are expected to end by reaching a deadline.
So, if scoring wins, draws, and losses results in a tie and we want to avoid scoring supply centers what can we use to break the tie? I believe the answer is game years played. A win achieved in 1909 is better than a win in 1910. A draw, likewise, is better the sooner it can be achieved. You could have had a four way draw and you didn't vote for it, but instead spent time dilly dallying and not eliminating any players to get the same result a few years later. Wasting time shows a lack of skill.
Eliminations, on the other hand should be rewarded for surviving as long as possible. The simplest system would be to add up -1 for each win or draw year, and +1 for each loss year, but this doesn't work well. Consider the following records in a three round tournament:
| Player 1 | Player 2 | ||||
| win | 1913 | -13 | 2draw | 1904 | -4 |
| loss | 1901 | +1 | 4draw | 1904 | -4 |
| loss | 1901 | +1 | 4draw | 1904 | -4 |
| -11 | -12 | ||||
These two records have the same YARS score. Player one had a win in 1913 (worse than 1904) and two losses in 1901 (the worst possible.) Player two had three draws all in 1904 (pretty good.) If you accept the fact that a win and two losses are equivalent to a two way and two four way draws then player two clearly had a better record of game years, but loses the tie breaker. The problem is that the negative points of wins and draws aren't really comparable to the positive points of losses. If losses count up from zero then wins and draws should start counting down from some positive number. Where they meet determines when two records are equivalent.
In games where you achieved a win or draw:
In games where you lost:
The value of X depends on how long you expect games to last. If your tournament does have a maximum game year you can use that as X. For example, if X is 14 then the following records are equivalent:
| Player 1 | Player 2 | ||||
| win | 1907 | 14-7 | 2draw | 1907 | 14-7 |
| loss | 1907 | +7 | 4draw | 1907 | 14-7 |
| loss | 1907 | +7 | 4draw | 1907 | 14-7 |
| +21 | +21 | ||||
Another factor in eliminations is the number of countries left at the end of the game. Player one is eliminated by a cohesive three way alliance, and the game ends in a three way draw. Player two is eliminated by a shaky three way alliance which breaks down after he is gone and the game ends in a solo victory. Player two showed less skill because he should have been able to break up the alliance with diplomacy while he was still alive. If you are eliminated, the more players that survive to the end of the game, the more players you probably had teamed up against you.
An alternative system would be:
In games where you achieved a win or draw:
In games where you lost:
In this case X would have to be modified to reflect the fact that losses now have some negative modifiers. A loss to a four way draw in 1907 would be equivalent to a loss to a solo victor in 1910, and if X were 10 would be tie break equivalent to getting a draw in 1906
The simple tie break system described above is as yet untested, but would probably work just fine under any circumstances. However, let me now describe an idea I've had for a system that might be more interesting. I think a draw year should be penalized more than a win year. If you could've gotten the win a year earlier you would have, but for a draw there is a conscious choice. "Should I vote for the draw now, or try for a better result?" An early draw could be the result of wisdom and decisive action. A late draw might have been the result of wasting time. I would also give more penalty to late high player draws. If you get a late two player draw you probably needed much of that time to eliminate the other players. If you get a late five player draw that shows a lack of something good.
Back to some examples:
| Player 1 | Player 2 | ||
| win | 1910 | win | 1910 |
| win | 1911 | win | 1910 |
| 4draw | 1909 | 4draw | 1910 |
Player one took an extra year to win, but got the 4 way draw one year earlier. By my logic, player two should be penalized more for the extra 4draw year than player one should be for the extra win year so player 1 should win the tie break.
Here is the system. Tying players line up their results from best to worst. This can be done for more than two way ties as well. For each line find a common denominator in the following way: Take the largest number of players and the smallest year. Pretend that all players were in this situation, and made different decisions. Some players may have eliminated opponents. For each opponent eliminated the player gets +1. Some players may have taken extra years. Each extra year gets a penalty based on the final outcome the player was able to achieve (not the denominator position).
Penalty Per Year
| win: | -.32 |
| 2draw: | -.35 |
| 3draw: | -.38 |
| 4draw: | -.41 |
| 5draw: | -.44 |
| 6draw: | -.47 |
| 7draw: | -.50 |
Another example:
| Player 1 | Player 2 | Common Denominator | |||
| 2draw | 1908 | 3draw | 1907 | 3draw | 1907 |
| 4draw | 1907 | 3draw | 1907 | 4draw | 1907 |
| 6draw | 1906 | 4draw | 1907 | 6draw | 1906 |
In the first line both players had a three player game in 1907. Player two decided to accept the draw. Player one kept playing and took 1 year (bad -.35) to eliminate 1 player (good +1.) In line two player two took zero years to eliminate one player (+1.) In line three player one accepted a six way draw and player two took 1 year (-.41) to eliminate two players (+2.) Player 1 gets -.35 +1 = .65, and player two gets +1 -.41 +2 = 2.59. Player two wins.
This system hasn't included losses yet. Once again comparing wins and draws to losses proves troublesome. We know that before the game started both players had the equivalent of a seven way draw in 1900 so that is the common denominator. The win or draw proceeds as usual counting the number of players eliminated against the number of years taken. The loss takes the following modifiers:
Comparing losses to losses finds a common denominator by taking the largest number of survivors and earliest year and proceeds from there with the modifiers above. Here are some final examples of tie break outcomes. A loss to X surviving players is written as an Xloss. Remember, the players' overall records from multiple rounds would have to be equivalent before the tie break would be counted.
Player 1 Player 2 Common Denominator
win 1912 4loss 1907 7draw 1900
6 eliminated +6 3 eliminated -3
12 years taken -3.84 7 years survived +3.5
------- ------
+2.16 +0.5
5draw 1905 1loss 1916 7draw 1900
2 eliminated +2 6 eliminated -6
5 years taken -2.2 16 years survived +8
------ ----
-0.2 +2
5loss 1906 2loss 1910 5loss 1906
3 eliminated -3
4 years survived +2
---- ----
0 -1
win 1909 3draw 1912 3draw 1909
2 eliminated +2
3 years taken -1.14
---- -------
+2 -1.14
Hopefully someone will use this tie breaker system, and let me know how it goes.
|
Robert Steinke (Robert.Steinke@colorado.edu) |
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