Round Robin: SODOS or Direct comparison?

[RobertJasiek]

In a context of an assumed possibility of up to an infinite number of games...?

topazg, please explain "symmetric [...] is not inherently the nature of winning and losing"! A win is defined to equal 1 tournament score point, a loss 0. Now I think one can derive symmetry. Would you want to assign other values instead, like +5 versus -1?

[topazg]

In a context of an assumed possibility of up to an infinite number of games...?

topazg, please explain "symmetric [...] is not inherently the nature of winning and losing"! A win is defined to equal 1 tournament score point, a loss 0. Now I think one can derive symmetry. Would you want to assign other values instead, like +5 versus -1?

Firstly, don't confuse a postulated hypothesis that I put forward with anything I want or a personal opinion of mine. To answer "would I want" would not give you the answer you are looking for.

Secondly, +5 and -1 is just as symmetrical as +1 and 0 - provided in the case of jigos each player receives 2 points. If jigos do not exist, the symmetry will obviously still remain.

It should be very clear some tiebreak systems do not view wins and losses as equal value. SODOS disregards losses completely, and rewards wins against higher performing players to a greater extent than wins against lower performing players. The symmetry here is completely broken - however, that doesn't make the system bad or unfair. If this is the metric of performance that the tournament rules organiser wishes to embrace, then it is the correct system for that tournament. Likewise, as you said, SOLOS will have a different effect, but again is not inherently bad.

A good tiebreaker is one chosen specifically for the advantages and disadvantages it has within the context of the aims of the scoring system, and provided it is explicitly explained to those participating in a way that can be understood (and some of all the scores of the people you beat isn't hard!) the choice can be considered correct.

I agree with you that additional games to break ties are a solution with minimal flaws, but I think they should be played at the same time controls as the main games in the tournament ideally, which takes considerable time (I think this is the only great disadvantage to the system).

[fwiffo]

You can arrive the same SODOS score by taking the sum of the scores of all of a player's opponents, then subtracting the scores of opponents a player lost to. When looked at this way, losses to a strong opponent appear to carry a large penalty, and losses to weak opponents carry a smaller penalty. This seems illogical.

I propose a trivial symmetric alternative - sum the scores of defeated opponents, then subtract an inverted score for the victorious opponents.

Something like:
TIEBREAK := sum(DOS) - sum(MAX-VOS)
( where DOS is each defeated opponent's score and VOS is each victorious opponent's score and MAX is the maximum possible score )

That can be rewritten as:
TIEBREAK := sum(DOS) - (NL*MAX - sum(VOS))
( where NL is the total number of losses )

Apply a little basic algebra:
TIEBREAK := sum(DOS) - NL*MAX + sum(VOS)

Since sum(DOS)+sum(VOS) is just the sum of ALL opponents' scores, it will be the same for all players, and we can just drop it. Likewise, MAX is a constant multiplier which is the same for everyone, so we can eliminate that factor as well, leaving just TIEBREAK := NL.

Thus, we can break ties in the number of wins by simply considering the number of losses. My system is awesome.

[willemien]

Sometimes Direct Comparison and SODOS give different results, and therefore the question arises What is the better tiebreaker?

This question is known as the ranking problem and has been flogged many times over in the statistical literature. A variety of tiebreaking methods have been proposed, including SODOS, sometimes iterated to infinite order. Others have suggested ordering the players so that the tournament minimizes the number of upsets; in some cases this would place a player with 3 wins behind a player with 4 wins. As Bill suggests, Bayesian methods are also possible.

About the only method that hasn't been proposed is the direct confrontation because it relies on the result of a only single game.

I have heard of the problems with tiebreaking in in swiss and mcMahon tournaments (especialy the combination McMahon/ sodos is fraud with problems)

but i always thought in Round Robin these problems don't exsist and that Sodos and DC always would give the same outcome,
And this was the first time i was confronted with that it for a 2nd place. (for first places DC and sodos always have the same winner , except that sodos sometimes doesn't break the tie)


Please post some links where the ame problem is described then i can have a look how they solve it.

[Javaness]

The OOF tiebreaker should be used

[RobertJasiek]

The Order of Finishing tiebreaker is a bad joke. Even worse, it contradicts the aim of the game to win every single game.

[willemien]

Did some more investigation into the "433221" graph
selecting all tournaments games where

A has 4 wins
B has 3 wins and won from C
C has 3 wins
D has 2 wins and won from E
E has 2 wins
F has 1 win
Then there are 12 different tournaments (8640 / 6!)

Of these 12
7 have C winning only against D E and F ( and thus having a sodos score lower than B)

The 5 others are:

Code: Select all

 1   a  b  c  d  e  f  | sc | sodos   
 A:  =  W  L  W  W  W  |  4 
 B:  L  =  W  L  W  W  |  3     6 
 C:  W  L  =  W  L  W  |  3     7
 D:  L  W  L  =  W  L  |  2     5
 E:  L  L  W  L  =  W  |  2     4
 F:  L  L  L  W  L  =  |  1 

 
 2   a  b  c  d  e  f  | sc | sodos   
 A:  =  W  L  W  W  W  |  4 
 B:  L  =  W  W  L  W  |  3     6 
 C:  W  L  =  L  W  W  |  3     7
 D:  L  L  W  =  W  L  |  2     5
 E:  L  W  L  L  =  W  |  2     4
 F:  L  L  L  W  L  =  |  1 


 3   a  b  c  d  e  f  | sc | sodos   
 A:  =  W  L  W  W  W  |  4 
 B:  L  =  W  W  L  W  |  3     6
 C:  W  L  =  W  L  W  |  3     7
 D:  L  L  L  =  W  W  |  2     3 
 E:  L  W  W  L  =  L  |  2     6
 F:  L  L  L  L  W  =  |  1 

 
 4   a  b  c  d  e  f  | sc | sodos   
 A:  =  W  L  W  W  W  |  4 
 B:  L  =  W  W  L  W  |  3     6
 C:  W  L  =  W  W  L  |  3     8
 D:  L  L  L  =  W  W  |  2     3
 E:  L  W  L  L  =  W  |  2     4
 F:  L  L  W  L  L  =  |  1 


 5   a  b  c  d  e  f  | sc | sodos   
 A:  =  W  L  W  W  W  |  4  
 B:  L  =  W  W  W  L  |  3     7
 C:  W  L  =  W  L  W  |  3     7
 D:  L  L  L  =  W  W  |  2     3
 E:  L  L  W  L  =  W  |  2     4
 F:  L  W  L  L  L  =  |  1 

All taking it together I find Sodos a fairer tie breaker than Direct comparison
(especialy seeing the 3rd example)

Sodos doesn't break the tie in the 5th example and i think that only then Direct comparison should be used.

[RobertJasiek]

"fairer" in which sense?

[willemien]

Hi Robert

"fairer" in which sense?
"fairer" in which sense?

Meaning that (for example in tournament 3 C is a better player than B


FRom an earlier post of you:

I have become thirsty of numbers:)

There are more interesting questions: What are the tie frequencies (esp. for place 1) after application of either of these tiebreakers?

- MutualGameScore iff 2 players are tied on a place
- Non-iterativeDirectComparison
- IterativeDirectComparison

What do you (exactly) mean by these tiebreakers

(Where does mutual gamescore differ from [non- iterative] Direct comparison , do give examples)

(Then i can build these tiebreakers into my program and give results

[RobertJasiek]

I understand that you prefer SODOS. I do not understand why you prefer it. Is, for you, "preference" the same as "fairness"?!

"MutualGameScore iff 2 players are tied on a place": It is applied if and only if exactly 2 players are tied on a place. For each of them, the MutualGameScore is 1, if he wins their game, 0 if loses their game, 0.5 if he ties their game.

"Non-iterativeDirectComparison / IterativeDirectComparison":
http://home.snafu.de/jasiek/DCqual.html
And if you don't understand that, read my more detailed explanations on RGG or SL.

[willemien]

The main reason why i prefer SODOS above DC is that SODOS makes a difference between winning against relatively stronger players and against relativly weaker players.
(what is defined by their number of wins)

as in example 3

Code: Select all

3   a  b  c  d  e  f  | sc | sodos   
A:  =  W  L  W  W  W  |  4 
B:  L  =  W  W  L  W  |  3     6
C:  W  L  =  W  L  W  |  3     7
D:  L  L  L  =  W  W  |  2     3 
E:  L  W  W  L  =  L  |  2     6
F:  L  L  L  L  W  =  |  1 

C won from A (the stronger player)
But lost against B

While B and C played the same against the other players.

In my view C deserves the 2nd prize trophy and B only the 3rd

But if you look to the game between them (or MutualGameScore, what is exactly the same)

B deserves the 2nd place and C the 3rd

The problem is that MutualGameScore only looks at the mutual game.


look also at the tie between D and E for the 4th and 5th place

D only won against E and F
While E won against B and C (stronger players measured by their number of wins)

Who should get the 4th place trophy?

[HermanHiddema]

http://home.snafu.de/jasiek/DCqual.html

Warning: the above linked document contains several falsehoods, the most glaring of which is: "Direct Comparison values do not contain any noise."

[Harleqin]

The main reason why I prefer SODOS over DC is that SODOS makes a difference between winning against stronger players and against weaker players.

The main reason why I prefer SOLOS over DC is that SOLOS makes a difference between losing against weaker players and against stronger players.

[RobertJasiek]

Herman, discussion of the various statements on my page, elsewhere and by others elsewhere can go on for some more years, I guess. It is not necessary to discuss that now though. I have linked to the page here just to point at the definitions.

[topazg]

C won from A (the stronger player)
But lost against B

While B and C played the same against the other players.

In my view C deserves the 2nd prize trophy and B only the 3rd

Of course, the flip side of this argument is that C only deserves 3rd because his loss was against a weaker player

I think the coin flips equally well both ways though, it's preference rather than intrinsic fairness to me