MDOS

MDOS - Mean of Defeated Opponents' Scores is a possible tiebreaker. It can be thought of as a measure of the strength of the average opponent a player was able to beat. In a Swiss tournament it resolves to the same thing as SODOS, but in a McMahon this is not the case, see the discussion around chosen origin of the McMahon bar. In short, people with different numbers of wins can be tied under SODOS in a McMahon, but not in a Swiss.

MDOS is calculated by adding together the MMS of each opponent you defeated and dividing that by your number of wins. (Should jigo occur then half the MMS of the opponent you drew with would be added. The win count would also be incremented by a half.) In this way it can be said to indicate the average 'performance strength' of the opponents a player defeated.

So far, MDOS is a theoretical tie breaker, it has not been used in any tournaments.

Discussion

pwaldron: The influence of the number of wins a player has is still there, but it's reduced. Consider player A, who starts with an initial McMahon score of 0 and wins 5 games, and player B, who starts with an initial McMahon score of 1 and wins 4 games.

If player A wins and player B loses their first games, then the two players will have the same McMahon score in round 2 and should be treated equivalently. If the two players meet equivalent opposition then those opponents should effectively cancel each other out in the tiebreaks. But player A's first opponent will also be in the mix, and on average he'll post a weaker score than the other opponents that A and B faced, which will in turn bias the results in B's favour.

Bass: To me it looks like this is exactly what the tie breaker is trying to measure. The player who wins early is likely to have an edge in MDOS, and the initial MMS does count as wins for any other purpose too.

isd: I calculated a few tiebreak scores for MDOS (or whatever you want to call it) and found that they generally gave the same result as SODOS. The one key advantage I see is that they don't suffer from the MM-bar origin choice problem that can cause different results. In the case of unequal number of wins, it will probably favour the player with fewer wins for the reason pwaldron gives.

Kaniuk Case Revisited

We have a shodan Alan winning 2 games playing:
Bob(1k)-, Cath(1d)+ Dave(1k)+

We have a 1kyu Juliet winning 3 games playing:
Karen(1d)+ Lionel(1k)+ Martin(1k)+

It is not necessary to show the entire tournament results table. Here is a summary of the key information for Alan's and Juliet's opponents' scores in UK and EU styles.

Alan's opponents:

                UK MMS         Euro MMS
Name       Wins initial final  initial final
Bob(1k)    2    -1      1      19      21
Cath(1d)   1     0      1      20      21
Dave(1k)   2    -1      1      19      21

Juliet's opponents:

                UK MMS         Euro MMS
Name       Wins initial final  initial final
Karen(1d)  0     0      0      20      20
Lionel(1k) 2    -1      1      19      21
Martin(1k) 2    -1      1      19      21

Suppose we use SoDoS as the one and only tie breaker. Then we can construct the portion of the final ranklist showing both Alan's and Juliet's position. The column MMSi is the initial McMahon Pairing score, and MMSf is the final McMahon Pairing score.

Then the final position of Alan and Juliet in the UK scale is:

           Wins MMSi MMSf   SoDoS   WHO CONTRIBUTES TO SODOS   MDOS
Alan(1d)   2     0   2      1+1=2   Cath+Dave                   1
Juliet(1k) 3    -1   2    0+1+1=2   Karen+Lionel+Martin         2/3

Alan and Juliet are ranked equal, and they split a box of chocolate.

However in the EU scale:

           Wins MMSi MMSf       SoDoS  WHO CONTRIBUTES TO SODOS    MDOS
Juliet(1k) 3    19   22   20+21+21=62  Karen+Lionel+Martin        20 2/3
Alan(1d)   2    20   22      21+21=42  Cath+Dave                  21

Now Juliet is ahead of Alan and gets all the chocolate!

In terms of who comes out ahead, MDOS is consistent across both examples.