# How to use SODOS in McMahon tournaments

In McMahon tournaments most European organisations advise against using SODOS because the tiebreak results depend on whether the zero score is placed at the top or at the bottom of the field:

- The British Go Association recommends that SODOS should not be used in McMahon tournaments. See SODOS/discussion.
- The EGF tournament rules expressly advise against using SODOS.

If SODOS is used after Swiss score, then this major problem of SODOS is avoided.

In North America, where ties are broken only between players who started off with the same McMahon score (i.e., players in the same division), SODOS is considered a standard and acceptable tiebreaker and is traditionally applied after SOS.

## The problem

(copied from the SODOS/discussion page)

Here is an example of a 3 round tournament illustrating my concerns about using SoDoS as a tie breaker.

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 Alan(1d) 2 0 2 1+1=2 Cath+Dave Juliet(1k) 3 -1 2 0+1+1=2 Karen+Lionel+Martin

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 Juliet(1k) 3 19 22 20+21+21=62 Karen+Lionel+Martin Alan(1d) 2 20 22 21+21=42 Cath+Dave

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

I am not worrying here about which result is better!. All I care about is that they are different.

Note that if you used SoS (Sum of all opponents McMahon scores) as the one and only tie breaker, then this effect does not happen because you are summing over all games, not just a selection.

## Solution

If SODOS is used after the Swiss score the SODOS tie break is not used. In the case above Juliet gets the chocolate because she has more wins. ( 3 vs 2)

Because after the number of wins only players with an equal number of wins have to be compared there is no problem where the zeropoint of the MM score anymore.

In the example the zeropoint for the UK is at 1 dan, for the EU it is 20kyu.

The difference between the SODOS of Juliet-EU and Juliet-UK is 3 times this difference for Alan the difference is is 2 times. this only depends on the number of wins.

If the players have the same number of wins then both their scores have the same difference.

**A resulting problem**

A resulting problem is that using Swiss Score before SoDoS score is that it in theory can lead to players trying to start on a lower rank (to have more chance of more wins) but because of the tiebreakers before this, final mcMahon score, SOS and SOSOS i guess this is only theoretical

### Mathematical Proof

**Definitions **

Pi :<==> Player i NWi :<==> number of wins by player i LOij :<==> the players j losing to Pi MMi is an MMscore with an arbitrary 0 point. MMi(Pj) is MMi of player j.

Let MM0 be an MMscore with an arbitrary 0 point. MM1 := MM0 - C; C is arbitrary, fixed, natural-number.

For each Pi: MM0(Pi) = MM1(Pi) + C

**Basic equations:**

SODOS(MMi, Pi) = Sum_j(MMi(LOij)) (SODOS is the som of the MM scores of al losing opponents)

NW(Pi) = NW(Pj) (*) (Both players have the same number of wins)

Step 1:

SODOS(MM0, Pi) = Sum_j(MM0(LOij) = Sum_j(MM1(LOij) + C) = Sum_j(MM1(LOij) + NW(Pi) * C = SODOS(MM1,Pi) + NW(Pi) * C

Step 2:

SODOS(MM0, Pi) - SODOS(MM0, Pj) = SODOS(MM1,Pi) + NW(Pi) * C - ( SODOS(MM1,Pj) + NW(Pj) * C ) = SODOS(MM1,Pi) - SODOS(MM1,Pj) + NW(Pi) * C - NW(Pj) * C = SODOS(MM1,Pi) - SODOS(MM1,Pj) + C * (NW(Pi) - NW(Pj)) (*) = SODOS(MM1,Pi) - SODOS(MM1,Pj).

QED

The difference between the Sodos scores of two players with the same number of wins is independent of the null point of the McMahon score and SODOS can therefore be used but only after the Swiss score tiebreaker