When someone does not start in the top group, he has weaker opponents.
So there will always be a difference in SOS.
So the score+SOS would not change anything.
I find it funny to say that someone has been unfairly robbed, and to propose a fairer system that leads to exactly the same result.

I'm not sure that the system of one Mcmahon point for 3 levels is better, for me it is too random.
Let's imagine that the 7,8 and 9k have the same macmahon score.
Then in a 4 round tournament an 8k who would win everything could have faced,
in the best case:

a 7k,a 4k with 0 wins, a 1k with 0 wins, a 1k with a win.
in the worst case:
a 9k,a 9k with 1 win, a 9k with 2 wins, a 9k with 3 wins.
for a less unlikely case:
a 9k, a 9k with 1 win, a 6k with 1 win, a 6k with 1 win(after round 3 in this system, 1v for a 6k is most likely).

3 very different cases, for the same player with the same results.


I totally agree with Pio2001 ,it's not possible to have a perfect system.


In my opinion, two things could be improved:
-in addition to a majority of McMahon tournaments, make sure there are a minority of Swiss tournaments(championships).

-Make sure that the ranking system works well. Specifically, we sometimes see players who after winning everything don't get the GOR of their strongest opponent. To me, these are the players who are penalized, because not only did they not play balanced games, but they still don't have the rank for which they performed.
The French ranking system use a specific calculation for this.
If a player wins more than X points at a tournament, the whole tournament is recalculated by giving this player his initial GOR+X.

Not only does the under-randed player win more points, but his opponents lose less

And i don't know why this system is not in the european rating system.

When I played in Canadian tournaments the practice was to cluster players into McMahon bands that were separated by 2 points. For tournaments of moderate size (~30-50 people) there tended to be breaks in the distribution of players that facilitated natural bands.

The effect of the larger gap between adjacent groups was that players tended to play more games within their group before being paired up.

I can't comments on the theoretical merits, but it certainly produced a good series of games and I never felt disadvantaged. Tournaments run like this tended to run over two days and have six rounds. I'm not sure what the practice elsewhere in the world.

Heh, my bad! I clearly meant SODOS in my head but typed SOS. I'm such a duh-brain. Sandy Taylor beat the winner Sai Sun and runner up Andrew Kay. S+SODOS respectively is 41, 43, 36 so it would put him in second place which seems more accurate to what the right result should be! 2015 london open https://britgo.org/results/2015/london



If a person wins consecutive games, then their McMahon score increases by the number of consecutive wins they've had up to that point plus one. So when they lose it resets to zero plus one and they need to win consecutive. If a person skipped a game then tthey'll still get the consecutive win points if all the games they last played were won games, but of course they would be behind by the number of rounds they skipped. This seems fairer.

Another solution might be to start of tournaments with a series of fast games worth fractional McMahon points, for the first third or so of the tournament by total time. For example, three fast games each worth a third of a McMahon point, and then two games three times slower each worth a whole point!

Your system work better in that particular case, but he is very unfair in a lot of other cases.
For example, in the Open London, if we have one undefeated player,the victory vs the 2th give him only a +8 S+SODOS score, so it means it is possible the only undefeated player of the top group lose the tournament if his early opponents do a bad tournament.
Not fair.



Not really.
The systems seems totally unfair for the top-group if there is no undefeated players.
For the non-top there can be other problems too.
We can imagine a 5k,after 3 rounds he did 1win-1 loss-1win. So he play like a good 5k.
His opponent can be a 10k who did 3 wins, and for the moment just play like a 7k.
So its not fair for the 5k player, with his results, he should have oportunity to play agaisnt stronger player,we give hima opponent who most probably is 2k lower.



As said Pio2001,we are limited by number of rounds. Yes, with a full blitz tournament we have better rankings.
But this mix-system seems totally bullshit.
What happens if the best 2 players of the tournament play each other at the 3th round, then have no defeat, we consider it is fair, to decide the result of a slow tournament on a blitz?

Well, if a player is undefeated only because they played weaker players, then I'm not sure how that's unfair! And also, if two players meet in the blitz phase of a tournament, that shouldn't necessarily stop them from meeting again in the slow phase.

Hmm, regarding the 10k and 5k, it can easily be set up so that among players with the same McMahon score, those with the most similar initial McMahon score--kyu/dan--are paired against each other. So it doesn't seem like a big deal! I guess.

I probably didn't make myself clear. So I'll detail an example to make it clearer:
Imagine a tournament:
We named the best player A, the second best player B. Both in top-group
At round 1, A won against C and D, B won against E and F.
A and B won all their others games until last round.
At last round, A won against B and A is the only player of the top-group to be undefeated.

With classic McMahon System, A won the tournament.
With your system(score+SODOS), if C and D do a worst tournament than E and F, it can be enough to have player B ranked first(B who lose vs A)
It means, a member of the top-group can't win the tournament even he won all his games, and won against his direct opponent.

This is not fair.

Same for blitz phase, even if you allow them to meet against, this may not be the case.

"those with the most similar initial McMahon score kyu/dan are paired agaisnt each other", it solve nothing,when you play a guy with same initial rank than you, it is the classic Mcmahon pairing. But with unfair ranking.
the 10k won against a 10k in round 1, a 10k in round2, a 10k in round 3, and a 9k in round4(no more 10k undefeated),so he will have a 1k Mcmahonscore?
And will be same score than a 3k,who did 2 won and 2 defeats against 3kyu?

All theses ideas seems to have very big flaws.

I think the McMahon troubles at the 2015 London Open shouldn't be blamed on the players that won most of their games. If it is anyone's fault then it would be the four players in the top group that won two or less games in a seven round tournament. Another reason is the small top group, this is a BGA policy that I can't explain. I know there was a simulation study into the size of the top group but I only remember that the results were controversial, not if they recommended a larger or a smaller top group but I think it was done before this event.

Basically, you can get into trouble because of the composition of the top group. If the top group is very big there is a good chance that the top players never meet in the tournament (too few rounds) and if the group is too small it is possible that it is much easier for a player in the second group to win every game in the tournament than it is for a player in the top group. It is also usual for the top players to draw on McMahon points unless a player wins every game. More trouble ensues when many players in the top group underperform, in this case the ranking of the top players may effectively be decided by a small number of games, assuming we can discount any games with the underperforming players as automatic wins but this is maybe only our perception in hindsight when we already know how these players performed.

Well, E and F will have to win over twice as many games as C and D overall, and being above the bar D must've won to have played A so they already have the same number of games after 2 rounds. I'm not sure if that's more or less likely than the situation I'm trying to solve, so. But if it's changed to S+(SODOS/2) or even recursively S+(SODOS/2)+(SODODOS/4) . . . then that seems to solve the problem

Pairing people with similar initial McMahons though was just off-the top of my head, and yes, it actually seems terrible. And yes, the original idea doesn't quite work, okay, but again with the same tweaking it may make things better.

So again, halving the weight so that a player gets an extra half a point for each consecutive win.
And instead of those points being added to the McMahon score, they are treated as a separate additional score for determining pairings that have no effect on the final score.

Now if we combine the two systems together, it becomes a lot more unlikely for E and F to win twice as many games as C and D since the difficulty of opponents increases the more you win, which makes counting S+(SODOS) or more preferably S+(SODOS/2)+(SODODOS/4) . . . much better!

And also this also makes the blitz phase more useful, although it seems it could always be useful!

I believe you can try out tournament systems with a rather simple statistical simulation.

_________________
A good system naturally covers all corner cases without further effort.

I think McMahon is a really good tournament system, it usually breaks when people don't understand how to implement it.

If you have a very small pool of players, sometimes it doesn't work that well, but I'd struggle to see what you could replace it with. For instance take a field like this over 5 rounds - 6D , 6D, 1K, 2K, 5K, 12K, 12K, 12K, 13K, 15K, 15K, 2K, 22K

It does seem interesting to experiment with using larger bands below the bar, but that's never caught on in the UK. I suggested using that to expand the number of players who could take part in the Candidates Tournament, but nobody agreed with me, and some people just mocked the idea that you could even think to do that.

_________________
North Lecale

Actually in Fenrings's example I don't think it's possible for A to end up with less points than B. D has at least 1 point, B has 2 points, A has 3 points, 1+2+3=6. E has 1 point at most, F 1 point at most, and B has 2 points. 1+1+2=4 points. So it's not possible for A to lose.

I'm also thinking of a completely different idea!

Use McMahon for pairing only, with the results determined based on the Sum Of Defeated Opponent's Score's: each players opponents are ordered from the player with the highest McMahon to the lowest. The McMahon of each opponent is divided by the number of their rank down the list, and then the divided scores are added up to get the adjusted SODOS which determines the winner!!!

edit: I definitely think that the correct way to use McMahon is the prioritise the SODOS and SODOSODOS and so on, ignoring SOS which is useless in McMahon but useful for Swiss. SOS doesn't make any sense whatsoever as a tiebreaker in McMahon. SOS is for Swiss tournaments. The entire point of McMahon is essentially to replace the need for SOS or SOSOS and use SODOS and SODOSODOS instead. The main function of McMahon is to preset people's SOS according to their rank that represents the previous average level of their play. Which means SOS is terrible at scoring for people who do very well on the day or have an off-day. SOS works for Swiss tournaments where you might randomly meet opponents of very different rank in the early rounds, but in McMahon it's the opposite so SOS has a negative effect in deciding the points. Here SODOS is the obvious choice instead, haha !

Inertia, I suppose. I think that such a change would need to be called for by some bid in the AGM, which requires some determined people that are backed by some EGF member countries.
The 2021 update of the European rating system took almost 2 years from proposal to implementation.