In some sports tie breaks are used when the match ends in tie after the main time. Extra time periods with possibly sudden death and penalty shoot outs are used for example in football and ice hockey. In go tournaments if there is extra time tie break games can be arranged. Tie break for top two players is straightforward. For three players it is not. Of course one may pick two players by other tied breakers or by lottery, pair them and let the winner play against the third, but this is slightly partial.

Here is my idea of the ultimate tie break:
Play round robin with the three players. In addition of winning and losing, the scores recorded as well. If there is still a new three way tie on wins, then the scores (win posiitve and loss negative) of players will be totaled. The player with lowest total will be eliminated and the two remaining will play a final.
If the last points are tied the two play an extra game to choose the finalist. If all three have zero score, then the two who played last play a game to choose a finalist against the remaining player.

Well any method can be used to break the tie, including drawing straws.

I used this example of a "non-go" solution to tie breaking precisely to draw attention to "margin of victory" not being part of the definition of the game of go.

Note that "the winner has more" does not actually require counting and arithmetic to determine (can be just pairing and "left over" --- which is what we do when not "countable", when we compare infinite sets).

Also, what is the margin of victory you propose to assign to any game that ended by resignation?

Interesting, but there is a rather severe issue of information asymmetry.

E.g suppose the first rounds are:

A defeats B by 4.5
B defeats C by 0.5

That means A is at +4.5, B is at -4 and C is at -0.5

Now, for the game A-C, both players are aware that while A can win outright with a victory, a win by C of less than 8.5 points still gives both players access to the final!

That means that, if A falls behind, he can play a super-defensive game to reach the final. And there is no incentive for C to take risks to maximize his score, as he cannot win outright anyway.

That is, in my opinion, a very severe downside to this system, so I don't think the system deserves a name like "ultimate tie-break"

I would not call this a flaw, but a property of the system. After B has lost to A by 4.5 points, he knows that to ensure a place in the competion he needs to win C at least by 4,5 points.

There's still information asymmetry. Let's say that the full series goes:
A defeats B by 4.5
B defeats C by 0.5
C defeats A by 7.5
A=-3,C=+7,B=-4.
In game 2, B needed 4.5 points to ensure a place in the final

But if we reorder the last two games:
A defeats B by 4.5
C defeats A by 7.5
B defeats C by 0.5

B goes into game 3 knowing that a 2.5 win is enough. So maybe rather than reaching for 4.5 and falling to 0.5, he's able to find a safer 2.5 win.

Additionally, the larger the margin in the 3rd game, the weirder things become. If we multiply those scores by 10, B needs to win his third game by 11.5 points. Let's say in the late middle game he's clearly up by around 5 points. Now he's kingmaker: if he wins by this margin it's A vs C, if he loses it's just C. A well designed tournament should minimize kingmaker scenarios.

That is a separate question. We can fix a number for example 38 points. Another option is let the person who resigns state the value of resignation, but then we need additional rules in case the winner disputes the points. Yet another possibility is to make players finish the game.

If one considers this kingmaker scenario a problem, then an additional game or two between the top two players would solve it.

I can't figure out how to get that to work. The most natural sounding scenario to me is:

"round robin, the two highest scoring players on wins, then sum of scores, continue playing until one has 2 wins over the other".

Now in the earlier scenario there's no way for B to influence the games between A and C.

Instead, you have:
A beats B by 4.5
C beats A 7.5
C vs B:

If C wins or B wins by 0.5, it's A vs. C
If B wins by 2.5-9.5, it's B vs C
If B wins by 11.5+, it's B vs A

Except that B doesn't want to face A (because A is up one win on him). So whoever is excluded from the final match has no hope of facing the opponent most advantageous to them.

I can't come up with any scenarios where somebody isn't either outright benefited from limiting their win margin in the last game, or given the ability to positively influence the outcome of the other match-up after being eliminated, or given the opportunity to choose their final opponent (thus promoting the inferior player).

edit: Which isn't an absolute knock on the system, this kind of thing comes up frequently in tournament design. But it does put question to the ultimate moniker

The information asymmetry problem would go away if the result of every game was unknown to the player who did not participate in that game. There would still be some dependence on the order of games though, as every player would have more information going into their second game than they had at the start.

Here is an updated version. Four or five rounds will be played.

three player round robin
If the win distribution is 2,1,0, the players with 2 and 1 win play against each other until one of them reaches three wins.
If the win distribution is 1,1,1, the players are ordered by score sum and ties are resolved by giving priority to the player with a later bye. A game 2nd vs. 3rd is played and the winner plays a final against 1st.

By now it may be best to change the thread title to "Fantasy Tiebreaker".

_________________
Dave Sigaty
"Short-lived are both the praiser and the praised, and rememberer and the remembered..."
- Marcus Aurelius; Meditations, VIII 21

If think my point "but that's a different game" is not being addressed.

Please, consider a match between two players, a match of nine games and a method of scoring games that does not allow for a tie game.

Method One: Whichever player has five or more victories wins the match. Note that there will be a winner. I contend that this is what we normally consider the game of go.

Method Two: Whichever player has the higher point total from the scores of the nine games combined. Note that we are no longer guaranteed a winner (tie is possible). Note also that player A could have won 8 of the 9 games (each by a couple points) but still lose the match. Is that what you consider go?

The point I am making is that there are games we score by method two (say poker, bridge, skat, etc.) but that go isn't one of them. There are also games that we could not score by method two (chess, for example, where there is win, lose, draw, but absolutely no measure of margin). That win or lose in go is determined by the point outcome of a game is a distraction.

Of course these two methods are different.

Tie breaks are often played by different rules than the main match. Let's consider the game before komi was introduced. It was known that black had an advantage. Suppose that two players were equal enough that black would win all the games in a match. The tie could be broken by having the players to play two games with alternate colours and adding the scores. The first game would not be exactly the same go we are used. Black attempts for a safe win and within the safety marginal maximize the score. White attempts to lose as little as possible or not to lose at all, but does not take any risks of losing big. The second game would be like normal go. The margin of the first game can be considered as komi for the second.

When we have three players and three games, each player has a pair of games that work this way. A player has incentive to win games. The more wins he gets, the less wins he needs afterwards. The scores of players total to zero, so a players with at least zero score will not be eliminated. Thus in his second game the players knows which score to aim.

Yes of course, including flipping a coin (break a tie election in some places). So we could decide a tied "football" match by having them shoot fouls (basketball).

But what we don't (usually) do is have the tie breaking method influence in any way the previous play. And what you are proposing does that, can change the way the games are played before there is a tie situation (in case there should be a tie).

Reality of go: The margin of victory is not an indication of how close the game was. A game where one player gradually falls 5-10 points behind and then the game proceeds solidly/steadily to its foregone conclusion is not a close game compared to one were there is a desperate life or death fight between two large groups going on for a sequence of 30-40 critical moves where both sides must find "the only move" (many of which far from obvious, some at first even seeming outlandish) that ends when the fight is decided (and the margin, if not resigned, would be >100 points).

This discussion is about tie break. The assumption is, that games played before have been played with the regular rules and have produced a three way tie, which we are trying to break with further games.
The scores are not used for indicating how close the game was. Do you assume that there can be only two meaningful results in a game? If there is non integer komi, that is the case, but if there is no komi or the komi is an integer there may be three different outcomes: black wins, jigo or black loses. In this case player attempts to win. If he sees that this goal is not realistic, he might strive for jigo instead of risking a loss when trying to win. He might also decided the opposite. The opponent has similar decisions to make. To me this game is also go.

Now let's analyse the tie break system:
1st game A vs. B:

Both players attempt to win. We call the winning marginal s1, positive if A wins and negative is B wins. Business as usual.

2nd game B vs. C:
We call the winning marginal s2.
s1<0

B's aims are s2+s1>0 and if that fails s2>0
C's aims are s2<0 and if that fails he prefers to foil B's primary aim

s1>0

B's aims are s2>0 and if that fails s2+s1>0
C's aims are s2+s1<0 and if that fails s2<0

3rd game C vs. A:
We call the winning marginal s3.
(s1<0 and s2>0) or (s1>0 and s2<0)

Three way tie on wins is excluded. Both players attempt to win. Business as usual.

(s1<0 and s2<0)

C's aims are s3>0 and if that fails s3-s2>0
A's aims are s1-s3>0 and if that fails s3<0.

(s1>0 and s2>0)

A's aims are s3<0 and if that fails s1-s3>0
C's aims are s3-s2>0 and if that fails s3>0.

This round robin sets up the stage for final games where players with two losses (from RR and subsequent games) get eliminated.

Even this stage is not business as usual, since the winning (losing) player has an incentive to win (lose) by as large (small) a margin as possible, and might well be willing to take risks to achieve that goal. So both players are likely to alter their strategy, making the game different than if played under normal rules where only win/loss matters. And compared to this, the strategy for later stages becomes absurdly convoluted. You might as well just say that the tie-breaker consists of one round of a new game called margin-Go, different from the game of Go played in earlier rounds.

You are right.I would not call it absurdly convoluted. In the second game players B and C know whether s1 is positive or negative. In both cases they have opposing aims, where ones primary aims foils both opponents aims, but both may achieve their secondary aim simultaneously. This is comparable with no komi game where jigo is possible.

The third game is not much more complicated. When B has 0 or 2 wins, its straightforward. Suppose B has one win. Both A and C have an incentive to win. With no wins one is out of competition. Having two wins is better than having one, because then one has two chances to win the tournament instead of one chance. If all three players get one win each, the player with highest score total has an advantage, while there is no difference between the second and third. Thus if B has negative score total, both A and C have the first place a stake, otherwise only one of them has.
One could say that.

I say we break the tie by awarding the win to the one with the fewest total ko captures at the end of the tournament. That way every time you think about fighting a ko, you not only have to count your threats, but think about how much you might hurt your chances in the meta-game.

Do you count only single stone captures or also if there are repeated captures of multiple stones?

Single or multi-stone captures would both count, but we are counting the moves not the number of captures. For the sake of making it easy, a ko capture can be any move that both captures at least 1 stone and creates a board position where there is at least one point the opponent may not immediately play due to a ko restriction.