Ultimate tie break for three way tie

[mitsun]

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

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.

[Matti]

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

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.

You are right.

And compared to this, the strategy for later stages becomes absurdly convoluted.

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.

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.

One could say that.

[Mef]

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.

[Matti]

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?

[Mef]

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.

[Matti]

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.

Let's call the number of ko captures NKC tie breaker. In this three way tie scenario I could use the NKC, if the players are equal by the score totals. However usage of NKC requires the games to be recorded.

[Matti]

I changed the subject to Penultimate tie break.

if we want to break a three way tie by playing we need at least two games for it. The question is how to do the paring. Assuming the players are equal streth the player who gets dirctly into the final has 50% chance of winning while the other two have 25% chances. Lets list some choices.

The penultimate tie break:

• a. Lottery
  b. Rating
  c. Prior order
  d. An extra round robin in between, if tied revert to a, b or c
  e. An extra round robin with point scoring tie break

In d. the chances of players winning are 6/16, 5/16 and 5/16. Do you find them equal enough?

In my opinion tie breaker e., when time allows is better than any of a.,b.,c.. It is more related to the playing skill. One coud compare it to extra time and penalty shoot out in football.

[ez4u]

The world has changed a bit since this thread began. I think that the interesting approach today would be to use an engine like Katago to determine a handicap/komi combination that gives Black a 66.7% chance of winning (let's guess it is something like a 2-stone game with a 17 point komi). So then...
1. Draw lots to give one player a bye (player C)
2. Players A and B play an even game (each has a 50% chance of winning).
3. The winner of this game (assume it is player A) takes Black against player C with the above mentioned handicap/komi kombo. This would result in both Players A (50% * 66.7%) and C a 33.3% chance of winning if I did my math right.

What do you think?! Maybe someone here could get Katago to spit out a sample configuration for the final game.

[Matti]

The world has changed a bit since this thread began. I think that the interesting approach today would be to use an engine like Katago to determine a handicap/komi combination that gives Black a 66.7% chance of winning (let's guess it is something like a 2-stone game with a 17 point komi). So then...
1. Draw lots to give one player a bye (player C)
2. Players A and B play an even game (each has a 50% chance of winning).
3. The winner of this game (assume it is player A) takes Black against player C with the above mentioned handicap/komi kombo. This would result in both Players A (50% * 66.7%) and C a 33.3% chance of winning if I did my math right.

What do you think?! Maybe someone here could get Katago to spit out a sample configuration for the final game.

Interesting.

[Waylon]

The world has changed a bit since this thread began. I think that the interesting approach today would be to use an engine like Katago to determine a handicap/komi combination that gives Black a 66.7% chance of winning (let's guess it is something like a 2-stone game with a 17 point komi). So then...
1. Draw lots to give one player a bye (player C)
2. Players A and B play an even game (each has a 50% chance of winning).
3. The winner of this game (assume it is player A) takes Black against player C with the above mentioned handicap/komi kombo. This would result in both Players A (50% * 66.7%) and C a 33.3% chance of winning if I did my math right.

What do you think?! Maybe someone here could get Katago to spit out a sample configuration for the final game.

Interesting idea.

Alternatively, you can make a kind of auction. The person who is willing to give the highest handicap in the final game, (stones or komi or whatever), goes directly to the final.

This way no luck is involved and they do not depend on the accuracy of a bot evaluation.
The players themselves decide, what kind of risk they are willing to take.

[Matti]

Interesting idea.

Alternatively, you can make a kind of auction. The person who is willing to give the highest handicap in the final game, (stones or komi or whatever), goes directly to the final.

This way no luck is involved and they do not depend on the accuracy of a bot evaluation.
The players themselves decide, what kind of risk they are willing to take.

How do you compare handicap and komi, for example 2 stones handicap vs. 20 points komi?

[Waylon]

How do you compare handicap and komi, for example 2 stones handicap vs. 20 points komi?

Results from pro games and bot games suggest that the advantage for the first move is close to 7 points. 2 handicap stones should be approximately equal to 21 points of komi.

The sorted list of possible handicaps for the auction is probably not the game-theoretic truth. I don't see this as a major problem, as long as the list is created by reasonably competent players and made public in the tournament conditions, to which all participants agree.

[Matti]

Next one needs to decide how to carry out the auction:

• from low to high alternate bids
  from high to low first bid
  simultaneous bids in envelopes

Each way has its own ramifications.

[Bantari]

Not to appear too flippant, but I like Mike's tongue-in-cheek suggestion - why not just draw straws?

With respect to the tiebreak being in any way connected to the strength, skill, or performance of the players - drawing straws is about as meaningful or as meaningless as what the OP proposes. The bonus of drawing straws is that it does not influence the strategy and/and tactics of preceding games.

If we really want to have a *meaningful* tiebreaker - I see no other way than to make them play games until a winner emerges. This might be inconvenient for the organizers, but such is the nature of this game.

Everything else is uncivilized.

[Bantari]

Oh... I forgot. And if we want something like penalty kicks in soccer to be a tiebreaker - we can get them to solve go problems and see who does it faster. This might be a good alternative, corresponding to penalty kicks. Just sayin....