A new formula for tiebreak score: WNT

[Mats Winther]

Until now, all tie-break systems (Buchholz, Sonneborn-Berger, etc.) have been designed around the principle of being easy to calculate by hand. The result is a coarse tie-breaking mechanism. There is no reason to hold on to that legacy today.

Tournaments often use both Buchholz and Sonneborn-Berger as tie-break systems, because Buchholz does not always succeed in separating players. Both the score and the Buchholz score may end up equal between players. Therefore, Sonneborn-Berger is added as a secondary tie-break.

Buchholz measures the strength of a player's opposition, or the difficulty of their pairing path: "How difficult was the player's route through the tournament?" Buchholz is completely blind to the actual outcomes of the games. A player receives full credit for an opponent's final score even if they lost to that opponent. Thus, it rewards players who have spent more time on the top boards.

Sonneborn-Berger, by contrast, measures performance or the quality of a player's results: "Which of these opponents did the player actually score points against?" It rewards players for scoring wins against strong opponents rather than against weaker players on the bottom boards. If a player has faced exceptionally strong opposition but lost all of their encounters with the top contenders, Buchholz captures the fact that they had a difficult draw. When two players have had roughly equivalent opposition, Sonneborn-Berger determines which of them actually performed better against the stronger players.

Some tournaments use this rather cumbersome tie-break sequence: Head-to-head result → Buchholz Cut-1 → Buchholz (full) → Sonneborn-Berger → Most wins / Most games played with Black. WNT resolves this more elegantly by combining strength of opposition and performance quality directly into a single measure.

To ensure that a player who faces strong opposition is not penalized for losing to the tournament leaders, the strength of the opponent must still be taken into account even in the case of a loss. The new algorithm is as follows:

Winther Score (WNT) = 100 × ∑ᵢ [ (Pᵢ / N) × (rᵢ + Pᵢ / (2N)) ]

∑ᵢ = the sum over all opponents you have faced

rᵢ = your result against opponent i (1, 0.5, or 0)

Pᵢ = opponent i's final score in the tournament

N = the total number of rounds in the tournament

Pᵢ / N = opponent i's score ratio (their final score expressed as a decimal between 0 and 1)

The key idea is that every opponent contributes in proportion to their final tournament performance. The result term (rᵢ) rewards points actually scored, while the additional strength term (Pᵢ / 2N) ensures that facing strong opponents retains value even when the game is lost. This combines the two dimensions traditionally captured separately by Buchholz and Sonneborn-Berger: strength of opposition and quality of results.

A player who both faces and scores points against the tournament's strongest competitors is rewarded quadratically and disproportionately more than a player who accumulates points against weaker opposition on the lower boards. As a result, the tournament strategy known as the "Swiss Gambit" becomes effectively useless.

In a five-round tournament, this has the following effect: losing to the tournament winner (40.5 points) is worth almost as much as defeating an average player (48 points), and it completely outweighs easy wins against the bottom of the field. Consequently, a player who has spent the tournament competing on the top boards will almost inevitably achieve a much higher total score than someone who earned their points primarily against opponents in the lower half of the standings.

Example: a five-round tournament

• A win against a strong player (4/5) yields the maximum return: 112 points. (Result bonus: 100 × 1 × 0.8 = 80. Opposition bonus: 50 × 0.64 = 32. Total: 112.)

• A draw against the same player yields 40 + 32 = 72 points.

• A loss yields 0 + 32 = 32 points. (In other words, you still receive 32 points simply for having faced top-level opposition.)

• A win against an average player (2.5/5) yields 62.5 points.

• A win against a player near the bottom of the standings (1/5) yields 22 points.

Thus, the system places a substantial premium on both facing and scoring against strong opposition. Wins against top performers are rewarded disproportionately more than wins against weaker opponents, while even losses to elite players retain significant value because opponent strength remains part of the calculation. This strongly incentivizes competitive performance on the top boards rather than point accumulation against the lower half of the field.

Unlike older systems, there are no unfair, abrupt threshold effects. Every half-point that a player's opponents earn during the tournament increases that player's WNT score. However, it is the encounters on the top boards that ultimately determine the prize standings.

In this way, WNT provides a continuous measure with no arbitrary cutoffs or discontinuities. Every result matters, but results involving the strongest players carry the greatest weight. The system therefore reflects both the overall strength of a player's opposition and their actual performance against that opposition, while ensuring that the decisive games at the top of the tournament receive the influence they deserve. Thanks to WNT's high mathematical resolution, the risk of two players finishing tied down to the exact same decimal is virtually negligible in practice.

Read more about it here: Chess Pairings: A Review of Monrad, Swiss, and Berger.

[JustinHorton]

What if a prior requirement for a tiebreaker is that it be easily comprehensible to the competitors and public

[John Upham]

What if a prior requirement for a tiebreaker is that it be easily comprehensible to the competitors and public

Something like rock, paper, scissors perhaps?

[Mats Winther]

What if a prior requirement for a tiebreaker is that it be easily comprehensible to the competitors and public

How is "Head-to-head result → Buchholz Cut-1 → Buchholz (full) → Sonneborn-Berger → Most wins / Most games played with Black" easily comprehensible to the competitors and public?

What they need to know is that the algorithm measures this: "Winning against a strong opponent is best, drawing them is second, but facing elite opposition is so valuable that even losing to a top contender earns points—consistently rewarding players who swim in the deep end over those who pick up wins against weak opposition."

The algorithm is simple enough that anybody with a calculator can verify their WNT. I have simulated several Monrad tournaments and it seems to work perfectly. The WNT is differentiated even if the tournament only lasts three rounds. The order in the result list tends to be close to what Buchholz generates, which is a good sign, despite its weaknesses.

Simulate a tournament yourself, with my online Monrad program. In the final result you can compare WNT with Buchholz and Cumulative.

[Roger de Coverly]

How is "Head-to-head result → Buchholz Cut-1 → Buchholz (full) → Sonneborn-Berger → Most wins / Most games played with Black" easily comprehensible to the competitors and public?

I think it's wrong to have tie breaks when you don't need them. Just share the prizes and report equal first etc. If there's a trophy or a qualifying place in a future tournament that's only when you need one. I think a tie break you can work out in your head is best as it can affect how you play in the last round, such as whether to accept a draw or not.

Complicated tie-breaks have become possible with the use of chess-results or similar. That organisers can use them doesn't mean they should, The same applies to rating prizes that are based on W-We.

[Ian Thompson]

Simulate a tournament yourself, with my online Monrad program. In the final result you can compare WNT with Buchholz and Cumulative.

Why simulate a tournament? Why not take a few real tournaments from chess-results.com where there were players tied at the end and show what the tie-break results would have been with your system compared to what they actually were. Try this one, for example. The tournament rules said monetary prizes would not be shared; they were all tie-broken. Therefore, for those players who finished tied for 2 - 5 places, the player who was 2nd on tie-break won €1700 while the player who was 5th got €700. Similarly, for those tied for 6 - 18 places, the player who was 6th on tie-break won €550 while the players in 16 - 18th places got nothing.

[Mats Winther]

I think it's wrong to have tie breaks when you don't need them. Just share the prizes and report equal first etc. If there's a trophy or a qualifying place in a future tournament that's only when you need one. I think a tie break you can work out in your head is best as it can affect how you play in the last round, such as whether to accept a draw or not.

Complicated tie-breaks have become possible with the use of chess-results or similar. That organisers can use them doesn't mean they should, The same applies to rating prizes that are based on W-We.

The strongest argument for using tie-breaks is that a raw tournament score is a coarse metric that treats fundamentally unequal paths as identical. In tournaments using open pairing formats, two players who finish with the exact same point total rarely faced the same level of opposition or endured the same level of difficulty.

A player who scores 5/7 by competing exclusively on top boards against the highest-rated seeds has accomplished something substantially more difficult than a player who scores 5/7 by playing catch-up against the lower half of the field. Tie-breaks correct for pairing anomalies by evaluating who you played, ensuring that surviving the toughest gauntlet is properly recognized.

Under pure score systems, taking an early loss or draw drops a player into the lower score brackets, where they face noticeably weaker opponents and can "coast" to late wins. Meanwhile, a player who starts strong is punished by facing the tournament favorites every single round. Tie-breaks eliminate the benefit of this "Swiss Gambit," ensuring that players who fight on the top boards throughout the event maintain an advantage over late surges from weaker tables.

In sports and games like chess, points are awarded in discrete chunks (1, 0.5, or 0). In a short 5- to 9-round tournament with dozens or hundreds of entrants, mathematical ties at the top are inevitable. A tie-break serves as a high-resolution lens over a low-resolution score, surfacing the subtle differences in overall tournament performance.

When only one player can lift a trophy, qualify for a championship, or claim a medal, an objective, pre-established mathematical tie-break provides an immediate, indisputable resolution that requires no extra rounds.

Points measure how many games you didn't lose; tie-breaks measure how hard those games were to win.

Without tie-breaks, open tournaments reduce an entire week of diverse, grueling competition down to a single blunt number, ignoring the quality of opposition, the courage of facing top contenders, and the reality of the tournament path.

Because WNT is such an objective and reliable measure of performance, one could even award a half-point or full-point bonus to any player who crosses a designated threshold.

[Christopher Kreuzer]

Sadly, your tiebreak does not help when someone swimming at the "deep end" (as you put it) is leapfrogged in the last round by someone who has played weaker opposition but ends up scoring more points.

It is a nice idea though. You might want to look into how complicated tiebreak systems have succeeded in getting adopted where others have failed.

[IM Jack Rudd]

Sadly, your tiebreak does not help when someone swimming at the "deep end" (as you put it) is leapfrogged in the last round by someone who has played weaker opposition but ends up scoring more points.

That's just inherent to the Swiss system, though. The only way around that is to have prizes based on things other than raw score.

[Roger de Coverly]

When only one player can lift a trophy, qualify for a championship, or claim a medal, an objective, pre-established mathematical tie-break provides an immediate, indisputable resolution that requires no extra rounds.

In my view that's the only time you need them.

What is the objection to sharing the money prizes? Pairings are arbitrary anyway, depending as they do on the initial starting rank which being based on published ratings could be objectively wrong.

[Mats Winther]

Sadly, your tiebreak does not help when someone swimming at the "deep end" (as you put it) is leapfrogged in the last round by someone who has played weaker opposition but ends up scoring more points.

It is a nice idea though. You might want to look into how complicated tiebreak systems have succeeded in getting adopted where others have failed.

It is far less complicated than current methods, which rely on a cascade of multiple, sequential tie-breakers. All you need is a single, straightforward formula, and it produces much better results. Anyone can test and verify it via the link I posted.

[IM Jack Rudd]

If you want a single tie-breaker that's unlikely to need follow-ups, sum of opponents' ratings usually does the trick.

[Mats Winther]

When only one player can lift a trophy, qualify for a championship, or claim a medal, an objective, pre-established mathematical tie-break provides an immediate, indisputable resolution that requires no extra rounds.

In my view that's the only time you need them.

What is the objection to sharing the money prizes? Pairings are arbitrary anyway, depending as they do on the initial starting rank which being based on published ratings could be objectively wrong.

One could award a higher share of the prize money to the player with the highest tie-break.

[Mats Winther]

If you want a single tie-breaker that's unlikely to need follow-ups, sum of opponents' ratings usually does the trick.

Rating is very unreliable among juniors.

[Christopher Kreuzer]

A player who scores 5/7 by competing exclusively on top boards against the highest-rated seeds has accomplished something substantially more difficult than a player who scores 5/7 by playing catch-up against the lower half of the field.

So, to change this slightly: "A player who scores 5/7 by competing exclusively on top boards against the highest-rated seeds has accomplished something substantially more difficult than a player who scores 5.5/7 by playing catch-up against the lower half of the field."

At what point does playing against a stronger field outweigh the points gained against weaker players? If chess tournaments were meant to be decided by strength of opposition faced, that would make it a different game.

A tie-break serves as a high-resolution lens over a low-resolution score, surfacing the subtle differences in overall tournament performance.

[...]

Because WNT is such an objective and reliable measure of performance, one could even award a half-point or full-point bonus to any player who crosses a designated threshold.

Really? Seriously?

There is a reason that all-play-all tournaments are considered a better measure of chess ability than Swiss tournaments.