In round-robin Chess tournaments, sometimes there are ties. This can, of course, be addressed by having the players involved play one another, but Chess games do sometimes end in draws, and time may be limited as well.

Thus, several systems are in use to provide an additional figure of merit for players in Chess tournaments.

The starting input into these systems is generally what will be termed each player's "conventional score": the player's score for the games played, based on 1 for each win, 1/2 for each draw, and 0 for each loss.

These systems all assume that the players for which a score is calculated have all played an equal number of games. Of course, in a round-robin all-play-all tournament, as opposed to an elimination contest, there is no need for any byes; but they are also used for shorter tournaments which use the Swiss System of pairing in order to have fewer rounds.

This system is the simplest to understand. The score is simply the average of the conventional scores of all the opponents one has played.

As this takes no account of whether one has won or lost, it cannot replace the conventional score.

This depends on the tournament being a Swiss System tournament instead of an all-play-all tournament, as it rests on the assumption that stronger players will be playing stronger opponents in the later rounds.

The cumulative total of a player's current conventional score after each game is then again totalled. Thus, the result is one times the score of the final game, two times that of the next-to-last game, three times that of the one before, and so on.

Here, what is taken is the sum of the conventional scores of the players against whom one has won, plus half the sum of the conventional scores of the players with whom one has drawn.

Inspired by both the Neustadtl Score, and the Harkness System of assigning players initial ratings, I will attempt to propose a system which gives a tournament score that accurately indicates what can be concluded about player strength from a tournament.

Let the tiebreak score of a player be the sum of the following three quantities:

- The number of games won, times (half the number of games played plus one-half the average of the conventional scores of the players against whom the player won).
- The number of games drawn, times the average of the conventional scores of the players against whom the player drew.
- The number of games lost, times one-half the average of the conventional scores of the players against whom the player lost.

To scale this score, so that it can be compared to the conventional score, it can be divided by the number of games played.

And, if this does not resolve a tie, calculate the tiebreak score for all the players, or at
least all the players who played against one of the players who is tied, and
then take the average of the conventional score and the *scaled* tiebreak score for each
player, and put that into the
formula for calculating the tiebreak score in place of the conventional score to calculate the
second tiebreak score.

Of course, no tiebreak formula of this general kind will resolve a tie if the two tied players won, drew, and lost against the same opponents (as well as possibly drawing, or otherwise having an even result, against each other). The Cumulative system, described above, is one of a different kind, since there the round in which one plays another player is significant, and, thus, it is available to be resorted to in this case. If it, too, produces a tie, the option is now available to replace the conventional scores of the players with their scaled tiebreak scores by this method, and then their scaled second tiebreak scores, and so on, as presumably that would differentiate the scores of some of the other players, leading to a difference in the scores of the tied players.

But if it is relevant to do so, a third tiebreak score can be calculated as follows:

- Calculate the tiebreak score for any remaning players for whom it had not needed to be calculated for the attempt to break the tie with the second tiebreak score.
- Calculate the second tiebreak score for at least those players, other than those involved in the tie, who played against one of the players involved in the tie.
- Calculate the third tiebreak score for the players involved in the tie by using the method for the tiebreak score, but with each player's conventional score replaced by the sum of 1/3 of the scaled second tiebreak score, 2/9 of the scaled tiebreak score, and 4/9 of the conventional score.

Following this pattern, a fourth tiebreak score would be calculated by replacing the conventional score of each player with 1/4 of the scaled third tiebreak score, 3/16 of the scaled second tiebreak score, 9/64 of the scaled tiebreak score, and 27/64 of the conventional score.

The pattern is that if there are N previous tiebreak scores, the (N+1)th tiebreak score is calculated with the conventional score replaced by:

1/(N+1) times tiebreak score N, plus

(1/(N+1)) times (N/(N+1)) times scaled tiebreak score N-1, plus

(1/(N+1)) times (N/(N+1)) squared times scaled tiebreak score N-2, plus

...

(1/(N+1)) times (N/(N+1))^(N-1) times the scaled (first) tiebreak score, plus

(N/(N+1))^N times the conventional score.