Article 16.2.3 – If three or more players/ pairs have won the same number of matches, ranking will be established by the difference between total games won and total games lost, with greater difference ranked higher. the question: how is the difference computed? is it the total points of the players in triple tie only or the total points of all 4 players in the bracket?
It is total games won minus total games lost, between everyone in the bracket. Let's say you have: Player A Won 3 matches 3-0. Won 2 matches in 2 and 1 match in 3. Player B Won 1 Loss 2. Won the 1 match in 3, lost 1 match in 3, and 2nd in 2. Player C Won 1 Loss 2. Won the 1st match in 3, lost the 2nd and 3rd match in 2. Player D Won 1 Loss 2. Won the 1st match in 3. Lost 2nd and 3rd match in 3. Player A is undisputed 1st as they didn't lose any matches. Player B is 2-1, 1-2, 0-2. So they are at -2 games. Player C is 2-1, 0-2, 0-2. So they are at -3 games. Player D is 2-1, 1-2, 1-2. They are at -1 games. Player D is 2nd, B is 3rd, C is 4th. If there was a 3 way tie, broken down to a 2 way tie, then the 2 way tie is then broken up by H2H result. If 3 way tie in games, goes down to points. If 3 way tie in points, you draw lots.
thank you for the response. so the game with the 4th player, irregardless if the 4th player is the top or bottom in the ranking, will still be used in the computation. some interpretations in local tournaments here have it that only the games of the 3 players are used in computing for the tie breaker.
