Kieran said:
masterclass said:
GameSetAndMath said:
Also, Andy is now #2 in the group pushing aside Kei to #3 (based on percentage of games won).
Hmm, not at this time. Kei is #2, and Andy is #3, because only he and Andy are tied, with same # of matches played and won, and if two players are tied, the first criterion is H2H. Kei beat Andy, so Kei is #2 at the moment.
Respectfully,
masterclass
Interesting. Hypothetically speaking, what if Raonic beats Kei and Roger beats Andy, both matches in straights, how will they separate the also-rans? Number of sets is tied, so it comes down to games won and lost?
a. Essentially, if a set of players have played
and won the same number of matches,
and
2 are tied, it is decided by H2H.
b. In the same scenario as a. but if
3 are tied, then it is decided by sets won. If one player has a better % than the other two, then he is ahead. If that leaves 2 still eligible, then their placing is determined by H2H. Anytime you have only 2 tied, it reverts to H2H.
c. In the same scenario as b. (3 tied), if all players have the same % of sets won, then it goes to game % won. They are ordered according to that, and then the same rules apply as described in b.
d. If 3 players all have the same set and game % won, then it goes to according to ATP ranking on Monday following Paris.
So in the scenario you describe, c. above will apply. Roger will win the group outright with 3-0 record. The other 3 will be 1 and 2 with equal sets (b). The player of the 3 with best game won % will be second in the group and the others will be eliminated. If two players have the same game won %, then the 3rd is eliminated and the player with that won the H2H will be second in the group. If they all 3 have the same game % then d will apply. The player with the best ranking will be second in the group and advance with Roger, in this case, Nishikori.
Respectfully,
masterclass
