What are Tiebreakers vs. West?
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Q. With the Pacers going for the best record in the NBA, what is the tiebreaker to determine who gets homecourt (advantage) in the NBA Finals? I'm sure head-to-head is one of the deciding factors, but what if each team ends up 1-1 against each other? Secondly, has any team with the best record in the NBA ever lost a first round series? (From Randy in Oshkosh, WI)
A. It boils down to this: the Pacers’ magic number for clinching the best record in the NBA is four. Any combination of victories by the Pacers and losses by the West’s top team that adds up to four gives Indiana homecourt advantage throughout the playoffs.
The first tiebreaker is head-to-head competition. Against the four Western Conference teams currently battling for the top spot – the Kings, Lakers, Spurs and Timberwolves – the Pacers have split three of the season series, but were swept by Sacramento. The second tiebreaker (as it applies to West teams) is the record against the opposite conference. While the Pacers have the best winning percentage in the league against the West (20-8, .714), they would lose this tiebreaker to any of those four teams. The worst record against the East among them belongs to San Antonio (22-8), which still has a higher winning percentage (.733) than the Pacers. So that’s where the debate will end.
Just for fun, the rest of the tiebreakers, in order (with the Pacers’ record), are: record within your own division (18-8), record against playoff teams in your conference (16-9), record against playoff teams in the opposite conference (currently 9-7, though it could drop to 8-8 if Portland winds up eighth in the West) and net points (the Pacers have outscored opponents by 418 points).
As for your second question, yes, the team with the league’s best record has been beaten in the first round – but just once since the playoffs were expanded to 16 teams. In 1993-94, Seattle (63-19) was stunned in five games by the eighth-seeded Nuggets (42-40). Denver won the final two games in overtime.