Does anyone know how they determine who wins in tie situations? What if Chicago, Washington, and Indiana all tie?

I'm pretty sure that if and when two (or more) teams are tied record-wise, the tiebreaker is division record. If that's tied, then it's conference record.

The tiebreaker is division record only if all 3 teams are in the same division, otherwise it is by conference record. You would think the head-to-head matchups would be the first factor. The Bulls definitely don't want to lose to the Raptors and play the Pistons.

Chicago would hold the tiebreaker over Washington and Indiana based on a better conference record.

Indiana holds the tiebreaker over Chicago based on their division record if they tied.

The tiebreaker is division record only if all 3 teams are in the same division, otherwise it is by conference record. You would think the head-to-head matchups would be the first factor. The Bulls definitely don't want to lose to the Raptors and play the Pistons.

Chicago would hold the tiebreaker over Washington and Indiana based on a better conference record.

Indiana holds the tiebreaker over Chicago based on their division record if they tied.
Thanks for clearing that up.

Odds are the Wizards are #5, Pacers #6, Bucks #7, and Bulls #8. Of course anything can happen when 4 teams are separated by half a game.

Odds are the Wizards are #5, Pacers #6, Bucks #7, and Bulls #8. Of course anything can happen when 4 teams are separated by half a game.


If the Bulls win tonight and the Pacers lose, the Bulls move to number 6 and play the Nets and the Pacers will be 7th and the Bucks will play the Pistons

If the Bulls win tonight and the Pacers lose, the Bulls move to number 6 and play the Nets and the Pacers will be 7th and the Bucks will play the Pistons


Anything can happen tonight. I just hope the Nets don't have to play the Pacers.

Anything can happen tonight. I just hope the Nets don't have to play the Pacers.



You better keep your eye on TWO GAMES tonight, the Raptors at Bulls and Orlando at Indiana