Dallas and OKC

[ChuckD]

I'd rather face Dallas. You have to remember OKC's record is deceptive, as they made a major upgrade midseason. Perkins just had 18 boards today and their record would be a lot better if they had this team all year.

To me, outside of their stars, the one player on OKC that gave us absolute fits was Jeff Green. I think he's like a 34% 3 point shooter (below NBA average) , but ing filled it up against us, and opened the lane for their guards. I'm glad he's gone. Perkins is a nice pickup, and a good defender and rebounder, but he's almost no threat to post up and score or step out for a jumper.

[MajorMike]

Wrong. A division winner is only guaranteed a top 4 seed, so if OKC and Dallas are tied for #2 (with the Lakers at #4), Dallas gets the #2 seed and HCA in the second round.

Good take the rest of the way, though. I still think the 6-8 spots could easily change, but I agree that 1-5 will end the way they currently stand.

Incorrect.


As per NBA.com:

In the event two or more teams are tied in the standings, a series of tiebreakers are applied to determine which team receives the higher seeding.

Two-Team Tiebreaker:

1.Division winner (this criterion is applied regardless of whether the tied teams are in the same division)
2.Better record in head-to-head games
3.Higher winning percentage within division (if teams are in the same division)
4.Higher winning percentage in conference games
5.Higher winning percentage against playoff teams in own conference
6.Higher winning percentage against playoff teams in opposite conference
7.Higher point differential between points scored and points allowed


HOWEVER, homecourt is determined by RECORD, not SEEDING. Therefore OKC would be seeded ahead of Dallas, but if they were seeds #2 and #3, which is still possible as per the scenario in my last post, that means Dallas would actually host OKC.

In other words, if OKC does get the #2, in effect, Dallas gets the #2 becaus ethey get HC over OKC.

[ChuckD]

Wrong. A division winner is only guaranteed a top 4 seed, so if OKC and Dallas are tied for #2 (with the Lakers at #4), Dallas gets the #2 seed and HCA in the second round.

Good take the rest of the way, though. I still think the 6-8 spots could easily change, but I agree that 1-5 will end the way they currently stand.

You need to take a fresh look at the tie-breaker criteria. Go to the nba standings by conference page and look below the standings. The first tie-breaker is "are you a division winner?". You can be as high as a #2 seed as a non-division winner, but you are at a distinct disadvantage if you allow yourself to fall into any ties in the top half of the draw. Head to head isn't even a factor if you beat the team 4-0 but lose your division. Dallas automatically loses tie-breakers to LA or OKC.

[Seventyniner]

Incorrect.

[snip]

HOWEVER, homecourt is determined by RECORD, not SEEDING. Therefore OKC would be seeded ahead of Dallas, but if they were seeds #2 and #3, which is still possible as per the scenario in my last post, that means Dallas would actually host OKC.

In other words, if OKC does get the #2, in effect, Dallas gets the #2 becaus ethey get HC over OKC.

Hrmm...thanks for pointing that out. I was only half-right, then; Dallas would get HCA over OKC in the second round but not the #2 seed in case of a tie. I thought Dallas would get both.

That division winner tiebreaker is silly, though. I thought they introduced the whole only-top-4-seed-guaranteed-for-a-division-winner because of the Spurs and Mavs in 2006 (best 2 records in the conference, but met in the 2nd round). Allowing division winners the tiebreaker over head-to-head goes against the spirit of this. I think it would make more sense for winning the division to be the 2nd tiebreaker (right behind head-to-head).

[ChuckD]

Hrmm...thanks for pointing that out. I was only half-right, then; Dallas would get HCA over OKC in the second round but not the #2 seed in case of a tie. I thought Dallas would get both.

That division winner tiebreaker is silly, though. I thought they introduced the whole only-top-4-seed-guaranteed-for-a-division-winner because of the Spurs and Mavs in 2006 (best 2 records in the conference, but met in the 2nd round). Allowing division winners the tiebreaker over head-to-head goes against the spirit of this. I think it would make more sense for winning the division to be the 2nd tiebreaker (right behind head-to-head).

You're not David Stern, trying desperately to prop up the aging Lakers, who play in one of the two weakest divisions in the NBA and are a shoo in to win it every year.

[MajorMike]

Hrmm...thanks for pointing that out. I was only half-right, then; Dallas would get HCA over OKC in the second round but not the #2 seed in case of a tie. I thought Dallas would get both.

That division winner tiebreaker is silly, though. I thought they introduced the whole only-top-4-seed-guaranteed-for-a-division-winner because of the Spurs and Mavs in 2006 (best 2 records in the conference, but met in the 2nd round). Allowing division winners the tiebreaker over head-to-head goes against the spirit of this. I think it would make more sense for winning the division to be the 2nd tiebreaker (right behind head-to-head).

I don't disagree.

I do applaud the NBA for allowing for a team that wins 60 games but failing to win their division on having home court even if they must play tougher teams because of lower seeding. The NBA has done this for some time. For instance in 2003 when Dallas had 60 wins and finished 2nd in the division to SA, Dallas was the #3 seed but played at home against #2 seed 59-win SACTO.

In 2005, before the went to the current 'top 4' formula, Dallas was 60 wins but the #4 seed, but still had HC in the WCF over #2 seed Phx (54 wins).