Will the Golden State Warriors Lose Before the Philadelphia 76ers Win?
The Golden State Warriors may never lose again.
At 13-0, the Warriors have looked pretty invincible so far this season. Their average point differential of 14.7 would set an NBA record if they can keep it up (the current record is held by the 1971-72 Los Angeles Lakers at 12.3). With each additional win, the Dubs seem more and more like they have a realistic shot at challenging the 1995-96 Chicago Bulls and their NBA record of 72 wins in a single season.
Golden State, as a team, has a 94.4 nERD. If you're not familiar with our metric, nERD is set on a scale from 0-100, with 50 as the league average. It's meant to be predictive of the team's ultimate winning percentage in a schedule-free vacuum based on performance-to-date. In other words, the Warriors are currently playing like a 77-win team (.944 winning percentage over 82 games).
That, in case you were wondering, is insane.
Honestly, if they were ever going to lose a game, it should've been last night on the road versus the Los Angeles Clippers. They were down by 23 at one point. With five minutes to go in the game, Paul Pierce hit a three and brought the Clippers' win probability up to 91.7%.Â
Well, as it turns out, the 8.3% that the Warriors had left was plenty for them to work with.
Meanwhile, the Philadelphia 76ers might never win again.
The Sixers are in what feels like year 10 of a slow rebuild (or total tank job, depending on how carefully you choose your words). They have been the antithesis of the Warriors, winless with a record of 0-12. Their average point differential of -14.0 is on pace to be the second worst of all-time and not far off the record setting -15.2 set by the 1992-93 Dallas Mavericks.Â
While the Warriors are eyeing the all-time win record, this year's Sixers look like they could take a run at the league record for losses, which coincidentally they set themselves in 1972-73 with 73 (poor Philly fans, man). Their team nERD currently sits at 9.5, meaning they're playing like an eight-win team (.095 winning percentage over 82 games).
One of the more intriguing storylines to emerge so far in 2015-16 is the possibility that we could end up witnessing both the best and worst regular season teams in NBA history in the same season. While 82-0 and 0-82 are obviously near impossibilities, it has become an interesting debate to try and guess which will happen first, a Warriors loss or a Sixers win.
Well, we ran the question through our algorithms and here's what we got:
| Situation | Probability |
|---|---|
| Sixers win first | 43.7% |
| Warriors lose first | 48.4% |
| Both events happen on the same day | 7.9% |
The two eventualities are fairly close, with the Sixers having a 43.7% chance to win first, while the Warriors have a 48.4% chance to lose before that happens (with a 7.9% chance that both events happen on the same day). So, if you were wondering what's
more likely to happen, the answer is that the Warriors are most likely to lose before the Sixers win.Â
A couple dates to watch for a Sixers win would be December 1st against the Los Angeles Lakers (27th in our NBA Team Power Rankings with a Team nERD of 26.1) and December 10th against the Brooklyn Nets (26th in our rankings with a nERD of 26.1). The Sixers might not be favored to win in another game this season, but they can't lose them all, can they?
Don't answer that.
Meanwhile, the Warriors don't face another one of the top-5 teams in our rankings until they host the Cleveland Cavaliers on Christmas Day. They'd be 28-0 if they made it that far. That's not likely to happen, but boy is it starting to look at least a little possible.
If all else fails, the Sixers and the Warriors will face each other on January 30th in Philadelphia. A Sixers win would certainly be poetic, but you don't really need an algorithm to tell you how unlikely that is to happen at this point.
