Joel Embiid will not start in the 2023 All-Star Game, the NBA announced on Thursday evening, snapping a streak of five consecutive starts for the Sixers’ big man in the midseason event.
The full lineups are as follows:
EASTERN
Frontcourt: Kevin Durant, Jayson Tatum, Giannis Antetokounmpo
Backcourt: Kyrie Irving, Donovan Mitchell
WESTERN
Frontcourt: LeBron James, Nikola Jokic, Zion Williamson
Backcourt: Luka Doncic, Steph Curry
The battle for the final spot is a fascinating look at how different types of voters approach All-Star campaigns.
Giannis was arguably the easiest guy to bump out of the lineup for Embiid on paper, trailing Embiid in most major categories and stats while playing essentially the same number of games at the time voting concluded. With Milwaukee and Philadelphia near-even in record, there was little separating these two from a team perspective, but the fan vote was overwhelmingly in Giannis’ favor, and that 50-percent share of votes is hard to overcome.
- MORE ON THE SIXERS
- Emotions run high in Sixers' win over Ben Simmons, Brooklyn Nets
- Instant observations: Sixers win offensive shootout over Nets
- M. Night Shyamalan hangs out with Sixers' James Harden in 'Knock at the Cabin' promo
Durant is nearly as popular as Giannis amongst fans and arguably more revered amongst his peers and the media even with a recent injury layoff knocking him out of the lineup. His numbers this year have been astounding on offense, with Durant shaking off a tough start to the year and turning in a spectacular stretch to lead Brooklyn’s turnaround. Their play without him says a lot about his impact, and his presence on the Nets is the main reason, perhaps the only reason, people have begun to consider the Nets as a contender.
And then there’s Jayson Tatum, who is having an excellent season in his own right despite probably being the “worst” player in this group (not exactly a diss of the Celtics star). The biggest point of separation for Tatum is the games played factor — he has been a lot more available than anybody else on the list, and when you factor that in with his high level of play and his team’s success, the case to give him the start was easy enough to make.
Missing out on the start is no real shame for Embiid, though it does sort of illustrate the absurdity of a position-locked ballot for the All-Star Game in the modern NBA. By any reasonable assessment, Embiid has been one of the most important, most dominant, and most exciting players in the league once again this season, clearly one of the 10 guys who should be starting in this game.
Embiid's teammate James Harden’s case for the All-Star game is a good one, with Philadelphia’s lead playmaker averaging nearly a full assist per game more than anybody in the league, balancing the responsibility of table-setting with late-game scoring. His return to the lineup in early December pushed the Sixers back in the direction everyone expected, with Philadelphia owning the No. 1 offense in the league since he rejoined the lineup.
Whether he makes the game comes down to a few factors. Will coaches ding him for missing time in November? Do they prefer to reward a young guy who hasn’t been there before over Harden? Perhaps most importantly, do coaches prefer to pick an All-Star from another team over giving Philly two?
In the end, I think Harden probably makes this game relatively easily, as he has the right blend of individual production and team success to gain consensus support. That’s not the case for everybody else he’ll compete with — teams like Atlanta (Trae Young), Miami (Jimmy Butler), and Chicago (DeMar DeRozan, Zach LaVine) have all disappointed, opening the door for Harden to walk in through.
As we wrote on the subject previously, this might end up being a good thing for Embiid, adding a bit more fuel on the fire for a guy who has previously vocalized concerns about how voters handle his case for honors and awards. This is good fodder for us, but when we check the Basketball Reference page in 20 years, you’re going to see that star next to his name and not blink twice.