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Report: Warriors’ David Lee passed over as Western Conference All-Star Game reserve

OAKLAND — Warriors forward David Lee was left off a list of seven All-Star reserves who earned selections Thursday to play for the Western Conference in the All-Star game in New Orleans on Feb. 16, according to a report from Yahoo! Sports’ Adrian Wojnarowski.

Lee had been hoping to be named an All-Star for the second straight season and third time overall. He has averaged 19 points and 9.8 rebounds this season.

There remains a chance Lee could make the team if he is named by the NBA commissioner as an injury replacement for a player unable to participate, with Western Conference starter Kobe Bryant being one such player. Lee was initially snubbed in 2010 while with the New York Knicks before being named to the squad as an injury replacement.

The Warriors will be represented in the game by guard Stephen Curry, who was voted by the fans as a Western Conference starter. Second-year forward Harrison Barnes has been named to the Rising Stars Challenge roster for a game to be played Feb. 14.

Diamond Leung

  • Grey Warden

    I’d rather have him use the All-Star break to rest his injured shoulder.

  • 808Oahu

    Their actually attacking the basket tonight and getting easier points.
    This team is so bipolar. Lol

  • Howard Benner

    This team has the weakest psyche of any potential playoff team in the league. For the most part they are outplaying the heavy legged clips, but here down the stretch they are re revealing their weakness as the clips have cut 10 points off the 25 point lead. . .Barnes & Harrison are rubbing off onto Crawford. . .the weakest mental makeup of any team in the league. . .

  • Howard Benner

    This blog broke down the prospective all star bench players. No way lee deserves to be on that team. . .it broke down the way is should. . .passed over? he should relish the days off. . .btw; the W’s should have played well enough in the fourth to allow curry to stay on the bench. . hope it doesn’t bite them on the a$$