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Lakers blog ~ The latest L.A. Lakers news, by the Orange County Register Sports staff

Kobe and Shaq on the same team?

January 6th, 2009, 7:00 am · 5 Comments · posted by JANIS CARR, OCREGISTER.COM

No need to rub your eyes,splash cold water on your face or pinch yourself. This is not a dream.

It’s not quite reality yet, either.

But there is a better-than-average chance Kobe and Shaq could be teammates again. No, no one is asking for a trade or opting out of their contract, but the former Laker duo could be playing on the same All-Star squad.

As the leading vote-getter in the West with more than 1.2 million votes, Bryant is certain of landing a spot in his 11th All-Star Game. O’Neal, however, will need the coaches’ help. Houston’s Yao Ming has more than double the number of votes Shaq, but the Suns center could get in as a reserve.

O’Neal is averaging 17.2 points, 8.9 rebounds and 1.5 blocks, and his 59.1 field-goal percentage ranks second in the league. Also, the game is being played in Phoenix, which could help Shaq’s cause.

The host city nearly always lands a player on the roster and Amar’e Stoudemire is the only other Phoenix player in the mix.

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    5 Comments

    5 Comments

    • deezy76 says:

      Wow, I never thought of that. I guess they would (or should) be back on the same team, if only for one night. Should be interesting…

    • titi says:

      this story was a feature yesterday on nba.com….ever think of at least give the link to the article that ‘gave’ you the idea of this ‘topic’?

    • JANIS CARR, OCREGISTER.COM says:

      youre right, titi, i should have linked it. thanks for reminding me.

    • Phoenix Stan says:

      This is the hands down winner of the lame ass attention grabbing headline of the year. Well done.

    • Chris says:

      Of course, the host city doesn’t always get a player on the roster. If I remember things correctly, when the All-Star game was in Oakland a few years ago, the Warriors didn’t have a single representative during any of the all-star weekend events.

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