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View Full Version : Random sidenote of the Celtics



Los Angeles
03-19-2006, 06:41 PM
I didn't know NBA players played for money.

Roy Munson
03-19-2006, 06:46 PM
He said West gets a paycheck for $89,000 every Friday, or at least he did last year.
..........

That comes to about 4.7 million per year. I don't think he makes that much.

GO!!!!!
03-19-2006, 06:49 PM
hmmmm

Ant
03-19-2006, 06:51 PM
Where do i sign up to be an Nba players best friend ?? :signit:

Roaming Gnome
03-19-2006, 06:52 PM
That comes to about 4.7 million per year. I don't think he makes that much.
Did you figure that times the number of weeks during the season? I don't think the players draw a paycheck durring the off season. Or did you figure it for 52 weeks?

Roy Munson
03-19-2006, 07:03 PM
Did you figure that times the number of weeks during the season? I don't think the players draw a paycheck durring the off season. Or did you figure it for 52 weeks?

I figured it for 52 weeks. I had heard that NBA players get 52 weekly paychecks per year, but maybe some arrangements are different.

Kstat
03-19-2006, 07:07 PM
Did you figure that times the number of weeks during the season? I don't think the players draw a paycheck durring the off season. Or did you figure it for 52 weeks?

West makes barely over a million dollars per season. NBA players draw bi-weekly paychecks to my knowledge, so say 13 paychecks in an NBA season.

West would get a check for about 77,000 every 2 weeks.

Ara
03-19-2006, 07:13 PM
That comes to about 4.7 million per year. I don't think he makes that much.

I wish I pulled that as my salary :D

Eindar
03-20-2006, 01:42 AM
It could also be that West was telling the guy how much he made per week, when endorsements (local) and other stuff was added in, not neccessarily his NBA check.

Or, he could be full of crap.

Chauncey
03-20-2006, 10:34 AM
They're paid bi-weekly.

ChicagoJ
03-20-2006, 09:27 PM
Kenny Anderson used to be paid in a lump sum, on the first day of the "season" (July 1).

Payment frequency can be contract-by-contract.