2007-07-13 12:56:15 · 12 answers · asked by Smitty 2 in Sports ➔ Baseball
Yes. He is the best player of our time, and fills the seats.
2007-07-13 13:00:10 · answer #1 · answered by Sam N 3 · 0⤊ 0⤋
This topic was actually the topic of extensive analysis in the excellent book Baseball Between The Numbers: Why Everything You Know About The Game Is Wrong.
The book analyzes A-Rod's impact on the team's record, as well as merchandise and ticket sales, with several economic models. It's a great piece that all baseball fans should read, but the short answer is that even the most optimistic analysis estimates that Rodriguez contributes about $19/million a year to team revenue.
Is A-Rod worth $25 million a year? Maybe not. But a lot of players make much more than they're worth. Once he becomes a free agent, his market value will likely be even higher than $25 million/yr.
Most teams overpay for talent. Rodriguez is probably overpaid, but he's a valuable producer and a very marketable player.
2007-07-13 13:07:32 · answer #2 · answered by ltjrawker 2 · 0⤊ 0⤋
Yes....he is worth every penny. If you dont believe that then you have been living under a rock for the last 10 yrs.
Think about it......he already is one of the top 10 greatest players ever. By the time he finishes his next contract (assuming he opts out ) He would have passed bonds for the all time home run record.
When he retires he will be called the greatest player of all time......regardless of if he wins a championship or not. Thats b/c of all the steroid use in baseball today.
30 million a year.........worth more like 40 million a year!!!!
2007-07-13 19:45:45 · answer #3 · answered by sup pimp2143 1 · 0⤊ 0⤋
Yes. A-rod is definitely worth his money. If they are thinking of giving Ichiro 20 million dollars a year then Alex Rodriguez is obviously worth 25 million a year. A-rod had 30 home runs at the mid-season mark and its possible for him to hit 74 home runs a season to brek Barry Bond's record for home runs in a single season.
2007-07-13 13:15:53 · answer #4 · answered by Mets Lover 4 · 0⤊ 0⤋
Yes! Are you worth what you get paid?
Gosh, I can't wait for the day fans get to vote for players salary like the All-Star game. They sure seem to know everything! I can't wait!
Just to let you know the comments aren't directed towards you (the poster), I just saw a question like this earlier and some people actually said no.
SO CAL CORY - So what you are saying is no matter what A-Rod needs to be able to play all 9 positions simultaneously? You just said he is responsible for everyone else's lack of play. He is the best player in the league. The teams dictate the market by what they are willing to pay. You are only worth what someone is willing to pay you. A-Rod gets paid what he gets paid, because he is worth it to the people who pay him.
2007-07-13 12:59:00 · answer #5 · answered by Veritas et Aequitas () 7 · 0⤊ 0⤋
NO! There's other players that produce almost as good as A-Rod and they are cheaper!!! Like many million $ cheaper! Only a Yankee fan would say he's worth what he makes now. NO PLAYER COULD EVER LIVE UP TO $30mil either!!
2007-07-13 14:12:44 · answer #6 · answered by Spike 3 · 0⤊ 0⤋
If an owner of a club says he is via a contract, then the decimal point is real. I never fault a player in receiving "big" contacts - if the owners didn't have the cash, they wouldn't be cutting the checks.
2007-07-13 13:26:56 · answer #7 · answered by Zombie Birdhouse 7 · 0⤊ 0⤋
Basebal isn't socialist. A player should be played what the market dictates.
2007-07-13 13:08:52 · answer #8 · answered by Craig S 7 · 0⤊ 0⤋
NO
25 million dollars a year!?!?!?!!?!?!?!?!?!?!?!?!?!?!?
let me rephrase this
25 million dollars a year
if your getting paid that much you better win MVP every years
and not let your team fall out of hand like the yankees have cuz you need to be a leader
2007-07-13 13:02:55 · answer #9 · answered by So Cal Cory 2 · 0⤊ 1⤋
No, he is getting overpaid! I think he's worth his $ if he can PRODUCE in the PLAYOFFS and NOT CHOKE!
2007-07-13 13:22:41 · answer #10 · answered by Nicky b 3 · 0⤊ 2⤋