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Marlins have five A-Rods (do the math)

We are headed into a world of Sabermetrics here that would be sure to send Bill James over the edge but here goes.

How come, we wondered, the Florida Marlins are 8-5 on the young baseball season and the New York Yankees are 8-7 when the Yankees have one player who is so terrific, Alex Rodriguez, that he makes more money than all the players on the Marlins combined?  Reportedly, the Yanks are paying A-Rod $28 million while the entire Florida roster makes less than $22 million.

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Well, we looked at the stats through the first few weeks of the season and here's the reason -- for less than what the Yankees are paying Ridriguez, the Marlins have five players whose average performance is just as good as the bazillion-dollar man.

A-Rod is having a decent start -- he's batting .300 (18-for-60) with three home runs and 8 RBIs.  Meanwhile, the Marlins who have a similar number of at-bats compared to A-Rod (48 or more) are batting a combined .282 (71-for-252) with an average of exactly three home runs (15 total) and 7.6 RBIs (38 total).  The five are Hanley Ramirez, Mike Jacobs, Josh Willingham, Jorge Cantu and Dan Uggla.  Not exactly Murder's Row but on average, about the same production as A-Rod.

And you get five of them.

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