Report: Alex Rodriguez won’t retire early, accept a buyout
With the 40-year-old Alex Rodriguez scuttling mightily at the plate this season, the idea has been forwarded that perhaps he conceivably could decide to take an early retirement before his contract is up or perhaps accept a buyout from the New York Yankees.
Evidently, neither of those scenarios are ever going to happen, according to someone in A-Rod’s camp.
FanRag Sports Network’s Jon Heyman writes in a recent column that the notion that Rodriguez would ever leave money on the table is a non-starter, despite the fact he’s quite financially stable, which should come as hardly a surprise. A-Rod’s money situation is further enhanced — to put it mildly — by how he’s dating Anne Wojcicki, the billionaire ex-wife of Google founder Sergey Brin.
But that all means little to Rodriguez, according to a friend, as it relates to the money still owed to him.
“He would never retire early. He would never take a buyout,” the friend said, per Heyma’s report. “You can nip that (idea) in the bud right now.”
The Yankees are on the hook for the balance of his $20 million salary for this season and another $20 million for the 2017 season.
To say Rodriguez is scuttling at the plate would be an understatement. He’s currently triple-slashing an abysmal .219/.257/.382 with only eight home runs and 27 RBI for the 39-39 Yankees. It’s clear the team needs much more out of the DH position if the team hopes to stay in the race for a spot in the postseason instead of being forced into seller mode at the trade deadline.
Granted, a buyout or early retirement isn’t the only way the Yankees could rid themselves of the A-Rod situation. The team could cut him outright, essentially paying him his salary to stay away, not an attractive scenario to say the least despite the Yankees’ immense wealth.
The other alternative would be for the Yankees to find a trade partner, although that’s as likely at this late stage of his career as the apparent notion of him not taking every cent the Yankees owe him.