July 1, 2019 is the ninth annual “Bobby Bonilla Day,” in which the New York Mets pay the former All-Star $1.19 million.
Every year on July 1, the baseball world celebrates “Bobby Bonilla Day.” This is the day that the New York Mets pay former MLB All-Star Bobby Bonilla $1.19 million.
I should explain.
In 1999, the Mets released Bonilla after just one season. However, they still owed Bonilla $5.9 million. Bonilla and his agent thus cut a deal with the organization, stating that Bonilla would defer the payment for a decade. The Mets would then have to pay Bonilla $1.19 million every year starting in 2011 and ending in 2035.
Yes, you read that right. A guy that hasn’t played for the Mets in 20 years is still getting more per-year than Pete Alonso is right now.
On Monday, Bonilla will receive his ninth installment of $1,119,248.20, with 16 installments left.
Bonilla played in the majors for 16 years, from 1986-2001. This included two separate stints with the Mets, one from 1992-1995 and the other in 1999. He finished his long career batting .279 with 287 home runs and 1,173 RBIs.
He was selected to six All-Star games in his career, won three Silver Slugger Awards and was a World Series champion in 1997 with the Florida Marlins.
So yes, happy Bobby Bonilla Day for the ninth time. There’s only 16 more!
As for the Mets, it’s probably a day they like to forget every single year. Let’s be real, it’s tough to pay a guy $1.19 million every single July when he hasn’t played for the team in 20 years. Moreover, it’s especially tough when the organization is struggling for any type of positive news lately.
