Jeff Passan of ESPN reports that the Los Angeles Angels and superstar center fielder Mike Trout are finalizing the largest contract in the history of professional sports. Per Passan, the deal is for 12 years and will be worth more than $430 million. It has yet to be revealed if the deal includes any opt-outs or a no-trade clause.
Trout, 27, will be making nearly $36 million per season which tops Zack Greinke‘s previous average of a $34.4 million AAV. Further, the $430 million-plus deal is more than 30% larger than the $330 million deal Bryce Harper signed on March 2.
Due to be a free agent following the 2020 season, many expected Trout to test free agency where he would be sure to set off an immense bidding war. Though, with many players scared by recent free agency trends, it’s not surprising to see him take the boatload of cash the Angels have offered him.
In 2012, when he won AL Rookie of the Year, the only reason he finished second in MVP voting was because Miguel Cabrera hit for the Triple Crown. Since then, he has finished lower than second in MVP voting once, and that was when he finished fourth because he played in only 114 games in 2017. He is remarkably consistent despite his elite-of-the-elite level of play, and he only seems to be getting better.
In 2018, Trout hit .312/.460/.628 (1.088 OPS) with 39 home runs, 79 RBIs, 24 stolen bases, 122 walks, a 191 wRC+ and a 9.8 fWAR. His on-base percentage and OPS both paced the Major Leagues, and he set new career highs in walks, OBP, OPS, and strikeouts — That is, he struck out less in 2018 (124) than any other season in which he played 140 games. He’s also a phenomenal fielder, posting eight DRS and a 4.0 UZR (5.6 UZR/150) in center field in 2018.
He is the best player in the game, so the fact that he’ll be paid more than anyone else makes total sense.
