LEAGUES News
Angels' Trout set to become MLB's highest-paid player in 2018
Here comes the money!
It seems only logical that the best player in any sport should be compensated with the highest salary. In 2018, that will be the case in Major League Baseball. Los Angeles Angels outfielder Mike Trout will see his annual salary jump from $20.083 million in 2017 to $34.083 million in 2018, according to Bob Nightengale of USA TODAY Sports.
On his current six-year, $144.5-million contract, Trout makes less per year than players like Clayton Kershaw and Zack Greinke - but that's because it's heavily back-loaded:
| Year | Salary (in millions) |
|---|---|
| 2015 | $6.083 |
| 2016 | $16.083 |
| 2017 | $20.083 |
| 2018 | $34.083 |
| 2019 | $34.083 |
| 2020 | $34.085 |
Unless one of the remaining free agents winds up signing an incredibly lucrative deal, Trout will get paid the most in base salary.
The 26-year-old has finished among the top five in MVP voting in six consecutive seasons, winning the award twice, and finishing as the runner-up three times. For his career, Trout owns a slash line of .306/.410/.566 with 201 home runs, 200 doubles, 40 triples, 569 RBIs, 692 runs scored, and 165 stolen bases in 925 games.
The Angels signed Trout to the contract to take him through arbitration and what would have been his first couple seasons of free agency.
His time atop the list will be short-lived, as the structure of Greinke's contract has him earning $34.5 million in 2019, so Trout will have to wait until after 2020 to reclaim his throne as baseball's richest man.
