With the 2018 regular season having ended award season is here. The 2018 awards will be given out following the World Series but it is never too early to look at who should win the most prestigious award: the most valuable player award. There is always controversy surrounding the MVP award as people struggle to define what the “valuable” in most valuable player means. Here we will define it as the best player in the American League in 2018.
Candidates
Throughout the season several players made a case to be included in this discussion including Matt Chapman of the Oakland Athletics, Alex Bregman of the Houston Astros and Francisco Lindor of the Cleveland Indians. In the end three candidates rose above the rest; Mookie Betts of the Boston Red Sox, the Indians’ Jose Ramirez, and Mike Trout of the Los Angeles Angels of Anaheim.
The Case for Mookie Betts
Mookie Betts was the best player on baseball’s best team in 2018 as he led the Red Sox to a franchise record 108 win season. Betts led the American League in WAR (10.9), batting average (.346), slugging percentage (.640), runs scored (129), and extra base hits (84). He was second in on-base percentage (.438), on-base plus slugging (1.078), doubles (47) and he was third in total bases with 333 (www.baseball-reference.com). The charismatic young star accomplished all of this while playing solid defense in right field while making his third all-star team in just his fourth full-season. His OPS+ of 186 means he was 86% better than the average hitter in 2018. He also had a 30-30 season (30 home runs and 30 stolen bases). There is no question that Betts had an outstanding offensive season. One of the few knocks on Betts is he missed 25 games due to injury. From a pure stat perspective Betts should run away with the award, but it one takes the “valuable” in most valuable player literally it is fair to question how valuable Betts was to a team like the Red Sox who likely would have won 100 game if Betts was just average this season.
The Case for Jose Ramirez
Jose Ramirez was key for the AL Central Champion Cleveland Indians. Playing third base and second base, Ramirez compiled the best season of his career. He had a batting average of .318 (4th in the league), a .387 on base percentage (6th), a slugging percentage of .552 (4th), an OPS of .939 (4th), and his 39 home runs and 105 runs batted in both ranked (4th). He was also 4th in the American League with a 7.9 WAR. (www.baseball-reference.com) While he did not lead the league in any of these categories, if the definition of valuable is taken literally one would be hard-pressed to find a player more valuable to his team than Ramirez was to the Indians. Take him away from the 91-win Indians and not only is it questionable if they make it to the playoffs, it is questionable if they even finish above .500.
The Case for Mike Trout
It is tradition at this point to include Mike Trout in most valuable player discussion. In fact Trout has never finished lower than second in MVP voting since he has been in the league since 2012 (He won in 2014 and 2016). Trout had another phenomenal season, batting .312 with 39 home runs, and an OPS+ of 199, in other words he was twice as good as the average hitter in 2018. (His OPS+ led the major leagues) His 10.2 WAR was second only to Mookie Betts in the American League. Trout led the league in on base percentage (.460), on-base plus slugging (1.088), and walks (122). He accomplished all of this while missing some time due to injury, playing in 140 games. (www.baseball-reference.com) Trout is criminally underappreciated, a modern-day Willie Mays playing for a mediocre team on the west coast and his Angels team went 80-82, well out of the playoffs. Playing for a west coast team that was not particularly competitive will hurt Trout compared to the big market appeal of Betts.
Who will win?
Jose Ramirez should finish in third-place, he had a great year but his season was a notch below the otherworldly production of Betts and Trout. Trout is almost a victim of his own brilliance, his season has been overlooked because he is so good year in and year out. Playing for a losing team will hurt his case too, as some will say the Angels can lose with him so they can lose without him. This leaves Mookie Betts, a young marketable star for baseball’s best team. He will likely win the award in part because of his big market appeal, but no one will complain. Betts certainly had an MVP-caliber year and he would be a well-deserving winner of the award.
Full Count 