​As the 2018 MLB season draws closer to its end, many people wonder who will receive the annual awards presented, one of them being the American League's Most Valuable 

Player. 


The race is tight, but one front runner for the honor is ​Red Sox outfielder Mookie Betts


Some say Betts isn't the best player in the league or even on his own team with teammate J.D. Martinez leading the league in home runs and runs batted in. However, there is one stat that works in his favor. 

​​It may be an oddly specific stat, but it bodes well for Betts to be compared to a Boston great in that manner. He already has three All-Star appearances and two Gold Gloves to his name.  He also finished second in MVP voting in 2016 behind Mike Trout, while finishing sixth last year. 


Betts currently has a .347 batting average, a .668 slugging percentage and an OPS of 1.103, all best in the MLB. He also 27 home runs, 59 RBI and 22 stolen bases. 

As a team, the Red Sox currently have the best record in baseball and stand in great position to win the AL East after a four-game sweep over the Yankees last weekend. They currently have the best record in baseball and are on pace to win over 100 games. 


They may be an improved team from previous years, but there's no way they get to the mark they're at now without Mookie Betts.