Cleveland Indians’ ideal lineup to close out the 2020 season

First Base, Switch Hitter

Although he doesn’t have the appearance of a lead-off batter, Carlos Santana is the best fit to be at the top of the Indians’ batting order, and he’s done it before.

Santana was atop the Indians’ order during the 2017 season, the season that recorded the best winning percentage in Cleveland over the last 20 years and ended in a divisional title. For Santana, it proved to be a successful season that resulted in his contract with Philadelphia before coming back to Cleveland in 2019.

As the first batter in 2017, Santana posted impressive numbers for the duration of the season. Despite drawing a career-low 88 walks, Santana still maintained a .363 OBP and an .818 OPS.

This year, Santana began the year with a near impossible walk rate that put him into some record books. Heading into Saturday night’s game against the Twins, Santana had drawn 38 walks over 45 games. That tally has helped him to the second highest OBP on the team currently at .356.

Now, Santana’s average hasn’t been the best at just .195, but his ability to get on base is what is important here. Although it might not be via hits, Santana gets on base at a high rate compared to the rest of the team and you need runners to score early.