CLEVELAND, Ohio (WOIO) - Cleveland Indians fans did it!
Carlos Santana was voted the starting first baseman for the American League All-Star team for the game on June 9.
The Cleveland Indians were asking fans to continue to get out the vote for Carlos Santana. The faithful responded.
Voting closed at 4 p.m. ET. The Indians announced that at 2:30 Santana was running away with it. Santana had 49.3 percent of the vote, ahead of the New York-AL’s Luke Voit (26%) and Minnesota’s C.J. Cron (24.7%).
The organization said Santana ran away with the AL first base vote, claiming 49.2 percent of the vote ahead of Voit (25.8 percent) and Cron (25.0 percent). Entering the day, he had the best 2019 numbers among AL first baseman finalists leading in average (.290) home runs (18), on-base pct. (.411), slugging pct. (.541) and OPS (.951).
The native of the Dominican Republic finished second in the Primary Election process, receiving 1,180,719 votes. He finished behind New York-AL’s Luke Voit (1,205,706 votes) and ahead of Minnesota’s C.J. Cron (1,045,120 votes).
In his 10-year Major League career, Santana has never been named to an All-Star roster. He would be the first Indian to start at first base in the Midsummer Classic since Jim Thome in 1999.
Santana said, “I’m excited to represent Cleveland alongside the best in Major League Baseball. To be able to start in my first career All-Star Game, and to do it at home in front of our fans, means everything to me.”
This is third All-Star Game in a row where the Tribe as boated an American League starter as Jose Ramirez got the nod at third base in 2017 and 2018.
Other members of the Tribe elected to start in the All-Star Games in Cleveland are OF David Justice (1997), 1B Al Rosen and 2B Bobby Avila (1954), and OF Joe Vosmik (1935).