Mike Trout Wins Unanimously

Los Angeles Angels

Los Angeles Angels of Anaheim center fielder Mike Trout was a unanimous choice as the American League winner of the Jackie Robinson Rookie of the Year Award, the BBWAA announced on MLB Network today.

Trout, 21, was listed first on all 28 ballots submitted by two writers in each league city for a perfect score of 140 points. He became the eighth unanimous choice in the AL voting, joining Evan Longoria, Nomar Garciaparra, Derek Jeter, Tim Salmon, Sandy Alomar Jr., Mark McGwire and Carlton Fisk. There have been 10 unanimous choices in the National League. Trout and Salmon, the 1993 winner, are the only Angels players to have been honored.

There was barely an offensive category in which Trout did not lead all rookies – batting average (.326), runs (127), RBI (83), hits (182), doubles (27), triples (8), home runs (30), stolen bases (49), total bases (315), extra-base hits (65), multi-hit games (56), walks (67), on-base percentage (.399) and slugging (.564). He is the sixth center fielder to win the AL award and the first since Carlos Beltran in 1999.

Texas Rangers pitcher Yu Darvish (16-9, 3.90 ERA) was the only other player on every ballot and was third overall in the voting behind Oakland Athletics outfielder Yoenis Cespedes (.292, 23 HR, 82 RBI).

Trout also became the youngest winner of the AL award. The previous youngest was the 1978 winner, Detroit Tigers second baseman Lou Whitaker, who was two months and 26 days older than Trout. The youngest winner overall was New York Mets pitcher Dwight Gooden, the NL recipient in 1984 at the age of 19.

The voting (5-3-1 scoring system):

Below is a breakdown of the 28 individual ballots, submitted by two writers representing each team in the American League. For more information on the voting, see our Voting FAQ.