Stephen Drew Starts at Shortstop Against Left-Hander Wei-Yin Chen in Series Opener With Orioles

The Red Sox will return home Tuesday night to face the Baltimore Orioles, and with the O’s going with a left-hander to open the series, it seemed to make sense to start Xander Bogaerts at shortstop to make his home debut.

However, it will be Stephen Drew in the lineup to begin the three-game series, which also opens a nine-game homestand. The left-handed-hitting Drew will get the start against southpaw Wei-Yin Chen.

Drew’s struggles against left-handers have been well documented. He enters Tuesday’s game hitting just .198 against lefties with a .620 OPS. He has 46 strikeouts in just 121 at-bats.

Bogaerts hit .298 against southpaws with a .926 OPS in 57 at-bats at Triple-A before being called up last week.

See the lineups for Tuesday’s game below.

Red Sox
Jacoby Ellsbury, CF
Shane Victorino, RF
Dustin Pedroia, 2B
David Ortiz, DH
Jonny Gomes, LF
Mike Napoli, 1B
Jarrod Saltalamacchia, C
Stephen Drew, SS
Will Middlebrooks, 3B

Felix Doubront, P

Orioles
Robert Roberts, 2B
Manny Machado, 3B
Chris Davis, 1B
Adam Jones, CF
Nick Markakis, RF
Matt Wieters, C
J.J. Hardy, SS
Danny Valencia, DH
Steve Pearce, LF

Wei-Yin Chen, P