Los Angeles Dodgers pitcher Yoshinobu Yamamoto gestures during the fourth inning in Game 6 of baseball’s World Series against the Toronto Blue Jays, on Friday in Toronto.
Brynn Anderson/AP
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Brynn Anderson/AP
TORONTO — Yoshinobu Yamamoto defeated Toronto for the second time in a week, and Mookie Betts delivered a two-run single in a three-run third inning as the Los Angeles Dodgers held off the Blue Jays 3-1 on Friday night, forcing the World Series to a decisive Game 7.
Yamamoto’s performance was not as dominant as in Game 2, where he pitched a complete game four-hitter. In this game, he pitched six innings and only allowed a third-inning RBI single by George Springer.
Rookie relievers Justin Wrobleski and Roki Sasaki combined for six outs, and starter Tyler Glasnow came out of the bullpen to secure the victory for the Dodgers.
Glasnow made a quick impact in the ninth inning, getting crucial outs to seal the win for Los Angeles. With runners in scoring position, he induced a pop-up and a game-ending double play to secure the victory.
Max Scherzer is set to start Game 7 for the Blue Jays, aiming to lead his team to a World Series title. He previously started a Game 7 in the 2019 World Series, which resulted in a championship for the Washington Nationals.
