Los Angeles Dodgers star hurler Yoshinobu Yamamoto will make one more appearance for Team Japan in the 2026 edition of the World Baseball Classic before he gets back to work with the reigning World Series champions.
Japan is playing Venezuela in a quarterfinal showdown this coming Saturday, and Yamamoto is scheduled to toe the rubber for the defending WBC champions at LoanDepot Park in Miami.
It will just be his second and last appearance for Japan in this tournament.
“Dodgers ace Yoshinobu Yamamoto will start Japan’s quarterfinal game against Venezuela and then will travel back to Phoenix to rejoin the Dodgers, manager Dave Roberts said,” wrote Bob Nightengale of USA Today on a social media post on X, formerly Twitter.
Yamamoto appeared in Japan's 13-0 drubbing of Chinese Taipei in Pool C play at the Tokyo Dome in Tokyo. He pitched for 2 2/3 innings and allowed zero earned runs and zero hits, while striking out two and walking three.
Yamamoto was also part of Japan's squad that ruled the 2023 World Baseball Classic, during which he went 2-1 with two earned runs and 12 strikeouts over 7 1/3 innings.
Japan manager Hirokazu Ibata does not appear to have any plans yet for who will start after Yamamoto, as the main focus remains on surviving the clash against a dangerous Venezuela side.
“Yamamoto is the starter. We haven’t decided anything beyond that. Since this is a game where losing means elimination, we need to pour everything we have into the next match,” Ibata said, via The Chosun Daily.
The Dodgers are scheduled to start their World Series defense on March 26 against the Arizona Diamondbacks at Dodger Stadium.
