
The 2026 World Baseball Classic has entered the knockout stage. The defending champion Samurai Japan will face Venezuela in the quarterfinal on March 15 at Beijing time, and the team has confirmed that Los Angeles Dodgers ace right-hander Yoshinobu Yamamoto will be the starting pitcher. Earlier, some U.S. media mistakenly reported that he would leave the team after this game to return to MLB spring training, but this was later confirmed to be false. Yoshinobu Yamamoto will remain with the Japanese team until the end of the tournament, striving fully for the championship.
This misinformation originated from a social media post earlier today by The Athletic reporter Fabian Ardaya. He initially cited Dodgers manager Dave Roberts, stating that Yoshinobu Yamamoto would leave the national team after the quarterfinal. However, less than an hour and a half later, Ardaya urgently corrected himself, indicating that based on the latest team information, Yamamoto will stay with Japan until the tournament concludes, clarifying the rumor of an early departure.
The 27-year-old Yoshinobu Yamamoto performed excellently in the group stage opener against Taiwan, pitching 2.2 innings with no hits or runs allowed, and was a key contributor to Japan's top-group advancement. He is currently preparing with bullpen sessions at LoanDepot Park in Miami. Speaking about the upcoming quarterfinal, Yamamoto said: "The teams reaching this round are all very strong. I will prepare thoroughly and take the field in my best condition."
Japan will face a stern test in the quarterfinal. The Venezuelan roster includes several MLB elite hitters such as Ronald Acuña Jr., Luis Arraez, and Gleyber Torres. Considering the tournament's pitching restrictions and his workload from last season, Yoshinobu Yamamoto is likely not to pitch again after this game, and teammate Shohei Ohtani also has no pitching assignments in this tournament. If Japan advances, their semifinal opponent will be the winner between Puerto Rico and Italy.