Yoshinobu Yamamoto reportedly agrees to join Dodgers on record 12-year, $325 million deal!

Yoshinobu Yamamoto agreed to a 12-year, $325 million contract with the Los Angeles Dodgers on Thursday, according to ESPN’s Jeff Passan. The deal followed Yamamoto’s posting and was made available to MLB teams in November. He was No. 2 behind Shohei Ohtani on Yahoo Sports’ list of the top 25 free agents this winter.

Yamamoto leaves Japan as the most decorated pitcher in Japanese professional baseball history. He made his debut with the Orix Buffaloes at age 18 and posted a 1.82 ERA over his seven-year career, reaching impressive comic heights over the past three years.

Yoshinobu Yamamoto

This performance earned Yamamoto the largest contract in MLB history. His $325 million was more than the nine-year, $324 million contract Great Cole signed with the New York Yankees in 2019. The Dodgers also paid $50.6 million to post in Yamamoto’s contract, according to Pasan.

According to Ken Rosenthal of The Athletic, unlike Ohtani’s 10-year, $700 million deal with the Dodgers, Yamamoto’s deal reportedly does not include any deferrals.

From 2021 to 2023, Yamamoto would win one Japan Series title, one World Baseball Classic title, three Pitching Triple Crowns, three Sawamura Awards (Japan’s equivalent of the Cy Young Award) and a three-time Pitching Triple Crown. -Three-time league MVP winner He is the first NPB player since Ichiro Suzuki to win the MVP award three years in a row, and the second player in history behind Hisashi Yamada.

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Yamamoto’s biggest truth became even bigger in his last initial appearance at NPB. He threw a 14-hit shutout in the Japan Series elimination game, breaking Yu Darvish’s streak record.

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