World Series 2018: Ex-Yankee killer Dave Roberts passing on Red Sox reunion before Dodgers' Game 2

by Felix Pantaleón

The Boston Red Sox and Los Angeles Dodgers meet in Game 2 of the 2018 World Series on Wednesday, October 24, 2018 (10/24/18) at Fenway Park in Boston, Mass. The Red Sox won Game 1, 8-4, behind Eduardo Nunez’s pinch-hit three-run homer and a 4-for-5 night from left fielder Andrew Benintendi.

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The Boston Red Sox and Los Angeles Dodgers meet Wednesday at Fenway Park in Boston, Mass, for Game 2 of the 2018 World Series.

Before the game, the Red Sox will honor their 2004 championship team, with several former players, including Hall of Famer Pedro Martinez, returning to throw out the first pitch, according to CBS Boston.

Dave Roberts will be there, too.

Just don’t count him in for the festivities.

That’s because the former Red Sox outfielder is now the Dodgers manager and will have more important things to worry about, like how to bounce back from the 8-4 loss in Game 1, per the New York Post:

“I got wind of [the ceremony] today and I don’t plan on partaking,” Roberts said Tuesday. “But I will be there to watch and see those guys throw out the first pitch.”

Asked why he wouldn’t participate, Roberts’ reply was expected.

“I am going to be getting ready for a game,” Roberts said. “So, I will have fun with those guys, but I won’t be part of it.”

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Roberts, you may remember has perhaps the most famous stolen base in postseason history. He entered Game 4 of the 2004 ALCS between the Yankees and Red Sox as a pinch-runner in the ninth inning, following a walk to Kevin Millar.

Yankees reliever Mariano Rivera was three outs way from moving New York into the World Series.

But Roberts stolen second, scored on a Bill Mueller base hit, and the rest is history, with the Red Sox winning Game 4, and completing an unprecedented comeback to reach the World Series.