There’s a first time for everything.
Including a Major League Baseball player playing for both teams in the same game. Danny Jansen is set to make MLB history Monday by doing just that.
How is that possible?
On June 26, Jasen started at catcher for the Toronto Blue Jays in a game against the Boston Red Sox, a game that was called in the second inning because of weather.Then, on July 27, Jansen was traded from the Blue Jays to the Red Sox.
On Monday, Jansen is set to start at catcher for the Red Sox in the resumption of that June 26 game at Fenway Park, according to Boston manager Alex Cora.The Red Sox and Blue Jays are playing a doubleheader Monday, with the first game, the completion of the June 26 game, beginning at 2:05 p.m. EDT (11:05 a.m. MST) in the top of the second inning with the score tied, 0-0.
MLB Playoffs standings:Who leads division, wild-card races for 2024 MLB postseason?The second game of the doubleheader is scheduled to begin at 7:10 p.m. EDT (4:10 p.m. MST).
“Definitely grateful,” Jansen told reporters about the opportunity to make MLB history by being the first player to play for both teams in an MLB game, via ESPN. “Honestly, when I heard about it, I didn’t think I would be the first. The game has been around for so long. It’s one of those oddities that happen in this sport. It’s extremely rare and cool.”