Dodgers in first World Series Game 7 since 1965

LOS ANGELES -- For the first time in 52 years, the Dodgers will play in Game 7 of the World Series, after drawing even with a 3-1 win over the Astros on Tuesday night at Dodger Stadium.

This is the sixth winner-take-all World Series Game 7 for the Dodgers, and the first since Sandy Koufax shut out the Twins in two days rest in the finale in Minnesota in 1965. He also shut out the Twins three days earlier in Los Angeles, in Game 5.

That’s the only Game 7 of the World Series for the Dodgers since moving to Los Angeles. They have had one other Game 7, in the NLCS in 1988. That was a 6-0 win over the Mets, behind an Orel Hershiser shutout and a five-run second inning, making the Dodgers 2-0 in Games 7 since coming west.

In World Series history this year will mark the 39th winner-take-all Game 7. Home teams actually have a losing record (18-20), including road wins by the Giants in Kansas City in 2014 and the Cubs in Cleveland in 2016, the last two Games 7.

The home team won the previous nine Games 7 of the World Series, from the Cardinals over the Brewers in 1982 through the Cardinals over the rangers in 2011.

The Dodgers’ history in deciding Games 7 in Brooklyn were all against the Yankees, with losses in 1947, 1952 and 1956 — the latter two at home in Ebbets Field — and a win in 1955 on the road at Yankee Stadium for the franchise’s first championship.

Their last four Games 7 have all been shutouts, including three wins by the Dodgers.

If the Dodgers are able to win on Wednesday night, they will be the 21st team to comeback from a 3-2 deficit to win the World Series in the current 2-3-2 format.