1956 NBA playoffs explained

NBA playoffs
Year:1956
Season:1955–56
Dates:March 15–April 7, 1956
Num Teams:6
Winners:Philadelphia Warriors
Count:2
Second:Fort Wayne Pistons
Semifinal1:St. Louis Hawks
Semifinal2:Syracuse Nationals
Prev Season:1955
Next Season:1957

The 1956 NBA playoffs was the postseason tournament of the National Basketball Association's 1955-56 season. The tournament concluded with the Eastern Conference champion Philadelphia Warriors defeating the Western Conference champion Fort Wayne Pistons 4 games to 1 in the NBA Finals.

It was the Warriors' second NBA title; their first was in 1947 back when the NBA was known as the BAA. They would have to wait until 1975 to taste championship gold again; by that time they had moved to the Bay Area and become the Golden State Warriors. Philadelphia's later team, the Philadelphia 76ers, would win the title in 1967.

This was the Pistons' second straight trip to the NBA Finals, but they would not make another appearance until 1988 as the Detroit Pistons. No team from Indiana would return to the NBA Finals until the Indiana Pacers did so in 2000.

The play-in game between the Syracuse Nationals and the New York Knicks was the last play-in game to determine a playoff spot until 2020.[1]

Notes and References

  1. Web site: August 11, 2020. How the Blazers, Grizzlies, Spurs and Suns make the West play-in. August 12, 2020. ESPN.com. en.