The 1948 BAA Finals was the championship round of the Basketball Association of America's 1947–48 season. The Philadelphia Warriors of the Eastern Division faced the Baltimore Bullets of the Western Division, with Philadelphia having home court advantage. The Bullets won the series 4–2.[1]
Baltimore was not the Western Division champion but advanced to the championship round by winning a four-team playoff among the Eastern and Western Division runners-up. Meanwhile, the Eastern and Western Division champions, Philadelphia Warriors and St. Louis Bombers, played one long series to determine the other finalist, a best-of-seven series that Philadelphia won 4–3. In the runners-up bracket, Baltimore and Chicago from the West had first eliminated New York and Boston from the East, then faced each other in a best-of-three series. The format was used only twice, in 1947 and 1948, and generated two champions from the runners-up bracket.
The six games of the final series were played in twelve days, with at least one day off except prior to the decisive game. Division champions Philadelphia and St. Louis had played the seven games of their semifinal series in fifteen days, March 23 to April 6, with at least one day off before every game. The entire playoff tournament extended 30 days.
As of 2023, this is the only BAA/NBA Finals which a now-defunct team won (as opposed to teams that changed names and/or moved to other cities in later years).
| Game | Date | Home team | Result | Road team |
|---|---|---|---|---|
| Game 1 | April 10 | Philadelphia Warriors | 71–60 (1–0) | Baltimore Bullets |
| Game 2 | April 13 | Philadelphia Warriors | 63–66 (1–1) | Baltimore Bullets |
| Game 3 | April 15 | Baltimore Bullets | 72–70 (2–1) | Philadelphia Warriors |
| Game 4 | April 17 | Baltimore Bullets | 78–75 (3–1) | Philadelphia Warriors |
| Game 5 | April 20 | Philadelphia Warriors | 91–82 (2–3) | Baltimore Bullets |
| Game 6 | April 21 | Baltimore Bullets | 88–73 (4–2) | Philadelphia Warriors |