This Day in Sports History: April 21, 1948

On April 21, 1948, the Baltimore Bullets defeated the Philadelphia Warriors 88-73 to win the Basketball Association of America (BAA) Finals series four games to two.

Baltimore’s victory that year was somewhat of a Cinderella story. The New York Knicks, Boston Celtics, and Warriors were the big three in the BAA.

The Bullets shocked the league when they came back to tie the Chicago Stag and Washington Capitols with a regular season record of 28-20 which lead to a Western Division tiebreaker for second place behind the St. Louis Bombers.

Back then, the division leaders would automatically advance to the Semifinals where they would play a best out of seven game series, while the second and third place teams from each division would play their way through the quarterfinals and would only have to play a best of three game series in the semis.

Chicago beat Washington in the first game of the tiebreaker for the West 74-70, but then lost the second game to Baltimore with the Bullets squeezing out a 75-72 win for second place.

In the quarterfinals, the Bullets played a three game series against the Knicks. They won game one in Baltimore 85-81 but then lost on the road in game two 79-69. Game three was played back in Baltimore and the Bullets advanced to the semifinals with a game three 84-77 final score.

With the Stag defeating the Celtics in the quarterfinals, Chicago and Baltimore met again in the semis with Baltimore winning two straight games to advance to the finals where they would meet the Philadelphia Warriors.

Philadelphia won the first game of the finals by 11, and in game two, the Bullets found themselves trailing 41-20 at the half. After only scoring 7 points in the second quarter, the Bullets flipped the switch and outscored the Warriors 20-7 and would end up winning game two 66-63.

Baltimore would win games three and four but drop game five before winning the title in game six.

References: Basketball-reference.com