Games on September 4, 1993

The games of September 4, 1993

The schedule for September 4, 1993 lists 14 games. The widest margin belonged to the Chicago White Sox, 11-2 winners over the Detroit Tigers. Runs came easiest when the Chicago Cubs beat the New York Mets 9-8. A shutout went on the board for the New York Yankees over the Cleveland Indians (4-0). Decided by a single run were the Atlanta Braves 3-2 over the San Diego Padres and the Cincinnati Reds 6-5 over the Philadelphia Phillies. Elsewhere, the Kansas City Royals topped the Boston Red Sox 4-2, the California Angels topped the Toronto Blue Jays 4-2, the Colorado Rockies topped the Pittsburgh Pirates 10-4, the Los Angeles Dodgers topped the Florida Marlins 9-4, the Montreal Expos beat the Houston Astros 7-5, the Texas Rangers beat the Minnesota Twins 6-4, the Baltimore Orioles beat the Oakland Athletics 6-3, the Seattle Mariners beat the Milwaukee Brewers 6-1, and the St. Louis Cardinals fell to the San Francisco Giants, 3-1. Only final scores are available for this date. Neighboring dates are one click away: the previous game day, Sep 3, 1993 and the next, Sep 5, 1993. The 1993 season page has the standings and leaders for the year.