Games on August 7, 1993

The games of August 7, 1993

On August 7, 1993, 14 games went into the books. The Milwaukee Brewers routed the Toronto Blue Jays 7-1, the widest margin of the day. Runs came easiest when the New York Mets beat the Pittsburgh Pirates 10-8. Decided by a single run were the Philadelphia Phillies 8-7 over the Florida Marlins, the Minnesota Twins 6-5 over the New York Yankees, the Pittsburgh Pirates 2-1 over the New York Mets, and the Houston Astros 6-5 over the San Francisco Giants. The rest of the slate: the Montreal Expos topped the Atlanta Braves 5-3, the Baltimore Orioles topped the Cleveland Indians 8-6, the California Angels fell to the Chicago White Sox, 6-4, the Detroit Tigers fell to the Boston Red Sox, 4-1, the Kansas City Royals beat the Oakland Athletics 5-2, the Los Angeles Dodgers fell to the Cincinnati Reds, 9-6, the Chicago Cubs fell to the St. Louis Cardinals, 4-1, and the Seattle Mariners beat the Texas Rangers 2-1. The Pittsburgh Pirates and the New York Mets met twice, as a doubleheader. Only final scores are available for this date. Neighboring dates are one click away: the previous game day, Aug 6, 1993 and the next, Aug 8, 1993. The 1993 season page has the standings and leaders for the year.