TONIGHT: Orioles-Indians should be close with clubs' knack for 1-run games


CLEVELAND -- It shouldn't come as a surprise that the teams with the worst two records in baseball also have the worst records in games decided by one run.

What might surprise a lot of people, however, is the fact that the Cleveland Indians (24) and Baltimore Orioles (21) play more one-run games than the other American League teams. It figures then, that these two teams should be closely matched in the three-game series that begins here tonight (7:35, Channel 2).

Despite winning a pair of one-run games in Kansas City over the weekend, the Orioles have only a 7-14 record in such slim decisions. The Indians also have lost twice as many as they have won at 8-16.

If there is an advantage in the series between the bottom two teams in the AL East, it would appear to go to the Orioles, who won two of three in the first series of the year between the teams. The Orioles won their last three games in Kansas City and have won six of their last 10. The Indians are working on a four-game losing streak and have dropped 10 of their last 11.

Righthander Jeff Robinson (3-6, 4.54) gets the pitching call for the Orioles in the opener tonight. He will be opposed by Rod Nichols (0-5, 3.55).

Tomorrow night it will be Jeff Ballard (4-7, 4.21) against Cleveland ace Greg Swindell (3-6, 2.70). Bob Milacki (3-2, 4.15), fresh from a 1-0 win over Kansas City, will close out the series for the Orioles Thursday night when he faces Jeff Mutis (0-2, 12.27).

Following the three games against the Indians, the Orioles return home for six games, three this weekend against the Red Sox followed by three against the Detroit Tigers.

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