Friday, Jun 22, 2018
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Bailey beats Indians

CLEVELAND - Homer Bailey pitched five innings for his first major league win in nearly two years and Jay Bruce hit his 18th home run for Cincinnati to help the Reds beat the Cleveland Indians 7-3 last night.

Bailey (1-0) was recalled from Triple-A earlier in the day to make his second start of the year and won for the first time since Sept. 20, 2007, against the Chicago Cubs.

He labored through five innings, giving up three runs on three hits and seven walks, giving him 13 in 91/3 innings this year.

Indians starter Tomo Ohka (0-2) needed 27 pitches to record an out in the second inning. By then, the Reds had a 4-0 lead on Bruce's solo homer, RBI doubles by Chris Dickerson and Jerry Hairston and a groundout from Joey Votto.

Those three, Cincinnati's first three batters, went a combined 7 for 15. Laynce Nix doubled twice and scored twice for the Reds.

Ohka, who also hasn't won since 2007, was making his third start for Cleveland.

Cleveland had runners in four of the five innings against Bailey, but Grady Sizemore's two-run single in the fourth was the only real damage. The Indians stranded seven runners through the first four innings.

Sizemore has eight RBIs in four games since coming off the disabled list on Tuesday.

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