Now Playing
Share Video
Link copied!
Pepiot (4-4) allowed five runs on six hits and three walks while striking out four batters over 4.2 innings in a loss to Atlanta on Saturday.
Pepiot looked to be on his game with four scoreless innings to begin his outing, but things fell apart for him in the fifth. In that frame, the right-hander gave up three home runs -- all with two out -- to account for five Atlanta runs. It marked the first time this season Pepiot has given up three home homers in a game, and the long balls prevented him from finishing five frames for the third time this season. The five runs Pepiot allowed Saturday were the most he's given up in an outing since his season debut against Texas on April 1.
