If the Indians are going to win this afternoon in Oakland, it would be helpful to score more than one run before the ninth inning.
On Tuesday it took a three-run ninth to win 4-1, and yesterday, there was no late rally, and the A’s evened this three-game series at one apiece with a 3-1 win that dropped Cleveland to 20-9 on the season and ended the Tribe’s seven-game winning streak.
Today Jeanmar Gomez is tasked with trying to get the Indians back on track before they head to Anaheim and continue running this gauntlet of imposing opponent pitchers.
The Tribe will be facing another such daunting starter today.
Oakland will send Brett Anderson (2-2, 2.95 ERA) against Cleveland today. That ERA is not a flash-in-the-pan, as Anderson went 7-6 with a 2.80 ERA in 19 starts for the Athletics last year.
In his last start, Anderson was roughed up by the Texas Rangers to the tune of nine hits and seven runs in just five innings, taking the loss in an 11-2 Texas win. But that poor outing was the exception rather than the norm in 2011 for Anderson, and once again, the Tribe needs to be ready for a potentially dominating starter.
In 39 2/3 innings, Anderson has allowed 40 hits, and has fanned 30 while giving up just 13 earned runs.
In his career, Anderson has faced the Indians twice. In those two starts, covering 12 innings, Anderson has gone 1-0 with a 0.75 ERA, allowing just one run and seven hits, and striking out 15.
For the Indians, Gomez (0-1, 6.23 ERA) needs to drastically cut down on the number of base-runners he allows. In only 13 innings covering three appearances — two of which have been starts — Gomez has allowed 22 hits and three walks, for an average of just about two runners per inning.
In his last appearance, Gomez got a no-decision against Detroit last Friday, surrendering 10 hits and three runs in just 5 2/3 innings, in a game that the Indians won in the bottom of the ninth on a Carlos Santana grand slam.
This will be Gomez’s first career appearance against Oakland.