Hey Bud: Don't Allow Us to Vote on the All-Star Rosters

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Hey Bud: Don't Allow Us to Vote on the All-Star Rosters

In 183 career games, Milwaukee's Ryan Braun has 54 home runs, 155 runs batted in, and has batted just over .300. In 2008 alone, he has 20 homers, 55 RBI, and is batting .287.

So far this season, Kosuke Fukudome has only five home runs, 20 RBI, and is batting .296. Alfonso Soriano has 15 home runs and 40 RBI, along with a .283 average.

Despite this statistical dominance, Braun is still in fourth place in All-Star voting for National League outfielders. Soriano and Fukudome are first and second, respectively, with Ken Griffey Jr. separating the two Cubs from the Brewer.

Soriano is hopeful that his broken hand will be healthy enough for action by the All-Star Game. If that is the case, and Soriano gets the votes for the start, Bud Selig will have to revise the way the All-Star teams are selected.  

He might consider letting people who know something about baseball choose the rosters.

It's great that Major League Baseball wants the fans to be involved, but the honor of being named an All Star shouldn't rest in the fans' hands.

First of all, decent players (Fukudome) and injured players (Soriano) on big-market teams like the Cubs will always get the nod over great players (Braun) who play for the smallest of all small markets, like the Brewers, even though Braun has clearly outplayed Soriano at the plate.

It's also important to have the best team possible on the field for the All-Star Game because of the home-field advantage for the Fall Classic is decided by this exhibition game; a ploy the commissioner thought up after the 2002 All-Star debacle.  

If Cubs fans are thinking about a World Series this year want home-field advantage, they better vote for the best available players to play this summer.

Even if that means voting in a division rival over one of their own. The Cubs’ fans should understand that Ryan Braun is the best option for the National League team in left field, and it's not even close.

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