BRENT LATHAM - Friday, December 4, 2009
After years of being up against some of the toughest the World Cup had to offer in the first round, Bob Bradley and crew will have no excuse in South Africa, having been drawn in arguably the easiest of the eight.

The United States will open play against colonial power England in Rustenburg on June 12th, the same venue in which they ran to a 3-0 defeat of Egypt in the Confederations Cup this past June.

The US will also take on Slovenia, making their second World Cup appearance, on June 18th in Johannesburg, before wrapping up play in nearby Pretoria five days later against African upstarts Algeria, in their first World Cup since 1986.

The Americans will also have the unique advantage of playing all three games in familiar surroundings, after having played matches in Pretoria, Rustenburg, and Johannesburg during the Confederations Cup. The Americans will also play all three games in a relatively compact area, while most teams will travel extensively around South Africa.

Friday's draw started off in a perhaps less than ideal manner for the Americans, as they were drawn into Group C with series head England, which has had an easy time in European qualifying under Italian coach Fabio Capello. Coach Bob Bradley had looked on as Mexico instead looked fortunate to be immediately drawn into Group A with hosts South Africa.

Luck turned rapidly, however, as the third team to be drawn into Group C was Algeria, likely the weakest team in Pot C, followed by the Slovenians, among the lower-regarded teams from the European pot.

The Americans also avoided a potential second round matchup with tournament favorites Brazil and Spain.

The top two teams from Group C will take on the first and second place teams from Group D - potentially the Group of Death - consisting of Germany, Ghana, Australia, and Serbia.