Panama vs. Cuba and Trinidad & Tobago vs. Haiti to determine Confederation’s final two entrants in historic tournament

Miami (Thursday, October 29, 2015) – The Confederation of North, Central America and Caribbean Association Football (CONCACAF) today announced details of the two play-in matches for the final two berths in the historic Copa America Centenario, to be played in the United States from June 3-26, 2016.

Two matchups – Panama vs. Cuba and Trinidad & Tobago vs. Haiti – will be played as a double header on Friday, January 8 2016, at the Estadio Rommel Fernandez in Panama City, Panama. The winner of each match will advance to the Copa America Centenario. Exact kickoff times will be announced shortly.

CONCACAF will have six representatives in the 2016 Copa America Centenario, hosted in the United States and marking the first time the South American championship has been played outside that continent. CONCACAF Gold Cup champion Mexico, 2013 Gold Cup champion and host United States, reigning Central American Cup champion Costa Rica and Caribbean Cup Champion Jamaica have already secured their places in the event.

The four playoff contenders set to compete in Panama earned their places in the play-in matches by virtue of their performances at the 2015 Gold Cup. Cuba, Haiti, Panama and Trinidad & Tobago were all quarterfinalists. Panama earned the right to host the play-in event as the best finisher of the group, advancing to the semifinals and eventually finishing in third place.

The U.S.-hosted Copa America Centenario was originally announced in May 2014, when it was determined the reigning UNCAF and CFU champion, as well as Mexico and the United States, would represent the CONCACAF region, along with two additional teams to be determined via play in matches after a ranking of the best finishers from the 2015 CONCACAF Gold Cup. That procedure was confirmed last week, along with the final details of the Copa America Centenario in the United States.