MANILA, Philippines – Former world champion Ricky Hatton of Britain will support Puerto Rican boxer Miguel Cotto when he takes on unbeaten Floyd Mayweather Jr. this Saturday in Las Vegas (Sunday in Manila), but admitted that the American boxer will most likely win.
In an interview with Boxing Scene, Hatton commended both fighters but said he will be cheering for Cotto during the fight.
"I would rather watch Cotto than Mayweather every day, but you have to give credit where it is due. Cotto is a wonderful fighter, but although I don’t really see eye-to-eye with Mayweather, he is an absolute boxing genius," Hatton said.
Hatton lost to Mayweather via a 10th round technical knockout when they fought in December 2007. Both boxers were undefeated going into the match, but Mayweather ended up giving Hatton the first loss of his career.
"I do think fighters like myself and Cotto bring the best out of Mayweather," Hatton said, adding that he expects this fight to follow the pattern of his own bout against the American.
"It will be hairy-scary for a few rounds, but I think Mayweather will take the sting out of (Cotto) and come through on a late stoppage or points," he said.
"I hope Cotto wins, but having shared a ring with Mayweather, I can tell you he is in a different class even to some of the very best," Hatton added.
Hatton also fought and lost to Filipino ring icon Manny "Pacman" Pacquiao, but he insists that Mayweather was the best boxer he ever fought.
"I didn't get a chance to find out how good Pacquiao was," Hatton said. Pacquiao brutally knocked out Hatton after only two rounds when they fought in May 2009.
"Floyd was really, really good. Fair play to Manny for what he did to me, but having see what Floyd can do for plenty of rounds, I have to side with him and say he is the best I faced," he added.
Hatton retired from boxing in July 2011 and has since devoted his time to being a promoter. He won championships in the light welterweight and welterweight divisions and finished with a record of 45 wins and two losses.