rios649

Roger Mayweather feels Rios has chance to beat Pacquiao

rios649By Chip Kuehn: Most boxing fans aren’t giving Brandon Rios (31-1-1, 23 KO’s) much of a chance this Saturday night in his fight against Manny Pacquiao (54-5-2, 38 KO’s), but Roger Mayweather sees Rios as a viable threat to beating Pacquiao if he can take the fight to him in their fight in Macao, China. Roger isn’t saying exactly how Rios will be able to beat Pacquiao, but it seems clear that he thinks he can do it by putting him under duress and forcing him to fight hard.

Roger told Hustleboss.com “They gave a good fight for him then. I don’t know if he can beat him or not, but if the guy is tough and he brings it to him; [Rios] may get a chance to whoop his *ss, because he’s been off for a while too.”

Pacquiao’s time off from the ring could be key in this fight if Rios is to have a chance to beat him. Before his 11 month time off from the ring, Pacquiao was already starting to show signs of age in his losses to Juan Manuel Marquez last December and his decision loss to Tim Bradley last year in June.

Pacquiao was starting to shown signs of age in his fight with Marquez in November 2011 in winning a controversial decision over him. From there, Pacquiao was beaten 7 months later by Bradley by a controversial 12 round majority decision. While it was a questionable decision, it was an off performance from Pacquiao, who looked sluggish throughout and incapable of fighting hard for the full 3 minutes of every round.

In Pacquiao’s last fight in December last year against Marquez, Pacquiao seemed to to freeze when Marquez nailed him with a right hand in the 3rd round that dropped him. Three rounds later, Marquez knocked Pacquiao out cold with a right hand in the 6th round. This Saturday night, Pacquiao will be returning to the ring after almost a year off from the ring. You can’t expect Pacquiao to be any better than he was in his last two fights because he’s been out of the ring a year, he’s a year older, and the rust will likely be there for him.