Marquez to ditch Pacquiao for Bradley, Rios?

MANILA, Philippines – Juan Manuel Marquez said there is no point in facing archrival Manny Pacquiao for a fifth time and has started looking for other possible foes.

In a report by the Latin American Herald Tribune, Marquez is now eyeing either Timothy Bradley or Brandon Rios as his next opponent.

“A fifth fight against Pacquiao would be pointless. We already saw (who is better). The goal we had has already been achieved. There’s nothing pending, everything’s been decided and (another bout) would be the last option,” Marquez said.

The Mexican indeed made a strong case as to who is the superior fighter between him and Pacquiao when he knocked out his Filipino nemesis last December to cap off their storied rivalry. In four fights, Pacquiao only managed to win two close, disputable decisions, with their first meeting in 2004 ending up in a controversial draw.

While Pacquiao is keen on facing Marquez for a fifth time in September, the same can no longer be said of the Mexican, who is now exploring other options.

Here’s where Bradley, who pulled off a shocker when he defeated Pacquiao in June last year, comes in.

“Bradley won a controversial decision over Pacquiao, so why not take him into consideration if he’s a good fighter,” Marquez said.

Bradley will defend his WBO welterweight title against Freddie Roach-trained prospect Ruslan Provodnikov this Saturday (Sunday Manila time), and a victory will put him in position for a showdown with Marquez.

Also in Marquez’s radar is the heavy-handed Brandon Rios, who is scheduled to face Mike Alvarado in a rematch later this month.

The Mexican counterpuncher added that they will continue to study their options.

 â€œIf there’s a green light to keep going, we’ll see what the best option is and who’s the best rival to fight next,” he said.

The best option, financially, will always be another fight with Pacquiao, as Marquez doesn’t stand to earn much against either Bradley or Rios than he will in a rematch with the Filipino icon.