PLEASE VOTE!! Not sure if these guys are going to fight this year or not but there's rumors. If it does happen, it'll likely be at the weight of 140. I give the fight to Juan Manuel Marquez via late round TKO. I think Hattons lunging style will be perfect for Juan's counter punching. It would be a hell of a fight though, with both sides being hurt. What do you guys think?
I would tend to agree. Would be an interesting fight though... Diaz seems to have a similar style to hatton except is a little slower and doesnt have as much power (but with a better chin); and Diaz gave Marquez some problems But still... Marquez is an elite fighter and has the better skill set.. so i would also go for a late round TKO
I would also favor Marquez but Marquez to me was significantly slowing down as a competitor at lightweight and Hatton was roadkill in his last fight so I'm not sure if this is a case of who is better or who is the least shopworn.
Yeah, but Diaz, while even in the process of being technically outclassed, was physically overwhelming to Marquez. Although in different ways, Diaz and Hatton are the same kind of fighter, who base everything on being physically overwhelming, rather than thinking. And, I'm not knocking it because they both were durable and strong enough in a total body sense (as opposed to punching ability), to achieve an awful lot with that style. Now Hatton is much bigger and more imposing than Diaz and Marquez is just very small for a 140 pounder. He was pushing it at 135. He can bulk up all he wants but it doesn't make his natural frame, with just his natural athletic weight any different.
Good post Ears. To me alot depends on where the fight is held. If Ricky is allowed to rough Marquez up, I think he'll wear him out. If Marquez is able to keep it a boxing match, I think he'll win on points.
Joke option. I was going to add elbows, wrestling, and low blowing too (Kostya fight). Haha someone actually picked it.
Before the Mayweather fight JMM was fairly widely considered P4P no 2/3. He has too much for this version of Hatton.