De La Hoya vs Mayweather both in prime who wins?

It was a close fight till the late rounds. I was laughing at the Floyd taking him apart comment. Floyd didn't take **** apart, and this was against a previously retired Oscar who already had a shoulder injury. Oscar was half what he was in his prime for that fight.

 

Floyd would win the same way he did 2007. Although Oscar wasnt prime, Floyd was so small compared to him and it was his first time at 154 pounds. A prime Mayweather at a comfortable weight against Oscar, Mayweather wins clearly.

 

And Floyd was half the fighter because Oscar weighed so much more and Floyd was on his first trip at 154 pounds. It was fair in my opinion. Floyd wins any fight against Oscar prime vs prime. That was definately not the best Mayweather we saw, and it wasnt the best Oscar either.

 

id go oscar in a close one

 

...At 154.

 

Younger Delahoya had all the tools to beat Mayweather . Could have potentially stopped Floyd as well . GoldenBoys respectable hand speed /knockout power could have gotten the job done . BUT i predict Floyd to lose on points 116/112

 

It was a prime Mayweather vs an old used up coming off injuries Oscar. Floyd is an average size welterwight at the least, yes he was smaller than Oscar especially at 154 but the size thing is blown out of proportion. Compare what Oscar looked like in his fights with Whitaker, Quartey and Trinidad to when he fought Floyd. He was sharper way faster and stronger in those fights. Floyd wouldn't be strong enough to hang with that version.

 

Yes he would. Floyd always look shite at 154

 

LOL so a split decision means it was a close fight?

No. Floyd won by such a big margin that they couldn't help but award him the decision. If it was an even fight the cards would read 118-110 for Oscar.

 

lol the same oscar that lost every big fight in his career?

 

Novice - biggest shame as a purported 'fan' of boxing. You obviously weren't a fan of boxing when DLH was setting his legacy.

 

Dlh beats the coward by split decision.