How good a win is JMM for Mayweather?

[BattlingNelson]

It means a little more than "****". But in essence you are right. Floyd has already beaten the best at 130 and 135. Why take a step backwards and do it again when you have so many unanswered questions at 147?

It means little. It's hopefully just a tune-up. Shame it is that the p4p no.2 is reduced to just that. It would have been better to take a WW contender instead.

[BattlingNelson]

He's fighting a 35 year old who is moving up 2 weight classses.

I'll give him a C

Now if JMM wins, I'll give JMM an A.

Absolutely. JMM is a brought in as a sacrificial lamb and should be given major props if he manages to beat Floyd.

[TheGreatA]

much like the victory of monzon over napoles IMO...

Napoles was more highly regarded than JMM but he was also smaller compared to Monzon than Marquez is to Mayweather.

Good comparison though. Some people think of Napoles as one of Monzon's best wins however, I doubt it'll be the case if Mayweather beats Marquez.

[rizkybizness]

Napoles was more highly regarded than JMM but he was also smaller compared to Monzon than Marquez is to Mayweather.

Good comparison though. Some people think of Napoles as one of Monzon's best wins however, I doubt it'll be the case if Mayweather beats Marquez.

the man has spoken

everyone shut the *** up, pack your bags and GTFO

[Doctor_Tenma]

I really don't know how to rate it, would it be fair if I rated it, a better win then Hatton?

[S.G.]

B.

would be A. but for Marquez being untested at 140/147

Marquez comes to win and he's a great boxer in his own right. Mayweather's in for a rough night

[UnDeniable]

JMM will be outweighed by 6 lbs but he will win

Pac will be outweighed by 10 lbs but he will beat Cotto

JMM - Pac 3

[Doctor_Tenma]

I'll say a B.

[Allucard]

If Floyd Mayweather beats Juan Manuel Marquez how would you rate that win?

Poll coming.

If the fans think the p4p #2 isn't a good win then there's something wrong with the fans...

[Doctor_Tenma]

Lets be real people, in no way is that an A.