When all is said and done...Who will hold the p4p crown in 2010

[P.WILL]

paul williams......

[dans]

John Duddy

[Stoppage]

Either Mayweather or Pacquiao. If Pacquiao continues fighting in 2010 against the top fighters and winning I think he'll remain number one. The ways Mayweather would get it is if he either beat Pacquiao, beat the man who beat Pacquiao, or if Pacquiao retires.

[Pullcounter]

marquez can not be #1 by beating Casamayor and Juan Diaz and possibly a Floyd whos been out of the game for almost 2 years..

c'mon!!! the case for jmm being p4p #1 is much much much greater than the case for cotto being p4p #1.

if the #2 cant be #1 by beating the former #1 and the current #1 losing, then something is ****ed up with the universe.

[Xyei]

I guess we'll have to wait until it's all said and done...

[TxHustla]

c'mon!!! the case for jmm being p4p #1 is much much much greater than the case for cotto being p4p #1.

if the #2 cant be #1 by beating the former #1 and the current #1 losing, then something is ****ed up with the universe.

Cotto would be knocking out the current p4p king so to speak.....marquez would be beating the former p4p king....i dont see how u have marquez becoming #1 over Cotto just with that alone..who cares about who floyd USED TO BE.PAC IS NOW.

Floyd to me was never #1..not by beating judah,old dlh and sorry ass hatton and baldemir.....but cotto beating mosley,judah,clottey and pac would be #1..come on dude..

[toysale]

I can't see Pac beating Floyd in all honesty, Mayweather is unbeatable. So if this fight ever does happen in 2010, Mayweather will be P4P king, but if Pac beats Cotto and May vs Pac never does happen, Pac stays P4P#1.

[vercingetorix3]

Hmmm..............Charlie Zelenoff, of course! Did you see his last "exhibition match" against the guy in his underwear? That body shot was incredible!

[Elie]

Floyd will beat pac in 2010. mark my words.

[TheSurgeonMDMPH]

PW, champ at 147-160 and being blatantly ducked by all those on the current p4p list!

The man is the epitome of p4p.