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Before Ortiz bout FMJ knockout percentage at 147 was 17%. Is Pacquiao higher?

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  • Before Ortiz bout FMJ knockout percentage at 147 was 17%. Is Pacquiao higher?

    Who has the higher KO percentage at welterweight?

    Pacquiao or Mayweather?

  • #2
    mayweather: 37.5%
    pacquiao: 25%

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    • #3
      Floyds KO% at 147 wasn't 17% genius; He KO'd Sharmba Mitchell and Ricky Hatton; decisioned Mosley,Zab and Baldomir and even if you want to include a ctachweight with marquez thats still not 17% its close to 40% learn how to do some math.

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      • #4
        Originally posted by dc3383 View Post
        Floyds KO% at 147 wasn't 17% genius; He KO'd Sharmba Mitchell and Ricky Hatton; decisioned Mosley,Zab and Baldomir and even if you want to include a ctachweight with marquez thats still not 17% its close to 40% learn how to do some math.
        Floyd's KO ratio is stated at 7:18 of video:

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        • #5
          Anything between 141-147 is WW

          Floyd at WW-
          Victor Ortiz- KO
          Shane Mosley - UD
          JMM - UD
          Ricky Hatton - TKO
          Baldomir - UD
          Zab - UD
          Sharmba Mitchell - TKO

          3/7 = 43 percent

          Manny Pacquiao at WW-
          Shane Mosley - UD
          Joshua Clottey - UD
          Miguel Cotto - TKO
          ODH - RTD (basically counted as a TKO)

          2/4 = 50%

          Huzzah for math

          When they said 17 percent, they didn't count one of his victories as a KO and basically had it 1 KO in 6 fights (because they said it while he was fighting Ortiz)
          Last edited by Hougigo; 10-25-2011, 10:53 AM.

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          • #6
            Originally posted by hougigo View Post
            Anything between 141-147 is WW

            Floyd at WW-
            Victor Ortiz- KO
            Shane Mosley - UD
            JMM - UD
            Ricky Hatton - TKO
            Baldomir - UD
            Zab - UD
            Sharmba Mitchell - TKO

            3/7 = 43 percent

            Manny Pacquiao at WW-
            Shane Mosley - UD
            Joshua Clottey - UD
            Miguel Cotto - TKO
            ODH - RTD (basically counted as a TKO)

            2/4 = 50%

            Huzzah for math

            When they said 17 percent, they didn't count one of his victories as a KO and basically had it 1 KO in 6 fights (because they said it while he was fighting Ortiz)

            couldn't have been said better....

            / thread.....

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            • #7
              Originally posted by dc3383 View Post
              Floyds KO% at 147 wasn't 17% genius; He KO'd Sharmba Mitchell and Ricky Hatton; decisioned Mosley,Zab and Baldomir and even if you want to include a ctachweight with marquez thats still not 17% its close to 40% learn how to do some math.
              Pac only had 3 fights at 147. DLH, Clottey and Mosley. So prior to the Mayweather Ortiz fight Pac had the higher percentage at 33%.

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              • #8
                Originally posted by texasboi15 View Post
                Floyd's KO ratio is stated at 7:18 of video:

                So just because Larry Merchant says it, that automatically makes it a fact? Come on man, actually do your own research before stating bull**** statistics. Floyd KO'd 3 guys at welterweight out of 7. Which is 43%. I guess if you exclude the Ortiz KO and count in the Oscar fight that brings his KO ratio at welterweight to 25%.

                Manny knocked out 2 out of 6 guys at welter and above. Which is 33%

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                • #9
                  Originally posted by hougigo View Post
                  Anything between 141-147 is WW

                  Floyd at WW-
                  Victor Ortiz- KO
                  Shane Mosley - UD
                  JMM - UD
                  Ricky Hatton - TKO
                  Baldomir - UD
                  Zab - UD
                  Sharmba Mitchell - TKO

                  3/7 = 43 percent

                  Manny Pacquiao at WW-
                  Shane Mosley - UD
                  Joshua Clottey - UD
                  Miguel Cotto - TKO
                  ODH - RTD (basically counted as a TKO)

                  2/4 = 50%

                  Huzzah for math

                  When they said 17 percent, they didn't count one of his victories as a KO and basically had it 1 KO in 6 fights (because they said it while he was fighting Ortiz)
                  The question was who has a higher percentage at 147

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                  • #10
                    Originally posted by hougigo View Post
                    Anything between 141-147 is WW

                    Floyd at WW-
                    Victor Ortiz- KO
                    Shane Mosley - UD
                    JMM - UD
                    Ricky Hatton - TKO
                    Baldomir - UD
                    Zab - UD
                    Sharmba Mitchell - TKO

                    3/7 = 43 percent

                    Manny Pacquiao at WW-
                    Shane Mosley - UD
                    Joshua Clottey - UD
                    Miguel Cotto - TKO
                    ODH - RTD (basically counted as a TKO)

                    2/4 = 50%

                    Huzzah for math

                    When they said 17 percent, they didn't count one of his victories as a KO and basically had it 1 KO in 6 fights (because they said it while he was fighting Ortiz)
                    What about the Margarito fight? That would make it 2/5 = 40%

                    Mayweather - 43%
                    Pacquiao - 40%

                    Not much between them TBH
                    Last edited by Pretty Boy1; 10-25-2011, 11:35 AM.

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