10 Reasons Conor McGregor Vs. Floyd Mayweather Is More Likely Than You Think

1. Mayweather Wants That 50th Win

Floyd Mayweather is seen during a WBC welterweight title fight against Robert Guerrero, Saturday, May 4, 2013, in Las Vegas. (AP Photo/Isaac Brekken)
Isaac Brekken/AP

Can he beat the curse?

The magic 50-0 number in pro boxing is an elusive number, though probably overrated. The story goes something like this: Rocky Marciano (heavyweight champ) retired after going 49-0, and ever since, fighters have been chasing his record. Those who come close always seem to fail. Ricardo Lopez managed to make it to 51-0-1, but that one draw came in his 48th fight. Larry Holmes got close, and probably started the talk of it being a cursed number, when at 48-0, he claimed Marciano couldn't have carried his jock strap. He then turned around and lost his 49th fight.

There's also the quality of competition argument: surely, someone could come around and crush 50 tomato cans, but who cares? Marciano was a legendary fighter, but he's not a top-10 all time sort of guy, so what does that undefeated record matter?

Yet for Mayweather there probably is an allure to hitting 50-0 while still being considered one of the best fighters of all time. The media at large, and the public, will focus on the number rather than the quality of the fight. And Mayweather may see an MMA fighter with no pro boxing fights as the easiest way to hit 50-0 - and because McGregor is a champ in two weight classes in his own sport, the fight will still come off as somewhat legitimate.

In reality, this is Randy Couture vs. boxing's James Toney, but in reverse.

Is this a fight you want to see? Add your thoughts below!

Contributor
Contributor

Primarily covering the sport of MMA from Ontario, Canada, Jay Anderson has been writing for various publications covering sports, technology, and pop culture since 2001. Jay holds an Honours Bachelor of Arts degree in English from the University of Guelph, and a Certificate in Leadership Skills from Humber College.