Thoughts? Or the brits would still buy the Pac-Khan fight if the fight is "boring" or ends too quickly.
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Will the AJ-WK affect the Khan-PAC PPV Numbers?
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Originally posted by Isaac Clarke View PostAnyone think Hearn is going to let Pacquiao-Khan be on Sky Box Office?
Khan is not a big draw in the UK and doesn't even fight on Sky that often.
With AJ v Klitschko and Brook v Spence happening on PPV in April and May, I doubt they'll squeeze in another one.
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khan v Pac won't be PPV, unless BoxNation/BT pick it up and try to make it into the PPV platform they're making.
Worth remembering too that Khan isn't really a big deal in the UK anymore. His lasts fights here weren't even sold out. The only way Khan will do big numbers in the UK again will be if he fights Brook.
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If you guys only pay $20 for a PPV over there in the U.K. I doubt it wil... If both were here in the USA on PPV I'd choose Joshua - Klit and stream PAC-Khan because they're $70 each. Both good fights to watch though...
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Originally posted by Jsmooth9876 View PostIf you guys only pay $20 for a PPV over there in the U.K. I doubt it wil... If both were here in the USA on PPV I'd choose Joshua - Klit and stream PAC-Khan because they're $70 each. Both good fights to watch though...
PPV, the PPV model would flourish but everyone involved is too greedy so the model as such will continue to fail.
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Originally posted by LacedUp View PostDon't think so.
Khan is not a big draw in the UK and doesn't even fight on Sky that often.
With AJ v Klitschko and Brook v Spence happening on PPV in April and May, I doubt they'll squeeze in another one.
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Pacquiao khan will be in Vegas
AJ Klitschko is in the UK
The whole card will be well finished hours before the Pacquiao khan card even starts
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Originally posted by English Lion View PostKhan is not a big draw in the uk? You must be absolutely livid this morning mate after your idol Brook's been upstaged for the second time in a row...how does it feel to know that even with the knockouts Khan will always be the bigger name compared to Brook?
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