Jake Paul claims Nate Diaz and Conor McGregor fights will happen “for sure”

Connor Bennett
Jake Paul talking to camera in white shirt against dark background

Jake Paul believes that he will fight Nate Diaz within the next two years, and should he win, it’ll lead to a superfight with Conor McGregor.

Since stepping into the world of boxing, Jake Paul has gotten himself involved in a number of high-profile clashes with different fighters and promoters. Though, his biggest have come with those involved in the UFC.

The YouTuber-turned-boxer has gone after the likes of Dana White, Conor McGregor, and Nate Diaz, with Jake wanting to fight both Conor and Nate at some point.

A fight with Nate is more likely, seeing as his UFC contract expires after his next fight – which comes on September 10 – and the fact that the pair have been teasing it for a few months now. Though, Jake sees a bigger picture too.

Jake Paul believes Conor McGregor fight will happen eventually

‘The Problem Child’ just UFC star ‘Sugar’ Sean O’Malley on episode 200 of the TimboSugar Podcast, where he revealed that a fight offer has gone out to a “big name” ahead of an event in October.

Following on from that, Jake was also quizzed on whether or not he and Nate will square off in the next year or two. “Yeah, that’s going to happen within the next two years for sure,” he answered.

“It’s huge, man, and it’s getting closer and closer to being real. Me versus Nate will probably be at 180lb and then, you know, I think it is like ‘okay, Jake Paul beats Nate’ and then it gets closer and closer to Conor not being able to duck me anymore.”

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While a fight between Jake and Conor may have seemed completely unrealistic when Jake first started boxing, the Irish superstar has engaged with the YouTuber’s jabs on social media.

McGregor still has some time to run on his UFC contract, and he’ll likely want to fight inside the Octagon a few times before doing something new, but him vs Jake would be very interesting. So, it would make sense for it to happen eventually.