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Tom Holland reveals if he thinks Spider-Man could beat Wolverine in a fight

The Marvel Cinematic Universe's biggest action sequences are essentially the $200 million version of a kid playing with their action figures, except in this case that kid is Kevin Feige and he's one of the most successful and powerful producers the industry has ever seen.

The Marvel Cinematic Universe’s biggest action sequences are essentially the $200 million version of a kid playing with their action figures, except in this case that kid is Kevin Feige and he’s one of the most successful and powerful producers the industry has ever seen.

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He took the idea of a shared mythology with overlapping storylines directly from the comic books, and as a result fans have been given crossovers, set pieces and smackdowns beyond their wildest dreams in live-action. Sadly, one of the fights we haven’t seen is Spider-Man vs. Wolverine, but Tom Holland has revealed how he thinks it would turn out.

After being asked by Geek Culture Explained, the actor put some serious thought into his response, and he wasn’t too proud to admit that the mutton-chopped mutant would probably make short work of the web-slinger.

“Are we talking Logan Wolverine or X-Men: Origins Wolverine? I’ll beat up that old man. With X-Men: Origins, he’s going to batter me. I think Wolverine beats Spider-Man every single time because there’s not that much battering Spider-Man can do. Plus, Spider-Man doesn’t kill people and Wolverine does, so that’s a huge advantage.”

If we boil it down the basics, then Wolverine in full berserker mode would simply slice and dice Spider-Man into mincemeat, especially when he’s nigh-on indestructible and isn’t against the idea of reducing his foes to nothing but a bloody pulp should they get in his way. At least Holland is aware of it, so he doesn’t have any ideas beyond his station of a teenager with some web-shooters taking down an immortal adamantium machine.


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Scott Campbell
News, reviews, interviews. To paraphrase Keanu Reeves; Words. Lots of words.