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Wolverine
Sabretooth(w/adamantium)
Deadpool(standard gear)

vs

Omega Red
Cyber(classic)

For the purposes of this fight, assume that Omega Red refrains from using his Death Spores unless/until Cyber gets taken out of the fight by the opposition.

Fight in a generic arena to KO or death.

Who takes it?

Got a question, is cyber at his prime or are we talking about the ill cyber.

Originally posted by carver9
Got a question, is cyber at his prime or are we talking about the ill cyber.
In his prime...

If this is cyber at his prime he could solo this given time. They have no way of hurting him, his poison claws bypass healing factors and he has a healing factor of his own (which he doesnt need). Cyber is a walking plot device.

Omega red is also a beast and took on the entire xmen by himself. They guy got a tank thrown on him by colossus and ripped through it like it was nothing. Not a good fight if you ask me.

Well, OR kills Cyber quite easily, though...

But his opponents lack means to put him down indeed. That is, until DP somehow showes a grenade down his throat. )

Originally posted by Survivor19
Well, OR kills Cyber quite easily, though...

But his opponents lack means to put him down indeed. That is, until DP somehow showes a grenade down his throat. )

Omega red and cyber is on the same team and what if omega red coils two of there opponents then that leaves one of them to face cyber and basically get killed.

Then cyber could just walk up to the people that omega red has in his coils and beat them to death. Remember this, iron man, rogue, and colossus failed to break free from omega reds coils.

None fight

Originally posted by carver9
If this is cyber at his prime he could solo this given time.

No. Logan has overcome his mental block.

Originally posted by carver9 They have no way of hurting him, his poison claws bypass healing factors and he has a healing factor of his own (which he doesnt need). Cyber is a walking plot device.

His face is still vulnerable.

Poison wouldn't bypass HF. Hallucinogen would, though.