Peter d-d-d-d-d-d-dies!?


by DocRyedale

22 years, 1 month ago

We all know that GB III hasn't been made, but a script has been written. Now, did I hear someone say that Dr. Peter Venkman supposedly dies at the beginning of GB III? yikes yikes yikes Does anyone know why/how does this happens (Other than the fact that Bill Murray isn't interested in being in it.)? Was it a car wreck, murder, or an accident on a bust? Do you think he'll come back as a ghost and get busted? Someone please tell me! I must know! yikes yikes yikes

by Ghostbuster_D

22 years, 1 month ago

Peter Venkman dying at the beginning and being a ghost throughout the rest of the movie, (from what I've heard), is the only way he'd do it, according to what he said.

by MasterSpider

22 years, 1 month ago

If I remember through RUMORS only, it was that he died in a busting accident…

by LordVego

22 years, 1 month ago

how could you die during a bust?!!? that must be gruesome, i mean what did he do? get hit by a proton stream?

by Scoleri_guy

22 years, 1 month ago

I don't really like that idea of Pete Dying. Peter is part of the team, not Ghost Peter. He should live.

by GBFreak

22 years, 1 month ago

I remember reading in Cinescape that Murray would only do the film if he dies at the beginning and then comes back as a Ghost. But how he dies and then how he is portrayed as a Ghost in the GB3 Script, is up in there air I think.

by roberto182

22 years, 1 month ago

nooooooo
peter is my favoryte character of the ghostbusters please tell me who im dream

by MasterSpider

22 years, 1 month ago

I'm guessing that came out of Bill's mouth just to create more friction on keeping it from being made…

by DocRyedale

22 years, 1 month ago

I've got it! He dies in a golfing accident with a Groundhog. . .

by roberto182

22 years, 1 month ago

noo would more best is day into an busting insident i got idea of this:
Peter: i think where busting
Ray. dont look the trap
Pete: haaaa
Ray: dont move
proton beam: fhuusssssh
and die for this this is only an idea no is true