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The wikipedia reference on this has a section on alternate scenarios which I am convinced is wrong.
 
It states that if after you have made your initial pick, the presenter forgets which door has the car behind it, and so goes for broke and luckily picks a door with a goat behind it, the odds are now 50/50 as to whether switching gives the car or not.  To me it shouldn't matter what the presenter knows or doesn't know, only the information he reveals once he has opened the door.
 
Granted, there is a 33% chance he will muck things up and open the door with the car, but after that stage, if he got a goat, nothing changes, and the odds should still be 0.66 if you switch, 0.33 if you don't.
 
The references are Granberg and Brown, 1995:712 and vos Savant, 2006 which I have not read so I could be misenterpreting them.> Date: Tue, 10 Jun 2008 10:52:35 +0300> From: [log in to unmask]> Subject: Re: Goat Problem> To: [log in to unmask]> > John McKellar wrote:> > Call me stupid, but I've just heard the 3 door game show on a> > discussion about probability on radio BBC4's Material World, and I think> > it was also discussed by Mervyn Bragg (sorry very UK-centric) this week.> >> > My problem is I don't believe the discussion. BUT I suspect it's an old> > topic here, so rather than generate the same again;> > Does anyone have a clear discussion on the problem from last time?> >> > John> > --------------------------> > John McKellar> >> >> > > Hi John,> > maybe this can help:> > http://en.wikipedia.org/wiki/Monty_Hall_problem> > > -- > Nikolaos A. Patsopoulos, MD> Department of Hygiene and Epidemiology> University of Ioannina School of Medicine> University Campus> Ioannina 45110> Greece> Tel: (+30) 26510-97804> mobile: +30 6972882016> Fax: (+30) 26510-97853 (care of Nikolaos A. Patsopoulos)> e-mail: [log in to unmask] 
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