I'm intrigued by a rather different goat question (some wd call it a
meta-goat):
What is it about this problem that makes it so appealing?
Is it the fact that it refers to a TV show that none of us (in the UK) have
heard of, or does it have some other intriguing feature?
If the latter, can we generalise - i.e. what other problems have this same
feature?
JOHN BIBBY
PS: BTW, the Wikipedia article changes many times each week. A couple of
years ago I tried to "correct" it, but I gave up because too many
dunderdasses were instantly re-editing my edits. (I find Wikipedia reliable
on many things and a great source - but not on Monty Hall)
-----Original Message-----
From: A UK-based worldwide e-mail broadcast system mailing list
[mailto:[log in to unmask]] On Behalf Of Breukelen v Gerard (STAT)
Sent: 10 June 2008 14:49
To: [log in to unmask]
Subject: FW: Goat Problem
Hi Allstat members,
It seems that the classical quiz master (goat) problem keeps several
list members quite busy. To me this once more shows the importance of
formalized instead of informal thinking. So below is my result, using
Bayes' theorem
(too simple not to have been produced by someone else somewhere else
before, I guess). See if you can find an error in it.
Kind regards,
Gerard van Breukelen
Maastricht University, The Netherlands
-----Original Message-----
From: Breukelen v Gerard (STAT)
Sent: dinsdag 10 juni 2008 15:40
To: 'Iain Third'
Subject: RE: Goat Problem
Hi Ian,
I am sorry to disagree with you. Using Bayes theorem shows that whether
the quiz master knows or does not know the right door, makes the big
difference between a posterior probability of 33% or 50% for the
candidate to have made the right choice. Below is the proof. See if you
can find an error there.
Best regards,
Gerard van Breukelen
Maastricht University, The Netherlands
Define the following events:
A = candidate chooses the right door (with the car behind it), P(A) =
1/3
a = candidate chooses a wrong door (with a goat), P(a) = 2/3
B = quiz master opens the right door,
b = quiz master opens a wrong door
Bayes theorem: P(A|b) = P(A) * P(b|A) / P(b)
with:
P(A|b) = posterior probability that candidate chose the right door,
given that quiz master has opened a wrong door.
P(b|A) = probability that quiz master opens wrong door,
given that the candidate chose right door = 1.
P(b) = prob that quiz master opens wrong door = P(A)*P(b|A) +
P(a)*P(b|a).
We already know:
P(A) = 1/3
P(a) = 2/3)
P(b|A) = 1
Crucial now is P(b|a) = prob that qm opens wrong door given that
candidate chose wrong door.
Assumption 1: qm knows right door and will never open it, so P(b|a) = 1.
This gives P(b) = 1 and so P(A|b) = {1/3 * 1} / 1 = 1/3,
So the posterior prob that the candidate chose rightly, given that qm
has opened a wrong door, is 1/3 only.
Assumption 2: qm does not know the right door and choses randomly
between the two non-chosen doors, so P(b|a) = 1/2. This gives P(b) = 2/3
and so P(A|b) = {1/3 * 1} / 2/3 = 1/2,
So the posterior prob that the candidate chose rightly, given that qm
has opened a wrong door, is 1/2 in line with intuition.
Now given a sequence of quizes in which the qm has never opened the
right door, assumption 1 is very likely to be correct, giving the
posterior probability 1/3 for the candidate to have chosen correctly.
In the very first quiz, however, we cannot tell which assumption is
correct.
But even then the candidate had better switch to the third door if he
chose the first one and the qm opens the 2nd one, finding a goat there.
After all, switching is wise if assumption 1 holds and does not matter
if assumption 2 holds.
-----Original Message-----
From: A UK-based worldwide e-mail broadcast system mailing list
[mailto:[log in to unmask]] On Behalf Of Iain Third
Sent: dinsdag 10 juni 2008 14:57
To: [log in to unmask]
Subject: Re: Goat Problem
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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