I'm no expert on this but;
If there are 3 doors and only one has a car then you have 2/3 chance of
choosing a door with a goat at the first step.
Now, at the second stage, when there are two doors, if you switch you
must either move "from goat to car" or "from car to goat".
If there is a 2/3 chance that you have a goat from the first round and
switching when you have a goat guarantees you get the car then there is
a 2/3 chance that to swap gives you the car and to stick gives you the
goat.
You should swap!
-----Original Message-----
From: A UK-based worldwide e-mail broadcast system mailing list
[mailto:[log in to unmask]] On Behalf Of Mario Ouwens
Sent: 11 June 2008 10:04
To: [log in to unmask]
Subject: Goat problem solution
Dear all,
I want to add a simple comment, which explains a lot to me. Suppose 3
boxes
A, B, C and suppose that a priori the chance is equal for the three
boxes.
Then
P(A) = P(B) = P(C) = 1/3.
Now suppose you choose A. Then the chance you are right is equal to 1/3.
If it is A, B or C is opened. The switch will end up in the wrong
result.
If it is B, C will be opened. The switch is to C and will result in the
right result.
If it is C, B will be opened. The switch is to B and will result in the
right result.
What happens is that in the second case and the third case, different
boxes
are opened. So, the chosen box when you switch is different, while in
the
case you do not switch, you stay in the first situation, regardless the
additional information. Hence, you use the information that the other
box is
not the right one.
Best wishes,
Mario Ouwens
-----Original Message-----
From: A UK-based worldwide e-mail broadcast system mailing list
[mailto:[log in to unmask]] On Behalf Of John Bibby
Sent: Wednesday, June 11, 2008 10:41 AM
To: [log in to unmask]
Subject: Re: Goat Problem
One way which I think is sound and shows that you should shift (although
it
does not provide any numbers) is the following:
Consider the choice as a two-stage problem:
D1: choose a door
D2: Should I shift, and if so to which one?
Suppose you decide D1 and D2 simultaneously on the basis that NO door
will
be opened.
Then 'clearly', D1+D2=shift has just the same chances as D1 alone (i.e.
D1+D2=noshift)
Now if a door is opened to show a goat. This HAS to increase the chance
of
getting the prize if you shift. So D2=shift is preferable to D2=noshift.
QED
JB
-----Original Message-----
From: A UK-based worldwide e-mail broadcast system mailing list
[mailto:[log in to unmask]] On Behalf Of Rob Fitzmaurice
Sent: 10 June 2008 09:15
To: [log in to unmask]
Subject: Re: Goat Problem
I didn't see the previous discussion but this offers some thoughts:
http://montyhallproblem.com/
Rob
>
> From: John McKellar <[log in to unmask]>
> Date: 2008/06/10 Tue AM 07:37:07 GMT
> To: [log in to unmask]
> Subject: Goat Problem
>
> 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
>
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