----- Original Message -----
Sent: Thursday, August 05, 2004 2:37
PM
Subject: Re: [DUG] Welcome back
problem
Definately!
(As long as I can lay an additional bet that the
result will be 50/50).
Ok, so the options are (assuming
GGC):
1. Picked a goat (door 1), host reveals the
other goat (door 2).
Probability that you are in this
situation: 1/3
Change to door 3 and get the car.
Value of this choice in this
situation: 1
Don't change, and get a goat.
Value of this choice in this situation:
0
50/50.
2. Picked a goat (door 2), host reveals the
other goat (door 1).
Probability that you are in this
situation: 1/3
Change to door 3 and get the
car.
Value of this choice in this
situation: 1
Don't change, and get a goat.
Value of this choice in this
situation: 0
50/50.
3. Picked the car (door 3), host reveals a goat (door 1).
Probability that you are in this situation:
1/6
Change to door 2 and get a goat.
Value of this choice in this situation:
0
Don't change, and get the car.
Value of this choice in this
situation: 1
50/50.
4. Picked the car (door 3), host reveals a goat (door 2).
Probability that you are in this situation:
1/6
Change to door 1 and get a goat.
Value of this choice in this situation:
0
Don't change, and get the car.
Value of this choice in this
situation: 1
50/50.
The probability of getting to any of the 4
possibilities is irrelevant, as they all end up with a 50/50
chance.
No, it's not irrelevant. Multiply
the probabilities of getting to each situation by the values of each
choice in that situation and sum them - the total value of changing is 2/3,
the value of sticking with your original choice is 1/3.
Essentially, the first pick, and the subsequent
door that the host opens, is irrelevant. You are left with two doors, one
of which has the car.
Essentially, in the first pick,
which you later change away from, you are choosing a door to
eliminate. There's a two-thirds chance you are eliminating one of the
goats. The host then eliminates the second goat for you.
Of course, there's a one-third
chance you blow it by eliminating the car in your first pick. Tough luck if
that happens, but the odds are on your side and if you play it right you're
more likely to win the car than not.
Regards,
Paul Matthews
ProSouth Technology Solutions
Phone: +64 3 470
1321
DDI: +64 3 470 1324
0800 PROSOUTH (0800 776 768)
249 Cumberland
Street, Dunedin, New Zealand
Visit us at
www.prosouth.co.nz
----- Original Message -----
Sent: Thursday, August 05, 2004 2:13
PM
Subject: RE: [DUG] Welcome back
problem
Try it - with
money. I volunteer to be the host.
-----Original
Message-----
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]] On Behalf Of Paul M -
ProSouth
Sent:
Thursday,August,05,2004 2:11 p.m.
To: [EMAIL PROTECTED]; NZ
Borland Developers Group - Delphi List
Subject: Re: [DUG] Welcome back
problem
This is not
correct:
> If you dont switch, you
have a 1/3 chance of car and a 2/3 chance of goat
> If you switch, you have a 2/3 (!)
chance of car and 1/3 chance of goat.
Why do you have a 2/3 chance of
the car, and a 1/3 chance of the goat, when
you only have two doors to choose
from?
All that you know is that if you
had chosen a different door (the one that
the host revealed to have a concealed
goat), you would definately have been
incorrect. Now you have two doors, and
hence a 50/50 chance.
Regards,
Paul Matthews
ProSouth Technology
Solutions
Phone:
+64 3 470 1321
DDI: +64 3 470 1324
0800 PROSOUTH (0800 776 768)
249 Cumberland Street, Dunedin, New
Zealand
Visit us
at www.prosouth.co.nz
----- Original Message -----
From: "Struan
Judd" <[EMAIL PROTECTED]>
To: "'NZ Borland Developers Group -
Delphi List'" <[EMAIL PROTECTED]>
Sent: Thursday, August 05, 2004 1:41
PM
Subject: RE:
[DUG] Welcome back problem
> As I looked it up here
goes.
>
>
First choice, 1 of three:
> 1/3 chance of car. (Case C)
> 2/3 chance of
goat. (Case G)
>
>
Remove a goat. (Always is assumed)
>
> Second choice, Switch or Don't.
> Case C
> Switch =
Goat
> Don't = Car
>
> Case G
> Switch = Car
> Don't =
Goat
>
>
If you dont switch, you have a 1/3 chance of car and a 2/3 chance of
goat
> If you
switch, you have a 2/3 (!) chance of car and 1/3 chance of
goat.
>
>
So you should always switch. The odd favour it.
>
> Of course I'd listen for the goat
:-)
>
>
Struan Judd
Mailto:[EMAIL PROTECTED]
>
Developer
Phone: +64 (9) 368 9368
>
eCargo
Fax: +64 (9) 368 9369
> Visibility - Communication -
Measurement Mobile: +64 (21) 685 335
> Private &
Confidential
http://www.ecargo.co.nz
> -----Original
Message-----
>
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
On
> Behalf Of Robert
martin
> Sent:
Thursday, 5 August 2004 1:22 p.m.
> To: NZ Borland Developers Group - Delphi
List
>
Subject: Re: [DUG] Welcome back problem
>
> Hi Deiter
>
> Im not sure I follow your logic
here.
>
>
> door 1, you get some new information only about door 1 and door
3.
> >
This
>
is
>
>
What new information do you ahve about door 3 that you do not have
about
> door
2?
>
>
> for door 3 is the sum of the probabilities of door 1 and 3 minus
the
> >
probability of door 1, it follows that the probability of door 3
is
> > 2/3
mins 0 equal to 2/3.
>
>
Isn't the probability of Door two 2/3 minus 0 ... 2/3 as well?
>
> I think the probability of door 2 is
the same as door three.
>
>
>
>
Rob Martin
>
Software Engineer
>
>
phone 03 377 0495
> fax 03 377 0496
> web www.chreos.com
> -----
Original Message -----
> From: "Dieter K�hler"
<[EMAIL PROTECTED]>
> To: "NZ Borland Developers Group - Delphi List"
<[EMAIL PROTECTED]>
> Sent: Thursday, August 05, 2004 12:35
PM
> Subject:
RE: [DUG] Welcome back problem
>
>
>
>
> >
>Doors 1 2 3
> > >
> > >I choose 2 - probability of Car = 1/3
> > >
> > >Host opens 1 =
Goat - probability of: 2 = car is 1/2. 3 = car is 1/2.
> >
> > The probability of
door 2 (1/3) does not change, and the sum of the
> > probabilities of door 1 and 3
(2/3) do not change. When the host opens
> > door 1, you get some new
information only about door 1 and door 3. This
> is
> > the reason, why the probability
for door 2 remains the same. For door 1
> > you get the information that
the probability is 0. Since the probability
> > for door 3 is the sum of the
probabilities of door 1 and 3 minus the
> > probability of door 1, it
follows that the probability of door 3 is 2/3
> > mins 0 equal to
2/3.
>
>
> >
Dieter
