Re: Monty Hall problem wrong?
- From: magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin)
- Date: Tue, 8 Apr 2008 20:24:31 +0000 (UTC)
In article <a05035e2-02d5-4d2b-89e8-a1ebaa02cf8e@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Maya <maya_souj@xxxxxxxxxxx> wrote:
If you have 3 doors, behind one of which is a car and behind each of
the other two of which is a goat, why is your second choice still 1 in
3, instead of 1 in 2 ?
You're already running into problems. The Monty Hall problem contains
some subtleties, and it is imperative to include all the pertinent
information if you are to do a proper analysis.
In the "classic" version of the Monty Hall problem, there are 3 doors;
behind one is a car, behind each of the other two is a goat. Monty
already knows where the car is. After you select one door, Monty will
open a door such that:
(i) Is not the door you picked; and
(ii) Monty knows the car is not behind the door he opens.
Monty will always open a door, and the door will never show the car,
and Monty will always know what he is doing. If you did not pick the
car in the first place, Monty's choice is forced. If you did pick a
car, Monty will open either of the remaining unchosen doors with equal
probability.
You are then given the choice of whether to stick with your door or to
switch to the remaining unopened door.
Under this conditions, why is the probability of your door still 1/3?
You can do the bayesian analysis if you wish, or you can ask yourself:
If instead of showing one of the unpicked doors and offering you
to switch to the remaining one, Monty instead said "You can either
stick to your choice, or you can switch and take whatever is
behind BOTH the other two doors"... would you switch? What is the
probability of winning if you switch under those conditions?
Now check that the game in which Monty knowingly opens a goat-door
under the restrictions given above is isomorphic to the the game in
which Monty does not open any doors but he offers you to switch from
the one-chosen door to the two-unchosen doors.
If you pick door 1, and it's the car, and then Monty opens door 2 and
shows you a goat
No; if Monty will ALWAYS open door 2 when the car is behind door
number 1 and you pick door number 1, then there is more information in
the game and the analysis is not quite right. (For instance, if you
were to pick door 1 and Monty opens door 3, then you would ->know<-
that the car is behind door number 2 and you would switch). You have
to be ->VERY CAREFUL<- in stating the hypothesis. Even a small
deviation may render the analysis useless.
and then asks you if you want to switch, why is your
chance of winning the car or winning the goat (losing the car) not 50%
at that point?
Go home and do nothing; ignore Monty. Then let him open whichever
doors he wants in whatever order he wants until he opens the door you
picked first and shows you whether you won the car or not. How come
him futzing around with the other doors changed your original 1-in-3
chance of winning and turn it into 50%?
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
.
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