Re: Oh no! Monty Hall problem again.....

From: Clinton C Zimmerman (clintonz_at_prodigy.net)
Date: 09/19/04 

Date: 19 Sep 2004 13:46:13 -0700

> >
> >
> >Maybe I read this wrong but how did you get three subcases out
> >of "changes decision" ( doesn't stay?). There are four.
> >Chooses one of the remaining doors wins or loses (2 cases)
> >chooses other remaining door and wins or loses (2 cases)
>
> No, there are not four options. This is the Monty Hall problem. After
> the contestant has made an initial choice, an unpicked door with a
> goat behind it is opened. There are thus three options. Perhaps this
> is more easily seen if there is a chicken and a goat (instead of two
> goats). After the initial choice, a door with a chicken or a goat
> behind it is opened.
>
> 1st subcase: He chooses the chicken at the beginning.
> 2nd subcase: He chooses the goat at the beginning.
> 3rd subcase: He chooses the car at the beginning.
>

You talk the same way as the original poster. Are you the
same?

> If he chooses the chicken, Monty will show the goat. The contestant
> switches and ends up by the car. (win)
> If he chooses the goat, Monty will show the chicken. The contestant
> switches and ends up by the car. (win)
> If he chooses the car, Monty will show either the chicken or the goat.
> Monty has no preference for either animal in this situation. The
> contestant switches and ends up with the animal that wasn't shown.
> (loss)

The last is two cases because either the door with the goat or
the door chicken could be opened. Just because the probability
of each of these two cases is less does meant they are not separate
and distinct events. We are getting bogged down in terminology here
and I'm not sure if I understand you 100%

>
> >If after Monty opens a door and the contestant says "I would like to
> >change
> >my mind, I'm not going to switch after all", the chances of success
> >cannot be determined without knowing the conditional probability
> >of Monty opening the door given the car is behind the firt chosen
> >door.
>
> This is also wrong.

What do you mean it is wrong?? Why? It is completely correct.

>
Its undefine
>
> >Do you mean if Conan makes the same decision as the original player
> >would have made? Switch?
>
> The OP states quite clearly that Conan does not have that information.

Then how do you know that Conan will choose each door with 1/2 probability?
He could choose door 1 100% of time. You can't even define the chance
of success for Conan. Your talking gibberish. The
fact that Conan winns 50% of the time by choosing randomly results from
the fact that when he choose door 1 he loses more than average and when
he chooses door2 he wins more than average. That does not mean that the
probability for each door is 1/2. If you take two tests and score 25%
on the first and 75% on the second your average score is 50%. That does
not mean your testing percentage for EACH test was 50%. You cant take
the average of two distinct event and then conlude that the probability
of each event is the average. That is Bizzare.

Yes, Marylin made this same mistake with her aliens argument. She was
wrong. When the aliens land after a door is eliminated their chance
of succes chosing randomly is 50% just like the chances of success for
the contestant would be if he had originally decided to switch 50%
of the time before the door was removed or after the door was removed.
But the chances of success for the "switch door" does not change for the
aliens just because they did not know through mental telepathy what
Monty was "thinking".

Relevant Pages

  • Re: Monty Hall problem
    ... Monty only open a door with a goat if the original choice was correct? ... I think *the player* must know, to get the usual 2/3 prob. ...
    (sci.math)
  • Re: Monty Hall problem
    ... Monty only open a door with a goat if the original choice was correct? ... I think *the player* must know, to get the usual 2/3 prob. ...
    (sci.math)
  • Re: Monty Hall problem
    ... Monty only open a door with a goat if the original choice was correct? ... I was thinking about "What if Monty lied and didn't know where the car ...
    (sci.math)
  • Re: James P. Hogan the left-winger
    ... This time Monty did open a door. ... or not he opens doors for other contestants, is just an easy way to ... a goat, B: Monty opens a door with a goat if and only if you chose the ...
    (rec.arts.sf.written)
  • Re: Monty Hall problem wrong?
    ... In the "classic" version of the Monty Hall problem, ... behind one is a car, behind each of the other two is a goat. ... After you select one door, ... Monty knows the car is not behind the door he opens. ...
    (sci.math)