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Subject: Re: New and final solution of the Monty Hall Dilemma

Author: Rolf Tueschen

Date: 04:59:32 09/28/02

Go up one level in this thread


On September 27, 2002 at 17:55:21, Antonio Dieguez wrote:

>On September 27, 2002 at 11:27:15, Rolf Tueschen wrote:
>
>>On September 27, 2002 at 11:18:44, Peter Berger wrote:
>>
>>>On September 27, 2002 at 11:10:52, Uri Blass wrote:
>>>
>>>>On September 27, 2002 at 11:03:56, Peter Berger wrote:
>>>>
>>>>>On September 27, 2002 at 10:43:02, Rolf Tueschen wrote:
>>>>>
>>>>>>Hehe. I want to state that I will prevent if we have here a united flight to the
>>>>>>moon tonight. I am the responsible and I will prevent such cases of despair.
>>>>>>
>>>>>>No, I can't accept. Formerly you had, in your example, one door with 1/1000000
>>>>>>and 999999 doors with 999999/1000000. Now you open, means take away 998999
>>>>>>doors, right? Then you have two doors, right? With - now - each 1/2.
>>>>>>
>>>>>>You know what I mean? It's not magic. The 998999 doors are away. So there is no
>>>>>>talking about such incredible chances.
>>>>>>
>>>>>>You see where your mistake was?
>>>>>>
>>>>>>Good weekend
>>>>>>
>>>>>>Rolf Tueschen
>>>>>>
>>>>>
>>>>>Come on - what is so difficult to understand here?
>>>>>Maybe 1.000.000 is too high a number and maybe cars confuse people - let's take
>>>>>a cookie and 10 cups.
>>>>>
>>>>>You know there is exactly one cookie and it is in one of the cups. You win when
>>>>>you find it.
>>>>>
>>>>>You make your first guess: cookie is in cup 1.
>>>>>
>>>>>OK, says the Monty guy - now I take away 8 loser cups from the 9 you didn't
>>>>>choose (either he knows or he is just lucky, doesn't matter at all).
>>>>
>>>>It is important
>>>>suppose for the discussion that he does not know and take cups 3-10
>>>>
>>>>There are 10 cases with probability of 1/10 for the place of the winner cup.
>>>>
>>>>In 8 out of 10 cases there is no game(because the winner cup is not in 1 or 2.
>>>>
>>>>In 2 out of 10 cases he really take 8 loser cups.
>>>>In 1 of these 2 cases the cup is in 1 and in the second case the cup is in 2
>>>>
>>>>The 2 cases have the same probability so the probability is 1/2.
>>>>
>>>>Hope my explanation is clear.
>>>>
>>>>Uri
>>>>
>>>
>>>Yes, it is. But this is not the game that is being played - it's all about the
>>>Monty taking away 8 cups and never failing to find the losers which is the basic
>>>Monty setup :).
>>>
>>>Your first cup's chances never improve at all - you are still with the 1/10
>>>chance you started with.
>>
>>
>>And this is exactly the delusion in the whole thing. You are - like many others,
>>perhaps most - thinking in the two camps theory of Marilyn vos Savant. But that
>>is nonsense for the event here with just 1 trial and nothing more. Look and
>>believe me at least a bit, say the candidate were on the toilette after his
>>first choice and then he came back after the host took away the 8 cups, ok? Just
>>relax, Peter!
>>
>>Now I ask you, what should the candidate do? He doesn't even remembers what cup
>>he has chosen a while ago!!
>>
>>How much chances he has? Between the two cups? Even if the host told him what
>>were his first cup? Does it really matters when the candidate can watch what the
>>hostis doing when he's unable to understand what the host has in mind?????
>>
>>Ahhhhhh, I'm tired now, but this was my last effort.
>
>It's incredible this thread is so large because you putting so much effort in
>your error.
>
>Let me try to convice you too.
>
>There are two doors.
>
>You are not choosing a door.
>You are choosing if you change the door you already have.
>
>If the door you already have is A then you lose.
>If the door you already have is B then you win.
>
>So, you must see wich is the probability of you having already a door type A or
>B.
>
>that is all.
>
>pd: it doesn't matter if the guy doesn't know probabilities or goes to the
>toilet. Here we can tell him that he should switch :)

You're right for either

- trials  n>3 in the original 3-doors problem

and

- when N of doors gets higher and higher than 3 in either 1 trial or more


Explanation: for 3-doors and 1 or 2 trials there is no defined probability.
 (I will quote here the explanation I wrote to Bruce in CTF.)

Rolf Tueschen



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