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

Author: Joachim Rang

Date: 13:31:55 09/25/02

Go up one level in this thread


On September 25, 2002 at 15:46:14, Rolf Tueschen wrote:

>On September 25, 2002 at 14:35:29, Joachim Rang wrote:
>
>>On September 25, 2002 at 12:38:06, Rolf Tueschen wrote:
>>
>>>Please take a look at my revolutionary solution of this confusing problem:
>>>
>>>http://hometown.aol.de/rolftueschen/monty.html
>>>
>>>
>>>At first I went into the net and collected all sort of data for my page. I
>>>wanted to show how important methods and methodology are for science and also
>>>statistics. In special the exact defining of the terms.
>>>
>>>Then suddenly I had the inspiration and in a few minutes whitewashed a million
>>>people who as pupils, students or even professors let them be proved wrong by
>>>Marilyn vos Savant who has an IQ of 228. For decades now the Monty Hall Problem
>>>is taken as example for conditioned probability, which is wrong!
>>>
>>>Hope you enjoy my revelations. Please tell me if you want to comment.
>>>
>>>Rolf Tueschen
>>
>>hm, I didn't read all your stuff (its simple too much), but if I understand you
>>correctly, you claim, that the probability is 50% in both cases (switch or
>>stick). Right?
>>
>>Than you're wrong ;-)
>>
>>Only a simple note:
>>
>>you wrote: the help of the host....(There is no help - Rolf Tueschen)
>>
>>actually there is help. Because the host can not choose to open a door _before_
>>you made your choice. He has to wait, which door you choose and than to open
>>from the left two doors the wrong one. This condition you may interpret as help
>>from the host.
>
>I like your reasoning. But it can't succeed. I am sure you saw that I already
>accepted that - sure - the host "helped" to bring the situation from 1/3 to 1/2.
>But unfortunately he didn't help more. But I'm open for explanations. Let me ask
>the following: Are you aware of the difference between a unique situation and
>the general question about the general probability in the long run? Because I do
>not deny that say a group of hundred people as a group have more wins if they
>switch! But the problem we have here, how you want to prove the increase above
>1/2 for a single unique case. I think that this is the crucial point of the
>whole problem. And I'm sure that all the experts who opposed Marilyn vos Savant
>at the beginning did it because they knew that for the particular case
>conditioned probability could not help. But then they were influenced by the
>rich vocabulary of the smart woman.
>
>Rolf Tueschen


Okay, let's try:

Assumptions:

1. There are 3 doors, each with a winning probability of 1/3
2. The host has to open a "wrong" door.

Setting 1:

The host anounces which door he will open _before_ you make your first choice.
Because he has to open a "wrong" door, after that the chances of the two
remaining doors are 1/2

Setting 2:

The host has to open a door _after_ your first choice. If you choose a "wrong"
door the host is _forced_ to open the only remaining "wrong" door. This changes
the setting similiar to one, when you have to choose between three doors with a
probability of 2/3.

Okay I can't explain it scientifically correct, but the "mystic" lays in the
dependency of your first choice and the second of the host.



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