Computer Chess Club Archives


Search

Terms

Messages

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

Author: Roy Eassa

Date: 14:33:00 09/25/02

Go up one level in this thread


On September 25, 2002 at 16:31:55, Joachim Rang wrote:

>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.


OK, I'll take my shot at wording the explanation:


"If n is the starting number of doors...

The door you choose always has (from your perspective) a 1/n chance of being the
winner.  All the unopened doors COMBINED (assuming there's at least one) always
have a (n-1)/n chance.  If Monty opens one of those doors that he KNOWS is not a
winner, that does not change the fact that all the (remaining) unopened doors
combined have a (n-1)/n chance.  But since there are now fewer of them among
which to divide it, each INDIVIDUAL unopened door's chance is (from your
perspective) higher than before.

Thus in the 3-door example, your door has a 1/3 chance and the one remaining
unopened door has a 2/3 chance of being the winner."




This page took 0 seconds to execute

Last modified: Thu, 15 Apr 21 08:11:13 -0700

Current Computer Chess Club Forums at Talkchess. This site by Sean Mintz.