Author: Rolf Tueschen
Date: 14:45:03 09/25/02
Go up one level in this thread
On September 25, 2002 at 17:33:00, Roy Eassa wrote: >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." Simply wrong maths. There is no "the one remaining unopened door" because there are at least two! And the candidate has only a single chance to make a choice. Now please try to explain why for such a case the two doors still closed shout get a different treatment. In the 1000 thought experiment I could make the following proposal. I return the formerly chosen - but still closed - door. Now Monty could go down until two doors were left. One must be the one with the car. The chances for the two are different??? I don't think so. 50% both. QED Rolf Tueschen
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