Author: Roy Eassa
Date: 14:50:46 09/25/02
Go up one level in this thread
On September 25, 2002 at 17:45:03, Rolf Tueschen wrote: >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 The door you chose at the start has a 1/n chance. The fact that Monty opens some irrelevant doors (which he KNOWS are not winners) does not change the fact that your door has a 1/n chance. Thus however many doors remain closed have the remainder of the chance, i.e., (n-1)/n. QED.
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