Author: Rolf Tueschen
Date: 15:51:28 09/25/02
Go up one level in this thread
On September 25, 2002 at 17:50:46, Roy Eassa wrote: >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. Roy, that is perfectly ok in the long run but not for trials n=1. Just take a look into the table at the bottom of my Monty-page. There you can see that trials begin with n=9. Result: already zero in the confidence interval for no switches. Go lower and you get negative signs. :) Rolf Tueschen
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