Author: Rolf Tueschen
Date: 05:09:13 09/28/02
In a discussion with Bruce on CTF I finally discovered the truth about the
irritating contradiction between the overall solution of 2/3 in case you have
only one or two trials. The candidate in a game show has only one single trial.
He can not profit of the "overall" probability. He has a win or lose situation
with no advantages of one side. It's notinteresting what the chances are from
three trial on upwards. He must make his unique choice.
But read the little discussion:
==========================================
Subject: Re: Mathematically proving the Monty Hall *RIGHT*
Posted by Rolf Tueschen (Profile) on September 28, 2002 at 07:32:01:
In Reply to: Re: Mathematically proving the Monty Hall *RIGHT* posted by Bruce
Moreland on September 27, 2002 at 21:33:42:
On September 27, 2002 at 21:33:42, Bruce Moreland wrote:
>On September 27, 2002 at 17:54:43, Rolf Tueschen wrote:
>
>>Since you are sure that you presented a proof here, let me doubt it. After the
>>first 1/3 you get 1/2 for the final decision - if and only if you think of one
>>single trial. Not two or three or 10000. Now could you show why it should be 2/3
>>for one trial?
>>
>>How will you prove that I can't have the car already? You know that I could have
>>it in 33% (not zero %) of the cases. Will you really prove a probability for a
>>sinular event? And you would _always_ switch?? No one in the show there should
>>stick? Excuse me but that is not correct.
>>
>>Rolf Tueschen
>
>*Why* is it 1/2? It's never 1/2, unless you are considering that 1/2 means that
>you can either win or lose.
>
>If an infinite number of people play this one time, and they all switch doors,
>2/3 of them will win.
>
>bruce
I'm sure that you will understand why I have a different position.
I concentrate on your wording above.
° What matters alone? Correct. That you as candidate win the car or lose, ok?
° A little intermediate question:
of 2 doors with 1 car only, how's about the odds? 50% for each door.
° Now we could invent a lot of finite or infinite situations.
We enter probability.
With a n>2 you come to the famous 2/3 chance. Bruce Simulation.
° Bruce, now try to concentrate further on this question:
You want a car or the 2/3 chance for a car? Remember you have 1 (one!) trial.
I agreed already, if you had more trials, then in the long run you win 2x
with "switch".
° What happens in 1 single case, if you think of the long-range 2/3?
° Gives it a certainty to get the car by switching? Answer? No! Ok?
° 33 versus 67. 3,3 versus 6,7 in 10 trials.
° What happens with our formula for 1 trial?? => 0,33 versus 0,67 => no solution
° Short discovery: You must at least run 2 trials to get for switch a solution
0,66 versus 1,34. 0,99 or 1 versus 2,01 or 2 for 3 trials. That's it.
° Conclusion: If you had three times the choice then you switch always and win
2 of the 3 cars.
° Back to 1 trial! Have you understood that it's not defined?
° Therefore you can either choose A or B and you will either win or lose.
° The paradoxy comes from the impossibility to define "always" for 1 trial.
° Your idea of several "1-trial" cases can't help the single candidate.
I sum up.
The single candidate has no clue or "possibly hidden advantageous" knowledge to
get the car for sure. Your and the mathematicians 'probability' does not
_guarantee_ you the win in a single trial. There you can only win or lose. But
this is 50% situation.
QED
Hope this helps now.
Add. In maths you have several of such paradoxies. You are not allowed to divide
by zero. You may remember the shit when you found a beautiful result and teacher
said oh no, that is wrong, look here, what if x is zero? Shit.
NB that all this is only ok for the situation of three doors. If you have one
million doors by definition you can't get the luck of having chosen the car at
first, so it's behind the other outsearched door - with near certainty.
(Yesterday I was a bit "unconcentrated" and thought that also here the 50% would
be correct which is wrong. See my confession in CCC.)
I haven't made the thread because I wanted to demonstrate how smart I were but
to show all in CChow difficult it can be if you pay not enough attention to the
definitions of your problem and the variables you want to examine. I think it
succeeded well included my own fault.
Hope you will find yourself not too much taught nevertheless. Thanks for the
patience and all contributions.
Rolf Tueschen
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