Re: Oh no! Monty Hall problem again.....
From: No Way (Not_at_real.com)
Date: 09/19/04
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Date: Mon, 20 Sep 2004 01:39:19 +0200
On 19 Sep 2004 13:46:13 -0700, clintonz@prodigy.net (Clinton C
Zimmerman) wrote:
>You talk the same way as the original poster. Are you the
>same?
No, but I do copy-paste. :)
>> If he chooses the chicken, Monty will show the goat. The contestant
>> switches and ends up by the car. (win)
>> If he chooses the goat, Monty will show the chicken. The contestant
>> switches and ends up by the car. (win)
>> If he chooses the car, Monty will show either the chicken or the goat.
>> Monty has no preference for either animal in this situation. The
>> contestant switches and ends up with the animal that wasn't shown.
>> (loss)
>
>The last is two cases because either the door with the goat or
>the door chicken could be opened. Just because the probability
>of each of these two cases is less does meant they are not separate
>and distinct events. We are getting bogged down in terminology here
>and I'm not sure if I understand you 100%
You initially pick the chicken and win. (one)
You initially pick the goat and win. (two)
You initially pick the car and lose. (three)
However, if you truly wish to split up the last case, be my guest
(although it only complicates things).
>> >If after Monty opens a door and the contestant says "I would like to
>> >change
>> >my mind, I'm not going to switch after all", the chances of success
>> >cannot be determined without knowing the conditional probability
>> >of Monty opening the door given the car is behind the firt chosen
>> >door.
>>
>> This is also wrong.
>
>What do you mean it is wrong?? Why? It is completely correct.
p = probabilty of car behind door 1.
q = probabilty of car behind door 2.
1-p-q = probabilty of car behind door 3.
The contestant randomly picks a door (which is his best strategy,
seeing he knows nothing about p and q). The probability that he picks
the door with the car behind it is 1/3. If he doesn't switch, he has a
chance of 1/3 of winning the car.
>Its undefine
No.
>> >Do you mean if Conan makes the same decision as the original player
>> >would have made? Switch?
>>
>> The OP states quite clearly that Conan does not have that information.
>
>Then how do you know that Conan will choose each door with 1/2 probability?
It is the best strategy he can take given the information he has. Of
course, Conan is a barbarian, which makes it a bit harder. :)
However, one of those things that is unknown about Conan is that he
has taken a probability course (to have a greater chance of ducking
arrows). He will take the best strategy.
>He could choose door 1 100% of time.
Yes, but that wouldn't be smart.
m = probability that car is behind left door.
1-m = car behind right door.
s = probability that Conan will pick left door.
1-s = Conan picks right door.
The chance that Conan wins is (if the game ends directly after he
chooses a door):
s*m + (1-s)*(1-m) = s*m + 1 - s - m + s*m
= 1 + 2*s*m - s -m
m can be anything between 0 and 1. Suppose m=0:
1 + 2*s*m - s -m = 1-s
So for Conan s=0 will be optimal. Unfortunately for Conan m might as
well be 1. In that case s=0 will be terrible. The fact that m is
unknown to Conan, leads Conan to the conclusion that he has to pick s
in such a way that his chance of winning is maximised in any
situation. That happens for s=1/2.
1 + 2*s*m - s -m = 1+m-1/2-m = 1/2
> You can't even define the chance
>of success for Conan.
I can't? I just did. Conan's succes rate is 1/2.
> Your talking gibberish. The
>fact that Conan winns 50% of the time by choosing randomly results from
>the fact that when he choose door 1 he loses more than average and when
>he chooses door2 he wins more than average. That does not mean that the
>probability for each door is 1/2.
I didn't say that the probability for each door was 1/2. I said it was
m for the one door and (1-m) for the other door.
> If you take two tests and score 25%
>on the first and 75% on the second your average score is 50%. That does
>not mean your testing percentage for EACH test was 50%. You cant take
>the average of two distinct event and then conlude that the probability
>of each event is the average. That is Bizzare.
Conan will win 1/2 the time. I don't claim anything else. Your
'analogy' isn't a proper one at all.
>Yes, Marylin made this same mistake with her aliens argument. She was
>wrong. When the aliens land after a door is eliminated their chance
>of succes chosing randomly is 50% just like the chances of success for
>the contestant would be if he had originally decided to switch 50%
>of the time before the door was removed or after the door was removed.
>But the chances of success for the "switch door" does not change for the
>aliens just because they did not know through mental telepathy what
>Monty was "thinking".
What is the chance of success for a door? And why is it something that
is connected to the door? Won't the original contestant, Conan/aliens
and Monty all have a different view of what the chance for success of
a door is? Please explain what you are thinking as to me you only seem
confused.
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