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22.
 
Re: Monty Hall
Dec 4, 2006, 17:34
Enahs
 
22.
Re: Monty Hall Dec 4, 2006, 17:34
Dec 4, 2006, 17:34
 Enahs
 
You have a 50% chance to get the car. Because the host always opens the door WITHOUT a car, that door is completely removed from the equation. Just look at it.

No matter what door you pick first, he opens a door. That door will never have the car. The door he opens is removed from the equation. He asks if you want to change doors. He is asking you to pick one out of two doors. One of them has a car. One of them doesn't. 1/2 the time you will win, 1/2 the time you will lose. Whatever door you picked first is completely irrelevant because he will always remove a wrong choice after you pick your door. These are not sequential probabilities; they are independent. Your first choice has *zero* impact on the second choice. So instead of it being a [1/3 2/3] problem you have a [[1/3 1/3 1/3] problem. With one wrong answer removed, it is no longer in thirds. He asks you again: which door do you want? There are now two doors, not three, to choose from. One of two doors has a car, one does not. Your first choice doesn't matter at *all*, because whether or not you chose correctly, he will remove a door WITHOUT the car from the equation.

You are wrong. It might seem that way, but it is not. I understand why you think that, but it is really wrong. It is what most people with out any knowledge of statistics think (and that is fine).

To do it the way the article did, but put it in English.

Doors are labeled A,B and C.
Case 1: The car is behind door A
Case 2: The car is behind door B
Case 3: The car is behind door C

Let us say that in every case you pick door A.

In case 1 Monty could open either B or C and you would win by staying with your answer.

In case 2 Monty must open door C, so you could win by switching to door B.

In case 3 Monty must open door B, so you could win by switching to door C.

Thus, in two of three equally likely cases, you would win by switching from A to the other door. In only one of the three equally likely cases would you win by staying with your original choice.

You therefor have a more statistical and probable chance of winning, if you switch. Most people think it is 50/50, it is not. If this was a one time event (irrelevant if you only get one chance), it would be 50/50. But it is not. That is how statistics and probability works. It might seem counter intuitive, but it is the truth.

But this question is over simplified and flawed, as there are other factors that must statistically be taken into account (Like for instance in this case, the majority of the time Monty only offered people the chance to switch when they picked the correct door first. There is a statistical value you must calculate for when Monty offered the chance when the person was right/wrong, and factor that into the previous values). There are also many other things. This article is just about math, not true statistics.


*edit*
Here, go here and try it your self to prove to your self:
http://math.ucsd.edu/~crypto/Monty/monty.html
Do it 10 times by keeping the same door, and 10 times by switching and see which way you win more.

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This comment was edited on Dec 4, 17:40.
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  Re: Monty Hall
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