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    Game Show Problem (From the movie 21)

    Suppose you’re on a game show and you’re given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you “Do you want to switch to Door Number 2?” Is it to your advantage to change your choice?

    Discuss.


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    Re: Game Show Problem (From the movie 21)

    If you're playing the odds you'll change. If you selected the car with your door and change you lose. But if you selected one of the two goats and change you win, so the odds say switch.



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    Re: Game Show Problem (From the movie 21)

    I've never been any good at statistics, but if IIRC, you should change doors, as the odds dramatically increase that the car is the door you did not pick


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    Re: Game Show Problem (From the movie 21)

    I would think it is in your advantage to switch because at the beginning, the player had a 1 in 3 of picking the car. After the goat was revealed, if the player picks again, he has a 1 in 2 of picking the car.



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    Re: Game Show Problem (From the movie 21)

    switch.




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    Re: Game Show Problem (From the movie 21)

    The odds always say switch. Your first selection had a 33% chance of being the car. If you change your selection after the door is open, you have a 50% chance of being a car.


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    Re: Game Show Problem (From the movie 21)

    Quote Originally Posted by CyinCo View Post
    The odds always say switch. Your first selection had a 33% chance of being the car. If you change your selection after the door is open, you have a 50% chance of being a car.
    Actually it'd be a 66% chance of being a car I believe. You're original selection still has a 33% chance, the other 66% has to be on the other door since we've ruled one out already.



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    Re: Game Show Problem (From the movie 21)

    [ame=http://www.youtube.com/watch?v=CcG8z2OTAWs]YouTube - The Simpsons - Mystery Box[/ame]



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    Re: Game Show Problem (From the movie 21)

    Quote Originally Posted by richey24 View Post
    Actually it'd be a 66% chance of being a car I believe. You're original selection still has a 33% chance, the other 66% has to be on the other door since we've ruled one out already.
    I believe you are correct. 66%, and not 50%.

    edit: Hmmm...Now I'm not sure about the 66%. Either way, you have better odds to switch because your original selection was from 3 doors. Now, when you switch, you are picking from two and not 3. Your odds got better.


    Last edited by CyinCo; 08-10-2009 at 10:19 AM.
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    Re: Game Show Problem (From the movie 21)

    Makes sense to me. If you choose the left and you konw there is a 67% chance of it being one of the other two and then you know one of them is not. Then there is still 67% chance of it not being the left one because of the original 33% on the left. My friend (a math major) explained it to me. So I would switch.


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    Re: Game Show Problem (From the movie 21)

    Numb3rs explained this really well with goats and a car.



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    Re: Game Show Problem (From the movie 21)

    Quote Originally Posted by cyismydog View Post
    Numb3rs explained this really well with goats and a car.
    Everything can be explained with goats and car.


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    Re: Game Show Problem (From the movie 21)

    So if you stay with your first pick wouldn't you in fact still be at 50%? If only doors 1 and 2 are left you are in fact picking to stay with door 1. Once door 3 is revealed you have a 50/50 shot no matter which door you choose in the second round of choices. I guess I thought of it as two distinct rounds and the first round has nothing to do with the second.


    Last edited by Clonehomer; 08-10-2009 at 10:31 AM.

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    Re: Game Show Problem (From the movie 21)

    Quote Originally Posted by cyfan964 View Post
    Suppose you’re on a game show and you’re given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you “Do you want to switch to Door Number 2?” Is it to your advantage to change your choice?

    Discuss.
    The normal rule is switch your choice. However, depending on what you mean by the bolded sentence above, the normal rule may not apply in this scenario. In other words, I think the "switch" approach depends on nobody having knowledge of what is behind any of the doors.

    Here, if the gameshow host has knowledge, then wouldn't he choose the door with the car to ensure that the contestant goes away with as little as possible? If he chooses a goat door, that implies that goats are behind both of the "unchosen" doors.



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    Re: Game Show Problem (From the movie 21)

    Quote Originally Posted by Clonehomer View Post
    So if you stay with your first pick wouldn't you in fact still be at 50%? If only doors 1 and 2 are left you are in fact picking to stay with door 1. Once door 3 is revealed you have a 50/50 shot no matter which door you choose in the second round of choices. I guess I thought of it as two distinct rounds and the first round has nothing to do with the second.
    66% is correct.

    Basically it boils down to that you had a 1/3 chance of your first pick being the car and a 2/3 chance of your first pick being one of the two goats.

    Good diagram here:

    Monty Hall problem - Wikipedia, the free encyclopedia
    Attached Thumbnails Attached Thumbnails Click image for larger version. 

Name:	montyhall.jpg 
Views:	80 
Size:	60.1 KB 
ID:	5645  


    Last edited by RayShimley; 08-10-2009 at 10:30 AM.


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