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Thread: Monty Hall Dilemma - Winning a GTI on a Game Show

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    Monty Hall Dilemma - Winning a GTI on a Game Show

    If you're ever on game show with, say, a GTI as a prize, you may want to know about this . . .

    Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a GTI; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to switch from door No. 1 to door No. 2?" Is it to your advantage to switch your choice?

    As the player cannot be certain which of the two remaining unopened doors is the winning door, most people assume that each of these doors has an equal probability and conclude that switching does not matter.

    In fact, all things being equal, the player should switch—doing so doubles the probability of winning the car, from 1/3 to 2/3.

    Worth knowing if you ever find yourself in the enviable position of being in the running to win a car!

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    Quote Originally Posted by Dubya View Post
    If you're ever on game show with, say, a GTI as a prize, you may want to know about this . . .

    Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a GTI; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to switch from door No. 1 to door No. 2?" Is it to your advantage to switch your choice?

    As the player cannot be certain which of the two remaining unopened doors is the winning door, most people assume that each of these doors has an equal probability and conclude that switching does not matter.

    In fact, all things being equal, the player should switch—doing so doubles the probability of winning the car, from 1/3 to 2/3.

    Worth knowing if you ever find yourself in the enviable position of being in the running to win a car!
    Schroedinger's cat could apply though.

    Your odds will be 1/2 after one of the doors is opened no matter if you stay with your original choice or go with the new door.

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    Quote Originally Posted by team_v View Post
    Schroedinger's cat could apply though.

    Your odds will be 1/2 after one of the doors is opened no matter if you stay with your original choice or go with the new door.
    Apparently not:

    http://en.wikipedia.org/wiki/Monty_Hall_problem

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    Quote Originally Posted by team_v View Post
    Your odds will be 1/2 after one of the doors is opened no matter if you stay with your original choice or go with the new door.
    Exactly correct.

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    Quote Originally Posted by team_v View Post
    Schroedinger's cat could apply though.

    Your odds will be 1/2 after one of the doors is opened no matter if you stay with your original choice or go with the new door.
    wrong.
    combinations are:
    (1) (2) (3)
    goat goat car

    goat car goat

    car goat goat

    you pick door 1, chances of car, 1 in 3

    the fact that the host shows you a goat doesnt change that.

    you double your chance of a car by swapping.

    I proved this to my sister using cards. She still thinks it's magic.
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    Quote Originally Posted by cme2c View Post
    wrong.
    combinations are:
    (1) (2) (3)
    goat goat car

    goat car goat

    car goat goat

    you pick door 1, chances of car, 1 in 3

    the fact that the host shows you a goat doesnt change that.

    you double your chance of a car by swapping.

    I proved this to my sister using cards. She still thinks it's magic.
    In reality though, you still only have a 50:50 chance of having the car behind the door you chose or the door you didn't choose becasue it has to be behind one of them as the other door revealed a goat.


    However, schroedingers cat apply's because you never know if you changing doors will change what was behind the doors.
    Last edited by team_v; 10-03-2010 at 12:41 PM.

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    Quote Originally Posted by team_v View Post
    In reality though, you still only have a 50:50 chance of having hte car behind the door you chose or the door you didn't choose becasue it has to be behind one of them.

    However, schroedingers cat apply's because you never know if you changing doors will change what was behind the doors.
    Not so, I have to say:

    From the outset you have a one in three chance of selecting the door with the car first off.

    This means that there is a 67% chance that the car is behind one of the two doors you did not choose and a 33% chance it is behind the one you first chose.

    These percentages do no vary after the door with the goat behind it is opened.

    We already knew there was a goat behind one of the two doors you did not select; now we just happen to know behind which of the two other doors the goat was hidden.

    As the two doors we did not choose initially still have a combined 67% chance of having the car behind them, with the knowledge that the GTI is not behind the door that has been opened, if you swap, your chances of the car being behind your second choice are 67%, not 33% or 50%.

    So if you swap, on average, you will win twice as often.
    Last edited by Dubya; 10-03-2010 at 12:50 PM.

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    the problem here is that each time time one door is eliminated, you start with a fresh set of variables and probabilities.

    so at first choice, your chance was 1 in 3, but once the first door is out of the equation, then its 50:50 for both doors, whether you change or not.

    Like tossing a coin, you have a 50:50 chance of getting a tails, but just becuase the first toss revealed a head, doesnt mean the next toss will reveal a tail.

    the next toss is still 50:50 for a tail vs a head.

    Probabilities and 50:50 outcomes are only realised over a very large sample.

    also, and more importantly accept that Murphys Law is the more powerful law stating that which ever door you choose will have a goat behind it
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    Quote Originally Posted by gareth_oau View Post
    the problem here is that each time time one door is eliminated, you start with a fresh set of variables and probabilities.

    so at first choice, your chance was 1 in 3, but once the first door is out of the equation, then its 50:50 for both doors, whether you change or not.

    Like tossing a coin, you have a 50:50 chance of getting a tails, but just becuase the first toss revealed a head, doesnt mean the next toss will reveal a tail.

    the next toss is still 50:50 for a tail vs a head.

    Probabilities and 50:50 outcomes are only realised over a very large sample.

    also, and more importantly accept that Murphys Law is the more powerful law stating that which ever door you choose will have a goat behind it
    Out of respect for the preferences of others, I make no comment as to whether the above comment is correct or not but refer the author to the discussion at:

    http://en.wikipedia.org/wiki/Monty_Hall_problem

    PS - I do not wish to imply that I have any issue with the comment about "Murphy's Law" except to say that, in my opinion, Murphy was an optimist!

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