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Tricky probability question. watch

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    There are three doors called A, B and C, one of which contains a prize that you want. you are to pick one of the doors, which you do. then, one of the doors, (not the one you picked) is eliminated. you are then asked if you wish to keep the door you originally picked, or to switch to the other door left. to maximize your chances of winning, what do you do, and why?
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    This is the Monty Hall Problem.

    You're best off changing, because there was a 1 in 3 chance of you being right initially (in which cases you should stick), but there is a 2 in 3 chance that you were wrong and that the prize is behind the remaining door.

    There's tons of this around on the internet.
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    dont pick any. because if you pick one door, the prob of u winning may go become 0. if u dont pick any door, it stays as 1/3.
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    Lol this is like philosophy not math!
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    This is maths at it's best! I love weird maths questions such as finding the possibilty of a coin landing on it's eadge, not head or tails, lol!

    Thanks for this. What should I search for on the net?
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    definitely switch.
    think of an extreme case, for example, you have to guess from an inverted pack of cards which is the King of Diamonds. after you pick your card, the rest of the pack except one are got rid of. to maximise your chances, would you keep the card you chose or swap it for the one remainding from the rest of the pack?
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    (Original post by MRLX69)
    Thanks for this. What should I search for on the net?
    Google "Monty Hall Problem" and "Marilyn Savant", as her explanation of it in a popular science mag produced much controversy - though she was right, she was lambasted by many people (some with PhDs) criticising the damage hre misinformation was doing.
 
 
 
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