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I cannot wrap my head around the Monty Hall problem.

inkblotinkblot
edited 2011-07-25 03:38:33 in General
000
:(

Comments

  • DynamodDynamod

    I read the wikipedia article, and the way they explain it with cards does it the most justice. suppose you have three cards. on is an ace of spades, and two are a three of clubs. the dealer shuffles them and gives you one card. you are not allows to look at it.


     your odds of having the ace: 1/3


     dealers odds of having the ace: 2/3


    then, the dealer looks at his hand and discards one of the three of clubs. this in no way affects the odds of him having the ace. but no matter what, he will always discard a three.


    your odds of having the ace: 1/3


    dealers odds of having the ace: 2/3


    since at the beginning of the game the dealer had twice as many cards at you, and gets to look at his hand and discard useless cards, he will always be twice as likely to have the ace.

  • MagusMagus
    I'LL STAY MAI HAUNDS...WITH YAU BLAHT
    Oh, thanks! I finally understand it!
  • Central AvenueCentral Avenue
    Wow, that actually makes sense now.
  • ClockworkUniverseClockworkUniverse
    OOOooooOoOoOOoo, I'm a ghoOooOooOOOost!
    The thing that always bothered me is: does it have any application outside game shows?
  • DynamodDynamod

    Not sure. it seems like it might be good to people whom study psychology, but I'm not sure at all.

  • Wicked223Wicked223
    We Played Some Open Chords and Rejoiced, For the Earth Had Circled the Sun Yet Another Year
    There have been several in-depth theses about it written by probability theorists, so I imagine it does.
  • inkblotinkblot
    000
    So what you're saying is that the dealer can only lose if he gave you the ace?
  • ClockworkUniverseClockworkUniverse
    OOOooooOoOoOOoo, I'm a ghoOooOooOOOost!
    Googling "monty hall problem applications" and "monty hall problem uses" is just giving me stuff about the implications of the fact that people get it wrong when first presented with it.
  • TheMightyAnonymTheMightyAnonym
    ~♥YES♥~! I *AM* a ~♥cupcake♥~! ^_^
    I figured it out about halfway through a test. At first I was ignoring that I had already heard the problem, and brushed it off as a semantics related thing because of how it would yield higher probability by choosing again or something dumb like that, but then I realized that a non-winning card MUST be revealed, and found the trick.

    Also:     http://www.philosophyexperiments.com/montyhall/images/monty002.jpg
  • NyktosNyktos
    a little muffled
    Yeah, a visual depiction like that is usually the best way to explain it to people who don't get it at first.
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