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Maths question II

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    #11
    Originally posted by suityou01 View Post
    Bellend.
    You shouldn't talk about yourself like that
    merely at clientco for the entertainment

    Comment


      #12
      Originally posted by EternalOptimist View Post
      When you make your guess, one of the other two doors is opened. If there is a car, you lose.
      This bit intends to mislead. It suggests a random game.

      If the game is played where the first door opened can be a car, then it doesn't matter if you switch or not. Its 50/50.

      However, if the game is played so that the first door opened by Monty is never a car (because he knows where it is), then you get a 2/3rd's chance of winning by switching choice.

      2/3rds of the time you initially picked wrong and Monty has to hide the door with the car. And during that 2/3rds he always eliminates the door with no car for you.

      Do I win a prize?

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        #13
        Originally posted by xoggoth View Post
        What is with these one line negative comments?

        Don't like a post? Close it. Why waste time commenting?
        Have a ,

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          #14
          Originally posted by xoggoth View Post
          What is with these one line negative comments?

          Don't like a post? Close it. Why waste time commenting?
          because they are sad.
          Always forgive your enemies; nothing annoys them so much.

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            #15
            Here it is explained.

            Understanding the Monty Hall Problem | BetterExplained

            The point is if Monty doesn't know where the car was he would sometimes open a door where the car is. Therefore overall it would be a 50/50, presuming that you made a choice without knowing what was behind the door which Monty opened, and that sometimes he opened a door with the car behind. In the cases where he opens a door and there isn't a car and you know there isn't a car, you have a higher probability of finding the car behind the other door.

            If there were a hundred doors, you pick one and then he opens the other 98 then the chances are almost certain that the car is behind the other door, because the door you selected only had a 1/100 chance of having the car, whereas the other door has a 99/100 chance. With three doors it isn't as clear.
            Last edited by BlasterBates; 4 October 2014, 13:21.
            I'm alright Jack

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              #16
              Originally posted by BlasterBates View Post
              Here it is explained.

              Understanding the Monty Hall Problem | BetterExplained

              Yes the key is Monty is opening a door where he knows the car isn't. Therefore the other door is more likely to have the car behind it.

              If there were a hundred doors, you pick one and then he opens the other 98 then the chances are almost certain that the car is behind the other door. With three doors it isn't as clear.
              Mr C has just explained it to me using 100 doors. It IS much clearer that way.
              "I can put any old tat in my sig, put quotes around it and attribute to someone of whom I've heard, to make it sound true."
              - Voltaire/Benjamin Franklin/Anne Frank...

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                #17
                Originally posted by cojak View Post
                Mr C has just explained it to me using 100 doors. It IS much clearer that way.
                I disagree. The number of doors is irrelevant to me. The only thing that 'solved' this for me was understanding that Monty must know where the car is, and he must play properly (i.e. never opening the door with a car) to ensure the player gets the decision to switch or not.

                If Monty doesn't know, and just plays a random game, even after opening 98 of the 100 doors without revealing the car, then you may as well keep your original choice. All he would have done is increase your chances from 1/100 to 1/2.

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