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Thread: 98%: A Peculiar Probability Problem

  1. #1

    98%: A Peculiar Probability Problem

    Consider 3 doors. Behind one door is a prize (a 'Let's Make a Deal' scenario) and behind the other two is nothing. Naturally, you wish to claim the prize. You are told to select a door. After selecting a door, say door number one, someone who knows where the prize is tells you it is not behind door number two and there is no chance he is attempting to deceive you. He then tells you that you may stick with your initial choice of door number one or switch your decision to door number three. Do you stick or switch? Does it make a difference at all? Why?

    My friend swears up and down that it is correct to switch but he has, as yet, been unable to provide me with a solid, mathematical reason. My intuition is that it makes no difference, but I am slowly convincing myself, after thinking through the situation a while, that perhaps it is correct to switch.

  2. #2
    Don Schlesinger

    Don Schlesinger: No posts on this, please

    With the conditions you stipulate, it is obvious that changing is correct, as it increases the probability of being right from the original 1/3 to a new, higher, 1/2.

    Please, folks, no posts on this here. It's been done and redone a million times.


  3. #3

    Oldster: Do not want to upset you Don, but

    I always have to draw it out, but of the 3 possibilities, 2 are winners if you change doors after Monty opens the door without the big prize.

  4. #4
    Don Schlesinger

    Don Schlesinger: Yes, of course you are right

    The easiest way for me to remember it is that if there are n doors, your first guess always gives you 1/n chance of being right, whereas, if all known wrong doors are eliminated, leaving only one more, and Monte (Monty?) "steers" you to that door, you obviously now have (n-1)/n probability of being right by switching.

    And that's my (and everyone else's!!) FINAL answer! :-)


  5. #5
    Jake { :>)

    Jake { :>): Re: Yes, of course you are right

    I think conditional probabilities is one of the concepts involved in the NRS formulas.



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