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Probability mind-bender

spurgemastur
This has probably been done before (and maybe even by me) but here goes anyway.

Suppose I make a game where you are to be shown three doors and behind one of those doors is some groovy prize. The game goes as follows: you'll indicate your choice of door and then I'll show you one of the other two doors that contains no prize. At that point, you may change your mind and choose the other door, or not.

Example: you choose door #1. I show you that door #2 contains no prize and give you the option of sticking with door #1 or switching to door #3.

What is the probability for you to choose the door with the prize if you change your mind? What is the probability for you to choose the door with the prize if you don't change your mind?

NOTE: The game is rigged so that whatever door you choose, you will ALWAYS be shown an empty door and given the option to change your mind.
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Comments

6 comments

Date Votes
  • reloader44mag
    Who do you think I am, MONTY HALL?
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  • Sith Lord
    50/50 chance.....I guess?
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  • Flyin_Paulie
    50/50
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  • Colt Super
    Ditto.

    D.
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  • callktulu
    I've forgotten how to prove it mathematically, but it can be done. Your odds of getting the prize are NOT 50/50, but 66/33. If you change to door 3 AFTER you are shown one of the bad prize behind Door 2, you increase your odds to 2/3. As long as there are 3 doors, your chances are 1/3 for each. When you remove one of those thirds, it doesn't just disappear, that third goes to the other door.

    Again, I can't remember the formula and reasoning to show you, but we did it in college statistics.
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  • callktulu
    OK, I wasn't exactly right in how I remembered it....

    Here's one explanation:

    http://www-cse.uta.edu/~cook/ai1/lectures/applets/monty/answer.html

    And yet another:

    http://staff.utia.cas.cz/vomlel/mh-puzzle.html
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