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A simple puzzle

  1. Jun 17, 2003 #1


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    Consider the standard "guess which one of three curtains the prize is behind" scenario. After choosing curtain 1, you're shown that nothing resides behind curtain 2. If allowed, would you switch to curtain 3?
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  3. Jun 17, 2003 #2


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    Of course!

    - Warren
  4. Jun 17, 2003 #3


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    I'm not surprised that you'd find this puzzle trivial, but you'd be surprised how many people get it wrong, and even when you explain it to them they don't believe it.

    Btw, remember that dumb stanford bookstore joke? I think the mixup occurred because you'd made the comment about the stanford bookstore when my user name was steinitz, but just before I teased you, I'd had it changed to my given name "jeff" because it was annoying being called by a last name and obviously that's not something you'd have noticed.
  5. Jun 18, 2003 #4
    First of all, my answer is just like chroot's.
    Here is the way i explain it :
    The chance of getting the right curtain from the first try is 1/3, after the second curtain is opened the chance of the prize being behin the third curtain will become 2/3, so i would choose the biggest chance.
    2-Trial and error:
    Write all the possibilites, you will see that if you change your mind about your curtain decision, you chance of winning will be the double of if you didn't (note that (1/3)*2=2/3 )
    Make the example bigger, suppose there are 100 curtains, and you choose 1 curtain, and 98 curtains are opened without the price behind them.
    Now, the only curtain that was not opened and was not chosen by you was not opeend for a good reason, which is that the prize could be behind it.
    There are two posibilited, either that the curtain which was not opened was not opened because the prize is behind it (among the 99 other ones), or to fool you because you had the right curtain from the first place !
    Now the chance of you having the right curtain from the first place is only 1/100, so 99/100 is in favour of "it was not opened because the prize is behind it", so change your choice to the 99/100 !
  6. Jun 20, 2003 #5


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    This is a very old (and very simple) problem. Even before it raised such a fuss in Marilyn Vos Savant's newspaper column a number of years ago, I found it as an exercise, in chapter 1, of an elementary probability text written 20 years ago.
  7. Jun 21, 2003 #6
    I get your last explanation... but for this one...

    I don't see why it wouldn't be 50/50 between the 1st and 3rd.

    Before: 1/3 ------ 1/3 ------ 1/3

    After: 1/2 ------- 0 -------- 1/2

    You say its: 1/3 ---- 0 ------ 2/3

    Why are you multiplying by 2?
    The 2nd curtain's chance is now 0. Its chances are spread equally between the other 2, so we add .5(1/3) to the 1st and 3rd curtains. Where do you get the "multiply by 2" thing?


    Some more explanation is needed, please. (details!)

    P.S. can someone please give the stanford bookstore joke?

    edit: typed a wrong number
  8. Jun 21, 2003 #7


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    Because there's no longer a uniform distribution of probability.

    In the beginning you see three curtains. You have no a-priori reason to think any is more likely than any other, so each of them has a one in three chance of having the prize.

    So you select a curtain. The odds are one in three that you were correct, and two in three that you were incorrect.

    Now the host opens one of the curtains to reveal there is no prize behind it.

    Does that give you any new information about your selection? Nope! You all ready knew that at least one of the curtains you did not select did not hide the prize.

    Here's another way of thinking of it. If the host had the freedom to open any of the three curtains he wanted, you'd then have a 50/50 chance when choosing which curtain you want... but if you tell the host "You cannot open curtain A", what then are the odds that the prize is behind curtain A or behind the other curtain he did not open?
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