A logic puzzle

[Deleted User]
[Deleted User] Posts: 7,658
edited May 2010 in The Clubhouse
Hello,
Here's an interesting puzzle:
There is a test subject and an experimenter and three closed boxes. One of the boxes has a $10.00 bill inside and the experimenter knows where it is. The subject doesn't, but if he can guess it correctly he can keep it. The subject picks one of the boxes and tells the experimenter. The experimenter opens one of the remaining two boxes (always an empty one) and asks the subject if he wants to now change his pick?
Should he do it and why?
Cheers, Ken
Post edited by [Deleted User] on
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  • disneyjoe7
    disneyjoe7 Posts: 11,435
    edited May 2010
    No it's now 50/50!

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  • bobman1235
    bobman1235 Posts: 10,822
    edited May 2010
    That's actually not true, specifically because the person running the experiment KNOWS which box is the correct one. You actually double your chances of winning by switching.

    Rather than try and explain it myself, Google the "Monty Hall Problem."
    If you will it, dude, it is no dream.
  • fatchowmein
    fatchowmein Posts: 2,637
    edited May 2010
    Yes, he should change his pick since one of the two remaining boxes was open leaving the one the subject picked and the other box that was empty. Therefore, the subject now knows which box has the money.
  • Knucklehead
    Knucklehead Posts: 3,602
    edited May 2010
    Just give me the 10 bucks. :)
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  • [Deleted User]
    [Deleted User] Posts: 7,658
    edited May 2010
    Keep in mind the original box that the subject picked is still closed. He can either stay with that choice or jump to the other, unopened box.
  • disneyjoe7
    disneyjoe7 Posts: 11,435
    edited May 2010
    Still feel it's 50/50

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  • tommyboy
    tommyboy Posts: 1,414
    edited May 2010
    bobman1235 wrote: »
    That's actually not true, specifically because the person running the experiment KNOWS which box is the correct one. You actually double your chances of winning by switching.

    Rather than try and explain it myself, Google the "Monty Hall Problem."

    Thanks. There's a good illustration on wikipedia about it, I'd never think of it that way. The power of math;)
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  • Tony M
    Tony M Posts: 11,141
    edited May 2010
    I agree...it's now a 50/50...I'd stay.

    How can it be any other way?
    Most people just listen to music and watch movies. I EXPERIENCE them.
  • disneyjoe7
    disneyjoe7 Posts: 11,435
    edited May 2010
    Now I'm not sure :( Monty Hall Problem.

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  • bobman1235
    bobman1235 Posts: 10,822
    edited May 2010
    I agree...it's now a 50/50...I'd stay.

    How can it be any other way?

    Think of it this way.

    You're choosing one of three boxes. So when you pick your one box, there is a 1/3 chance you're correct, and a 2/3 chance it's one of the other two boxes, right?

    Well, the person conducting the experiment knows which box is the correct one. He opens one that he KNOWS not to be the correct box. Your odds haven't changed - there was a 1/3 chance you were correct, and a 2/3 chance that it was one of the other two boxes. It's still a 2/3 chance that it's the one REMAINING box.

    Tricky.
    If you will it, dude, it is no dream.
  • [Deleted User]
    [Deleted User] Posts: 7,658
    edited May 2010
    When the subject picks one of the boxes he has a 1/3 chance of being correct. The remaining two boxes represent a 2/3 chance of being correct. Once the experimenter shows that one of the remaining boxes doesn't have the $10.00 the remaining box still represents a 2/3 chance, but concentrated in that single box. The original choice still has the 1/3 chance, so he should switch, every time. Having a 2/3 chance is better than a 1/3 chance.
    The vast majority of people, who take this test, will always say remain with the original choice.
    Ken
  • [Deleted User]
    [Deleted User] Posts: 7,658
    edited May 2010
    Sorry, Bobman, while I was typing my response you were already posting.
    Ken
  • Willow
    Willow Posts: 10,999
    edited May 2010
    When the subject picks one of the boxes he has a 1/3 chance of being correct. The remaining two boxes represent a 2/3 chance of being correct. Once the experimenter shows that one of the remaining boxes doesn't have the $10.00 the remaining box still represents a 2/3 chance, but concentrated in that single box. The original choice still has the 1/3 chance, so he should switch, every time. Having a 2/3 chance is better than a 1/3 chance.
    The vast majority of people, who take this test, will always say remain with the original choice.
    Ken

    Very interesting, thanks Ken. I would say it's not worth the headache for only 10$ ;)
  • concealer404
    concealer404 Posts: 7,440
    edited May 2010
    This puzzle was in a movie i watched fairly recently....
    I don't read the newsssspaperssss because dey aaaallllllllll...... have ugly print.

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  • [Deleted User]
    [Deleted User] Posts: 7,658
    edited May 2010
    I think what most people figure is that once one of the other two has been shown to not be the correct one, their odds go from 1 in 3 to 1 in 2. It started off with three boxes, now there are only two.
    Many researchers feel that the way we view the world as a series of interrelated events. One action following another causes us to see the reduction of the number of boxes as changing the original odds.
    Ken
  • tommyboy
    tommyboy Posts: 1,414
    edited May 2010
    When the subject picks one of the boxes he has a 1/3 chance of being correct. The remaining two boxes represent a 2/3 chance of being correct. Once the experimenter shows that one of the remaining boxes doesn't have the $10.00 the remaining box still represents a 2/3 chance, but concentrated in that single box. The original choice still has the 1/3 chance, so he should switch, every time. Having a 2/3 chance is better than a 1/3 chance.
    The vast majority of people, who take this test, will always say remain with the original choice.
    Ken

    Yes, the probability doesn't change to 50 after you find out cause your original choice was made before the box is revealed and thats where the biggest confusion is.
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  • tommyboy
    tommyboy Posts: 1,414
    edited May 2010
    I think what most people figure is that once one of the other two has been shown to not be the correct one, their odds go from 1 in 3 to 1 in 2. It started off with three boxes, now there are only two.
    Many researchers feel that the way we view the world as a series of interrelated events. One action following another causes us to see the reduction of the number of boxes as changing the original odds.
    Ken

    Haha beat me to it
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  • Jstas
    Jstas Posts: 14,808
    edited May 2010
    I would open the box holding my gun. Then the test conductor gives the $10 and all the cash in his wallet as well! :p


    Oh and BTW, technically that's not a logic question but a statistical odds question.
    Expert Moron Extraordinaire

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  • bobman1235
    bobman1235 Posts: 10,822
    edited May 2010
    And if you think THAT'S confusing, try figuring out the Birthday Paradox (not really a paradox, but it sounds cooler that way).
    In a group of at least 23 randomly chosen people, there is more than 50% probability that some pair of them will have the same birthday.... For 57 or more people, the probability is more than 99%
    If you will it, dude, it is no dream.
  • lumpy
    lumpy Posts: 113
    edited May 2010
    sorry guys - the ability to reselect the box starts you off from a new slate. 1 out of 2 no matter how you slice it. People choose to keep the original box because there is nothing positive about the descision that would make them change thier mind - they are comparing one box to another matching box. In their minds their have already chosen the box, and putting it down to pick up the other one means twice the descision making cost ='s they have to select two boxes for the price of one.
    pop

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  • RuSsMaN
    RuSsMaN Posts: 17,987
    edited May 2010
    Ken, try this one.

    There is a room. In that room are three lights. Each of those three lights is connected to one of three switches outside the room. The room has a door, but seals tight. You may only open the door and enter the room once, and you may only have one switch on when you do.

    How can you tell which light goes to which switch?

    Cheers,
    Russ
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  • bobman1235
    bobman1235 Posts: 10,822
    edited May 2010
    lumpy wrote: »
    sorry guys - the ability to reselect the box starts you off from a new slate. 1 out of 2 no matter how you slice it. People choose to keep the original box because there is nothing positive about the descision that would make them change thier mind - they are comparing one box to another matching box. In their minds their have already chosen the box, and putting it down to pick up the other one means twice the descision making cost ='s they have to select two boxes for the price of one.

    You're right, all the statistics classes I've taken that have used this example were taught by idiots! Lumpy has revolutionized math!
    If you will it, dude, it is no dream.
  • concealer404
    concealer404 Posts: 7,440
    edited May 2010
    The movie was "21." Variable Change and all that jazz.
    I don't read the newsssspaperssss because dey aaaallllllllll...... have ugly print.

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  • Jstas
    Jstas Posts: 14,808
    edited May 2010
    RuSsMaN wrote: »
    Ken, try this one.

    There is a room. In that room are three lights. Each of those three lights is connected to one of three switches outside the room. The room has a door, but seals tight. You may only open the door and enter the room once, and you may only have one switch on when you do.

    How can you tell which light goes to which switch?

    Cheers,
    Russ

    According to most building codes, the light switch closest to the entry point should be the one operating the lights inside the room.
    Expert Moron Extraordinaire

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  • Tony M
    Tony M Posts: 11,141
    edited May 2010
    bobman1235 wrote: »
    Think of it this way.

    You're choosing one of three boxes. So when you pick your one box, there is a 1/3 chance you're correct, and a 2/3 chance it's one of the other two boxes, right?

    Well, the person conducting the experiment knows which box is the correct one. He opens one that he KNOWS not to be the correct box. Your odds haven't changed - there was a 1/3 chance you were correct, and a 2/3 chance that it was one of the other two boxes. It's still a 2/3 chance that it's the one REMAINING box.

    Tricky.
    When the contents of one of the boxes was exposed, the odds just changed. there is no 1/3 mystery placed on each box now. It's a 1/2 or 50% mystery now.;)
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  • hearingimpared
    hearingimpared Posts: 21,137
    edited May 2010
    Ken the way my brain is scrambled these days, I would have subdued the tester and opened all the boxes for the $10, LOL!!!!
  • Tony M
    Tony M Posts: 11,141
    edited May 2010
    Jstas wrote: »
    I would open the box holding my gun. Then the test conductor gives the $10 and all the cash in his wallet as well! :p


    Oh and BTW, technically that's not a logic question but a statistical odds question.

    Statistics.;) that's what I say.
    I picked one. didn't win...game over.
    If I'm allowed to play again then the game should state in the beginning I have 2 tries and THEN it becomes a 2 out of 3 or 2/3rds chance of winning.

    Funny, I use to loose no matter what, now I can not pay attention and win.
    Except these Karmas on this forum:(
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  • concealer404
    concealer404 Posts: 7,440
    edited May 2010
    Ehhh....

    The guy that showed you the box gives you the answer.

    If he's going to show you a box, he won't show you yours off the bat, because then he wouldn't be able to let you change your mind anyways.

    You have boxes A, B, and C.

    You choose B.

    He can't show you B, because then the scenario is all moot.

    He shows you C, which doesn't have a box. He DOESN'T show you A for one of two reasons:

    1) It has the item
    2) Neither A nor C have the item, B does. We can't determine this right now.

    Yes, there IS a 50% chance between the two remaining boxes, but looking at the bigger picture I.E. the THREE boxes, box A has the 2/3rds chance.
    I don't read the newsssspaperssss because dey aaaallllllllll...... have ugly print.

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  • [Deleted User]
    [Deleted User] Posts: 7,658
    edited May 2010
    That's a head scratcher, Russ. Let's see what I've come up with:
    1). Let's number the switches 1,2 and 3 and the lights as A,B and C.
    2). On a piece of paper I put 1, 2 and 3 at the tops of three columns. I placed an A, B and C underneath each column. Indicating that each switch could be connected to each of the lights.
    3). If we throw switch 1 and look in the door and find light B lit we can cross off A and C from the 1 column. We can cross off B from column 2 and 3.
    4). This is where I run out of logic steam. It seems to me that switch 2 could be connected to either light A or C. And the same can be said for switch 3, either A or C.
    You only get one switch throw and one room look, so only one event can be known. I don't know.
    Ken
  • Tony M
    Tony M Posts: 11,141
    edited May 2010
    RuSsMaN wrote: »
    Ken, try this one.

    There is a room. In that room are three lights. Each of those three lights is connected to one of three switches outside the room. The room has a door, but seals tight. You may only open the door and enter the room once, and you may only have one switch on when you do.

    How can you tell which light goes to which switch?

    Cheers,
    Russ

    I used to know that one.Now I don't..

    I really love the one with 3 men splitting a hotel room, each pays something and later the clerk gives back some refund to one man and when he splits it up, a dollar dissapears. Now that was fun to tell in my younger years when we'd sit around for an hr. and laugh about it trying to rationalize while mentaly going in the other direction:D
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