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    The Riddle Thread

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    Post by Creepsville Wed Apr 28, 2010 9:52 am

    And to add mine to the mix:

    On my way to the fair, I met 7 jugglers and a bear, every juggler had 6 cats, every cat had 5 rats, every rat had 4 houses, every house had 3 mouses, every mouse had 2 louses, every louse had a spouse. How many in all are going to the fair?
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    Post by CheeseJam Wed Apr 28, 2010 10:41 am

    Hotwire wrote:You have on a black hat. The other two are confused, because they each see one red and one black hat. Leaving each of them to only guess at to what color their hat is.
    Nope.

    To Creep:

    Is the answer just one because you are the only one going to the fair? Otherwise, that is a pretty long calculation.
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    Post by Creepsville Wed Apr 28, 2010 10:45 am

    CheeseJam wrote:
    To Creep:

    Is the answer just one because you are the only one going to the fair? Otherwise, that is a pretty long calculation.

    Hahaha. That's correct, Cheese. Like a Star @ heaven Like a Star @ heaven Like a Star @ heaven
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    Post by Hotwire Wed Apr 28, 2010 11:05 pm

    Creepsville wrote:

    On my way to the fair, I met 7 jugglers and a bear, every juggler had 6 cats, every cat had 5 rats, every rat had 4 houses, every house had 3 mouses, every mouse had 2 louses, every louse had a spouse. How many in all are going to the fair?

    Nice work, ever see Die Had with a vengeance?
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    Post by CheeseJam Thu Apr 29, 2010 10:43 am

    Anyone on the hat question? Smile
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    Post by NiceMarmot Thu Apr 29, 2010 11:34 am

    CheeseJam wrote:Anyone on the hat question? Smile
    Is it red? Everyone is confused because they all see red hats. Each person sees the other two with red hats on, so they are just as confused as the reader.
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    Post by CheeseJam Thu Apr 29, 2010 10:23 pm

    Yep, correct enough! If you had a black hat on, and guy A saw guy B raising his hand, he would quickly realize that he is wearing a red hat. So the answer has to be red.

    You are up.
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    Post by NiceMarmot Thu Apr 29, 2010 10:33 pm

    A man had twelve toothpicks in front of him. He took one away. Now he had nine in front of him. How is this possible?
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    Post by CheeseJam Fri Apr 30, 2010 10:28 am

    I think I have heard that one before. Doesn't he spell out the word or something with he 11 toothpicks?

    N+I+N+E=NINE
    3+1+3+4=11
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    Post by OUTLAW Fri Apr 30, 2010 12:38 pm

    Yeah you got that cheesejam
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    Post by Reed Fri Apr 30, 2010 3:40 pm

    I don't understand that one scratch
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    Post by OUTLAW Fri Apr 30, 2010 4:51 pm

    WIth the picks you have left you put them in positions that make it speel the number nine...

    like cheers leaders kinda
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    Post by Reed Fri Apr 30, 2010 4:53 pm

    oh I see
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    Post by CheeseJam Fri May 07, 2010 10:32 am

    Sorry for not keeping up guys, here's another riddle.

    There used to be this game show called Lets Make a Deal. In the show there were 3 doors. Two doors with a bad prize and one with a good prize. Lets say the good prize is a car, and the bad prizes are both goats. What you do in the game is pick a door. Then the host of the show, Monte Hall, would reveal one of the two doors you didn't choose. Since he knew what is in every door, he would always reveal one of the bad prizes. Then you have the choice to switch doors or keep the one you have. You get whatever prize that is behind the door you end up with. So lets say you are playing this game. You choose a door. He reveals one other door which has a bad prize(goat) inside. So now you have the option to either keep the door you have or switch doors. Which gives you the best chance of getting the good prize(car), switching doors or keeping the door you have? EXPLAIN
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    Post by Hotwire Sat May 08, 2010 3:02 am

    Doesn't matter either way, you have a 50/50 chance of winning the prize at this point. As a matter of fact you always had a 50/50 chance of winning, because no matter what you choose, one of the bad prizes is going away and you will have a chance to choose between the winner door and the looser door. Changing your answer now or leaving it the same doesn't change you odds at all.
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    Post by CheeseJam Sat May 08, 2010 3:28 am

    Believe it or not, 50/50 is NOT the correct answer. Read the whole question, use some probability, and you will get the right answer! Smile
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    Post by Red_Dead_Demon Sat May 08, 2010 3:39 am

    you already lost the game pick which goat you want.
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    Post by Bitter Marston Sun May 09, 2010 12:55 am

    THe host knows whats behind the doors so he would never pick the door with the good prize in it so you switch
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    Post by Red_Dead_Demon Sun May 09, 2010 1:09 am

    i think your right
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    Post by Bitter Marston Sun May 09, 2010 4:16 am

    I hope so
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    Post by CheeseJam Sun May 09, 2010 3:28 pm

    Bitter Marston wrote:THe host knows whats behind the doors so he would never pick the door with the good prize in it so you switch

    The question says he never picks the door with a prize in it. So, when there are two doors left, with a goat and a car left, why would you switch? You need to use probability to get the riddle right.
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    Post by NonDairy Soup Sun May 09, 2010 4:09 pm

    Switch why would you keep the door with a goat?
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    Post by Red_Dead_Demon Sun May 09, 2010 4:11 pm

    Cause i am love with her! soon after gay marriage we will be able to wed animals. pig
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    Post by CheeseJam Sun May 09, 2010 5:25 pm

    NonDairy Soup wrote:Switch why would you keep the door with a goat?
    Read the riddle again. You pick a door, and don't know what's inside. Then the host reveals one of the other doors that has a goat in it. So then, there are ALWAYS two doors left, one with a goat and one with a car. Should you switch or stay with your door?
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    Post by Blaze Wed May 12, 2010 11:24 am

    lol i knew this one
    you should always switch, because it'll increase your chances, but why, i dont remember xD

    i also got a riddle, if you don't mind, this way there'll be two, so it'll double the fun(?)

    It's not right, but it's definitely not wrong.

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