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jump to last post 1-3 of 3 discussions (6 posts)

A Really Hard Riddle...(I think)

  1. Jackson Riddle profile image51
    Jackson Riddleposted 7 years ago

    Ok, so yes it is ironic that Riddle is my last name but someone has given me a riddle to solve and I can't solve it at all (I highly doubt it is a stupid answer, rather one that is rather logical when worked out correctly). So if anyone wants to give it a shot here it is:

    There are 4 people in a room. IN the room is a bag containing 8 stickers, 3 blue and 5 white.

    3 of the four people then close their eyes, randomly pick 2 stickers and then stick them to their forheads. They then open their eyes and can see the remaining two people and what stickers they have on their head. It just so happens that the first person (lets Call him A) has a blue and a white sticker on his head, person B also happened to choose a blue and white sticker, person C also chose a blue and white sticker (randomly of course).

    The person without any stickers then proceeds to ask person A if they can tell what the colour of his two stickers are. He can answer with yes or no. If he can work it out then he will win $10,000 but if he does not he must pay $5000. However, if he says no, the person will then ask person B if they can work out which colour stickers they have on their head, if yes they win if no; person C is asked the same question. This is classified as one round of questioning. If person C answers "no" a new round of asking the same questions begins.

    After each answer the prize money in the pool decrease and the payment of an incorrect answer decreases therefore each person wants to work out the answer as quickly as possible and the game is ended if someone incorrectly states what stickers they have on their head.

    Therefore, using rational thinking can you discover which person correctly identifies what colour stickers they have and during what round of questioning does this take place?

  2. Jackson Riddle profile image51
    Jackson Riddleposted 7 years ago

    Haha, anyone?

  3. sunforged profile image75
    sunforgedposted 7 years ago

    Its a logic puzzle , not a riddle.

    I used to do them in 2nd grade

    create an ABC grid - such as this one and have fun -

    http://pages.prodigy.net/spencejk/solvelps.html

    well, maybe its something else - but it looks like a logic puzzle to me. The exact language of the puzzle is important and you will have to infer your own rules, unlike the one I used as an example

    1. Jackson Riddle profile image51
      Jackson Riddleposted 7 years agoin reply to this

      Haha, that was the my first thought of how to solve it lol (I too remember doing them when I was young). However, I couldn't work out how the person figures out what they have when all they can see is 2B and 2W stickers and hear the other people say no they can't work it out either.

      Thanks anyway.

      1. cre8tive profile image84
        cre8tiveposted 7 years agoin reply to this

        This is bringing back memories. My Dad used to do a version of this puzzle only they were prisoners who could get out of prison if they guessed and they 4th person was the gaoler.

        Can't remember the exact solution but think it all stems from Person A who can see 1 B & 1 W on each of the other 2's foreheads.

        Person A also knows that persons B & C can also see the b&w on each others foreheads but has neither of them are jumping up and winning person A knows that he cannot have 2 B stickers which means his only options are 1B & 1W or 2W.

        Person A then goes off down the strack of treating it like a logic puzzle and working out the permutations ~ which I haven't got around to yet.

        I do seem to remember that it didn't take long to do though.

        On the other hand I could also have started this with a completely illogical assumption ~ so for now I think I'll retreat back to my coffee cup.

        1. Jackson Riddle profile image51
          Jackson Riddleposted 7 years agoin reply to this

          Yeah that is pretty much the same puzzle. I believe if they guess it right they are free'd and if they are wrong they are executed.

          That's as far as I get aswell, how do they know if they have one of each or two of the same colour?

 
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