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friendofgunnar
Honorable
Legendary Hero
able to speed up time
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posted July 18, 2006 08:37 PM |
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IRISCANNOTPOST
Hired Hero
J/K, DON'T DELETE ME PLEASE!
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posted July 18, 2006 10:01 PM |
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NO YOU DIDN'T! NO YOU DIDN'T!
OKAY, WHAT IS IT? 
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friendofgunnar
Honorable
Legendary Hero
able to speed up time
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posted July 18, 2006 10:09 PM |
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Aerie
You've come back to me
Okay the last guy in the line, if sees two reds or two blues then he knows what color his hat is and the game is over and everybody goes free.
He says nothing though.
So the guy in the middle automatically knows that the guy in front of him is wearing a different color hat then he is.
He tells the warden and they are all released, free to enact whatever mayhem they please on the countryside. Meanwhile the warden is fired for giving the cons such an easy opportunity to go free.
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EDIT
this is in answer to Iris's clarification of the riddle (right below):
The guy in back looks at the first two guys.
1.If the first two guys are wearing red hats he knows he is wearing a blue hat and everybody goes free.
2.If the first two guys are wearing blue hats he remains silent.
3.If the first two guys are mixed he remains silent.
If the guy in back doesn't speak, the second guy knows it's either 2. or 3..
he looks at the guy in front (the one facing the wall)
1.If the guy in front has a red hat he knows he is wearing a blue hat. He speaks up and they all go free.
2. If the guy in front is wearing a blue hat then he doesn't know what his own hat is so he is silent.
If the first guy and the second guy don't speak then the third guy can logically deduce that he is wearing a blue hat.
In any case, they all go free.
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Iris
Responsible
Supreme Hero
of Typos
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posted July 19, 2006 03:32 PM |
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Good job, Foggy!
...Unfortunately, I told the riddle wrong. HA HA HA HA HA HA I suck at this. x_X
Okay, so the correct version (even though you've pretty much solved it already )...
Quote:
There are three prisoners standing in a single file line, with the first prisoner facing a wall and the other two behind him. These people are wearing blue hats and red hats. They do not know how many blue or red hats there are, just that both colors are out there. An officer comes and tells the prisoners that if a/the person with a blue hat can speak up, he will let all of them go free. How do the prisoners solve this puzzle?
Aerie will most likely pay a visit again at some point. HAHAHAHAHAHAHA oh coffee. ^_^
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted July 20, 2006 03:00 PM |
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I love riddles. I want to join!
Quote:
So if Mr.P would know the "4", he would also say "I donīt know what it is", and if Mr.Q would had "clubs", wouldnīt he also say "I know that u donīt know what it is", because there are 3 numbers of clubs which appear at least 2 times?
And the answer is NO because as Iris said:
Quote:
The line "I know that you don't know what it is" --> "In the suit that I know, there are only cards that has 2 or more cards with the same number."
Spades and Clubs both have card numbers that only appears once, which eliminates them.
Perhaps a more intuitive way of this sentence is (after all it was translated):Quote:
I knew that you don't know what it is
or (explaining the above): I knew your answer before you give it to me because every card in this suit appears at least 1 more time (either in the same suit {which is not the case} or in another suit).
I really enjoyed this riddle and am looking forward for more!
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The empty set
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted July 20, 2006 03:34 PM |
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Edited by dimis at 15:36, 20 Jul 2006.
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TV-show riddle
This riddle has evolved to my personal favourite over the years (hopefully there will be little flaming - I mean it ...). Here is the case:
There is a TV-Show where one player is picked (by the TV-Showman - or whatever the name is) among the crowd to play a game with boxes.
Contents of the boxes:
We know that boxes contain the following:
--> One is empty.
--> One contains a mere cent (another variation of the riddle is we have 2 empty boxes).
--> One contains the BIG PRESENT; you can place inside anything that fits to your imagination. 
Rules of the game:
1) Initially the player is presented with 3 different boxes.
2) Boxes' contents is a permutation of what is described above.
3) TV-Showman knows the contents of each box.
4) The game is split into 2 phases.
5) On the first phase, the player picks a box.
6) Now, the second phase starts. TV-Showman, opens up one of the unselected boxes and reveals its contents. However, he is not allowed to reveal the BIG PRESENT in any case.
7) The player is allowed to switch boxes.
And the question is:
What should the player do if he is (obviously) "hunting" the BIG PRESENT? Should he stick on his initial choice? Should he switch the box which he initially picked with the one that remains? Does even matter what the player does (stick/switch)?
And of course, why?
Hope you have fun!
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The empty set
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friendofgunnar
Honorable
Legendary Hero
able to speed up time
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posted July 20, 2006 09:21 PM |
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I remember Marillyn Vos Savant in Parade magazine posed this question and there was a back and forth with the readers about it for a long time because the correct answer is so counter-intuitive that it's hard to understand.
The immediate answer is "it doesn't make a difference, stick with the box you picked."
The correct answer is "pick the other box"
I have to think about this some more to be able to explain why.
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Iris
Responsible
Supreme Hero
of Typos
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posted July 20, 2006 09:34 PM |
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I've heard this riddle too, that's why I didn't answer it. It's been a while though, and I don't really remember the explaination...
I think the answer was something like this:
During phase one, you have a 1/3 chance of getting the big present. However, after the host takes away one of the smaller boxes, your chances are now 1/2 for phase two.
Even though out of the 2 boxes left, you have a 50-50 chance of getting it, the fact that you chose the first box when the chance was 33-67 suggest that you should switch.
Or something like that. ^_^*
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friendofgunnar
Honorable
Legendary Hero
able to speed up time
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posted July 20, 2006 09:39 PM |
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I'll have a go at explaining it.
There's three scenarios.
1. You pick the correct box
2. you pick the first incorrect box
3. you pick the second incorrect box
The host reveals one of the incorrect boxes.
let's update the scenarios:
1. You pick the correct box and the other box is incorrect.
2. You pick the first incorrect box and the other box is correct.
3. You pick the second incorrect box and the other box is correct.
At this point there's two scenarios that get you the prize, which is double the probability of the scenario where you get the incorrect prize. All you have to do is switch boxes.
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted July 21, 2006 09:25 AM |
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Edited by dimis at 09:26, 21 Jul 2006.
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TV-show riddle / Extension
Ok, correct answer is switch because you are expected to win on 2 out of 3 occasions. Proofs will follow after the following extension to the riddle.
Extension:
Now, suppose that for a pretty long series of episodes the amount of times a player wins the BIG PRESENT is about 50%. Can you give a reasonable explanation of this 50%-50% that is observed? 
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The empty set
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted August 03, 2006 02:08 PM |
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do we have any suggestions for the extension of the riddle or should I let it go?
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The empty set
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JoonasTo
Responsible
Undefeatable Hero
What if Elvin was female?
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posted August 04, 2006 11:01 PM |
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The chances of getting the big present is 50-50 when there is only two boxes left because even if it was 67 to 33 in the beginning it doesn't matter anymore since probabilities don't coun't things that had happened in the past.
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DON'T BE A NOOB, JOIN A.D.V.E.N.T.U.R.E.
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friendofgunnar
Honorable
Legendary Hero
able to speed up time
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posted August 04, 2006 11:50 PM |
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Quote:
Extension:
Now, suppose that for a pretty long series of episodes the amount of times a player wins the BIG PRESENT is about 50%. Can you give a reasonable explanation of this 50%-50% that is observed?
What if two thirds of the people that get an offer to switch actually stay with the box that they picked originally? That would explain why the final result is still 50/50.
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EDIT
For Joonasto and anyone else who can't understand this riddle:
1. Get out a piece of paper and a six sided die.
2. Mark three columns and 12 lines.
3. Roll the die and put an X in one of the columns for each row.(this is the random placement of the big prize)
4. After you fill in each row, go back and roll the die again for each row. (this is your random guess).
5. Scribble out the empty space for each line if you did not circle the X. Scribble out randomly one of the two spaces if you did circle out the X. (this represents the removal of one box)
6. You will see that if you switch boxes you will win 2/3 of the time.
(the more lines you start with the closer you will get to the 2/3)
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friendofgunnar
Honorable
Legendary Hero
able to speed up time
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posted August 06, 2006 10:27 AM |
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I found some riddles in "Make" magazine
Coins on the Table
Your roomate challenges you to a game with the quarters from each of your sock drawers. Whoever wins the game gets to keep all the money. (assume you each have an unlimited supply of quarters).
You sit down at the perfectly round kitchen table and each person takes a turn, placing a quarter down anywhere on the table. No quarters can overlap and the entire quarter must rest of the table surface. The first person that can't put a quarter down on the table loses. You each have plenty of quarters and won't run out during the game.
Your roomate wants to go first. But you think about it for a while and realize you would much rather go first because you have figured out a surefire method to win (without cheating). After some convincing, your roomate allows you to place the first quarter.
Where do you place it, and what is the winning strategy?
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Iris
Responsible
Supreme Hero
of Typos
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posted August 06, 2006 10:34 AM |
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Somewhere in the middle. And you make sure you fill up empty spaces.
Say there is enough surface to fit 1 quarter, you put it in the space.
If there were enough to fit 2 quarters, you put it in the middle so that your roommate can't finish. You win.
Now say you have a big table. You two put down quarters (filling up blank spaces) until there's only room to fit 1 or 2 more. If it's 1, you win. If it's 2, you put it in the middle, and you also win.
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friendofgunnar
Honorable
Legendary Hero
able to speed up time
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posted August 06, 2006 10:40 AM |
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What if there's two separate spaces, each only big enough for one quarter, and you unfortunately have to fill up one of the spaces first?
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Iris
Responsible
Supreme Hero
of Typos
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posted August 06, 2006 10:55 AM |
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Edited by Iris at 11:14, 06 Aug 2006.
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Quote:
What if there's two separate spaces, each only big enough for one quarter, and you unfortunately have to fill up one of the spaces first?
Uhhhh, make sure that doesn't happen?
Let me think about it for a bit...
Edit: Okay, how about this: You start off by putting a coin in the center. Then afterwards, you use the middle coin as a rotation point. Wherever your roommate puts the next coin, you put your coin at the same place rotated 180 degrees. This will form a... hmm... not a mirror image, but some sort of rotated image so that you will never run into a situation where there's 2 empty spaces.
Edit 2: Haha, after reading my own edit, I realized that it would be a lot easier to just draw a line of reflection down the middle coin. This will create a mirror image. The rest is the same.
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kookastar
Honorable
Legendary Hero
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posted August 06, 2006 12:04 PM |
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Yeah I think I agree with Iris - although I don't really understand what she is saying heehee {my ignorance  }
The first coin goes in the centre of the table, because each circular 'row'surrounding this coin will have an even number of coins in it - which will mean that each 'row' is finished on your go {the guy that started it}.
I didn't have any quarters, but it works with Aussie 10c pieces
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JoonasTo
Responsible
Undefeatable Hero
What if Elvin was female?
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posted August 06, 2006 01:40 PM |
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Edited by JoonasTo at 13:42, 06 Aug 2006.
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Thanks for clarifying FoG
I got messed up because I didn't notice the thing that the announcer's choice isn't random and doesn't change probabilities. That messed me up
In the next problem I think Iris is right about the mirror image thing. The only problem would be really placing the quarters right.
Here is my problem for you:
There are three boxes and every box has two sentences on it.
The sentences are both true in one box, both false in one and in one box there is one false and one true sentence.
There is also a diamond in one of these boxes and you have to know in which one.
_____Golden____________Silver____________Steel
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DON'T BE A NOOB, JOIN A.D.V.E.N.T.U.R.E.
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Iris
Responsible
Supreme Hero
of Typos
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posted August 06, 2006 05:40 PM |
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Edited by Iris at 17:54, 06 Aug 2006.
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This might be a little hard to follow...
Gold is both True --> in Silver box
Silver (both false) --> in Gold box
Inconsistent. Correct clues are not on the Gold box.
Silver is both True --> in Steel box
Gold (both false) --> not possible
Gold (1 true, 1 false) --> in Steel box
Steel (both false) --> not possible
Inconsistent. Correct clues are not on the Silver box.
Steel is both True --> in Gold box
Gold (both false) --> in Gold box
Silver (1 true, 1 false) --> not possible/inconsistent
Silver (both false) --> in Gold box
Gold (1 true, 1 false) --> in Gold box
Consistent.
The correct clues are on the Steel box, the false clues are on the Silver box, and the Diamond is in the Gold box.
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