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 Colour-blinded LOGIC - Not Adequate for Real-Life Situations Goto page 1, 2  Next
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mole
Subterranean Member

 Posted: Thu Mar 21, 2002 9:49 am    Post subject: 1 I mean, what kind of a group of evil barbarians is this? They set such an easy puzzle! Take any 129 tokens, turn them all over, and call that Group 1. Call the rest of the tokens Group 2. If there were x red tokens in Group 1 originally, there are 129 - x red tokens after they're all turned over. The 129 - x red tokens in Group 2 remain unchanged, so they both end up with the same number. [This message has been edited by mole (edited 03-21-2002 04:52 AM).]
Lepton
1:41+ Arse Scratcher

 Posted: Sat Mar 23, 2002 4:20 am    Post subject: 2 Why not flip them all over, and then flip two back. Put one with all the others in group one, and the other coin by itself in group two. So that you can at least have real numbers. [This message has been edited by Lepton (edited 03-22-2002 11:21 PM).]
BearTalon
Guest

 Posted: Sat Mar 23, 2002 2:23 pm    Post subject: 3 The problem, Lepton, tells you that there are 129 RED showing. It doesn't say how many green, only that there are hundreds of the bone chips. You're also blindfolded, so how can you tell which are red chips to flip? There's no indication if the barbarians dropped the chips randomly, or placed them in an orderly fashion. I'd remove their order by flipping them all myself, giving me a good chance that they will fall in a truly random fashion, so that 50% land on reg and 50% on green. And to throw a wrench into my own solution, what if there is an odd number of chips?
i_h8_evil_stuff
Daedalian Member

 Posted: Sat Mar 23, 2002 7:20 pm    Post subject: 4 there is very little chance of my solution working.... count the chips. if there are 129, turn an odd number over, and seperater the rest into 2 piles. if there is more than 129 chips, mole is right. ------------------ The name explains it all. [This message has been edited by i_h8_evil_stuff (edited 03-23-2002 02:22 PM).]
El Blobbo revived
Daedalian Member

 Posted: Sat Mar 23, 2002 10:28 pm    Post subject: 5 Bury 'em. Never said to use all of them.
The Cheshire Man
Not a pussycat

 Posted: Sun Mar 24, 2002 4:24 am    Post subject: 6 Bah! It wasn't until I read Mole's solution that I realized there were more than 129 coins on the ground. Phooey. ------------------ Now you see me, now you don't... but keep smiling!
El Blobbo revived
Daedalian Member

 Posted: Sun Mar 24, 2002 4:52 pm    Post subject: 7 mole got it of course: code:``` |green| | red | |green| | red | | | red | |green|``` See? [This message has been edited by El Blobbo revived (edited 03-24-2002 06:56 PM).]
Marmoset Hunter
Guest

 Posted: Sun Mar 24, 2002 7:36 pm    Post subject: 8 I don't understand the solution here at the bottom, and I'm not sure if someone else has already posted the solution--but as a variation to burying (and with a lot less sweat) one could simply stack the bones sideways (facing the same way as cookies face in the package). There shouldn't be any problems with balance in a cave with walls. Making two groups evenly, one can either count each chip or no chip as having a red side.
cha
Guest

 Posted: Mon Mar 25, 2002 12:15 am    Post subject: 9 This really does seem to be one of the easier puzzles, which always worries me... But if it were me, I'd count out 129 bones into one group, and leave the rest in the second group. I would then flip each bone in the first group. Eh?
babw44
Icarian Member

 Posted: Mon Mar 25, 2002 12:19 am    Post subject: 10 I have two solutions, neither of which seem physically practical, but possible. 1> Seperate the tokens into 2 STACKS. Then only 2 tokens will actually showing, improving your odds depending on the total number of tokens. 2> ok, this one is out in left field, but so is burying them. The tokens are made of bone, so if they are rubbed together, the color should rub off.
CrystyB
Misunderstood Guy

 Posted: Mon Mar 25, 2002 12:42 am    Post subject: 11 You second CORRECT answer, cha! You should really consider registering and joining the fun most of use get from constantly 'hanging out' here!
ande
Guest

 Posted: Mon Mar 25, 2002 3:54 am    Post subject: 12 here is a wild guess. how about you just divide the chips into two piles without caring about the color. now think of the following....this is a barbaric tribe and chances are that they don't buy their dies at a craft store. the green color might have come from crushed leaves and the red color could have come from the blood of others (those who didn't make it). in either case the color of these "dyes" would over time become a dull brown. the tokens are also on the ground in a tent. i'm thinking that this tribe doesn't have electric lighting so it is fairly dark inside that tent. even with a torch, the light would be dim. so with the dull, faded colors and the bad lighting i think it would be hard to tell which color is which. there is a clue in the title too, "Color Blinded". if you are color blind it is hard to tell the difference between color (uhhhh...yeah). this may be hinting that some of the tribesmen are slightly color blind. overall, i think it would be hard for someone to do this even without the blindfold. another thing- if you are traveling in uncharted lands wouldn't you be carrying some sort of weapon such as a gun? your walking through a strange land with no knowledge of what lies ahead. chances are you are going to bring a weapon for protection or just to get some food. i say, that even if you don't get the puzzle right, at least make a scene and go out shooting some of their best hunters or even the cheif. you could at least make a distraction with your cell phone or flash light. depending on the knowledge of this tribe, you could possibly convince them that your their god or some holy spirit. just use your imagination..............or bring a magnum -beware of the yellow faces-
i_h8_evil_stuff
Daedalian Member

 Posted: Mon Mar 25, 2002 9:24 pm    Post subject: 13 k, i got a solution if its an even number of bones.... split them up into 2 equivilent groups, and you explain that if you take off ur blindfold, the man behind u will "ensure that they turn a bright crismon rather fast," meaning all the chips would be red. Since there are all/2 red chips in each group, he has to let u free, meaning he shouldnt have just killed u, which confuses the whole tribe. Then, you take off ur blindfold and kill them all while they're discussing it, and leave, taking some of their resources with u (ie. food). If there's an odd number, bury/eat/destroy one chip, and use the above method. ------------------ The name explains it all.
Mark
Icarian Member

 Posted: Mon Mar 25, 2002 9:48 pm    Post subject: 14 ^if there is an odd number you could just turn one over. There would still be the same amount of red in each just a green in one aswell.
El Blobbo revived
Daedalian Member

 Posted: Tue Mar 26, 2002 12:31 am    Post subject: 15 Nonono, there are at least 200.
mr_kilroy
Icarian Member

 Posted: Tue Mar 26, 2002 2:47 am    Post subject: 16 Split the bones and ask the king what color each top is,one by one.When the king answers,ask the guard if the king is a liar.The quard will never call his king a liar and answer no if the king told the truth and be silent if the king is a liar. Hummm (scratching the back of my neck)
Qball
In the Quorner Pocket

 Posted: Tue Mar 26, 2002 3:48 am    Post subject: 17 No, no, no. You only need three weighings to do it. jeez.
Buddy
Guest

 Posted: Fri Mar 29, 2002 12:23 am    Post subject: 18 how about splitting up as evenly as possible and turn over one bone(on one side or other). they mite have put them all on onecolor side.
petero
Guest

 Posted: Fri Mar 29, 2002 6:12 pm    Post subject: 19 You missed the correct answer posted by Cha a while ago. Gather 129 tokens into a pile. The remainder (unknown how many) are in the second pile. Flip over all the tokens in the pile with 129. This will give you 2 piles, each withthe same numbe rof red tokens. It works.
Fayelea
Guest

 Posted: Mon Apr 01, 2002 3:51 am    Post subject: 20 Lesse...Alright...each chip has two sides...one green, one red, right? The man should first count all the chips to see how many there are. Listen though: The cannibal guy said that he must seperate the bone chips into two equal piles of each color right? lets do this: Multiply 129 by two. That equals 258, right? Seperate the chips into two equal piles of 129. Flip over all the chips in one of the piles. there is a slight chance you will die... Wait a sec. Did any of you ever think of this as a trick question? The guy could be lying or is gonna kill you anyway because you woulod have to take off the blindfold.
Pyramid
Guest

 Posted: Mon Apr 01, 2002 5:54 pm    Post subject: 21 Snap each bone in half to make two piles.
fetus
Daedalian Member

 Posted: Tue Apr 02, 2002 3:56 pm    Post subject: 22 so far as i can see, pyramid wins.
Victor_Vinfroi
Icarian Member

 Posted: Fri Apr 05, 2002 5:49 pm    Post subject: 23 Pyramid, this is a nice idea, but I dont think we are allowed to 'divide' the tokens in this way. English is not my mother tongue, so I am not sure what exactly is 'packed earth', but it is not to be red? So that, by rubbing BOTH faces of a token on the ground, your token is 'showing red' whatever the face up? If it the case, rub all the tokens, and share the lot into two equal parts (provided the total number of token is an even number). If it is an odd number, flip the last token, and you get 50% chance to survive...
CrystyB
Misunderstood Guy

 Posted: Sat Apr 06, 2002 1:37 am    Post subject: 24 what if there is an odd number of tokens? Burry the extra red half-tokesn? BTW Why does everybody use "bones"? They're (quote)circular discs(/quote), or (quote)reindeer bone tokens(/quote) Heh, i noticed the second mention of "improving your odds to 50%". How's 100%->50% gonna improve any odds? [This message has been edited by CrystyB (edited 04-05-2002 08:40 PM).]
cha
Daedalian Member

 Posted: Sat Apr 06, 2002 3:05 am    Post subject: 25 I'm bored, so I'll 'splain... Pull out 129 tokens into a separate pile. Call this pile 1. Call the other pile 2. Pretend you now have two red tokens face up in pile 1. This means there are 127 red tokens face up in the pile 2, right? See? Now, turn over all the tokens in pile 1. Now pile 1 has 127 red tokens face up, follow? Same number of red tokens showing in each pile. There will also be the same number of green tokens face up in each pile. No need to know how many there are total. What is madness but nobility of soul at odds with circumstance?
PhoenixMan
Guest

 Posted: Sat Apr 06, 2002 10:04 pm    Post subject: 26 Cha is right, 'cept for one thing. You'll have the same amount of red chips in each pile, but not green. Suppose you have a million chips. The reds will be equal but there will still be at least 999,871 green chips. If that's the case, you'll have none in the 129-chip group. Only the reds will be equal.
Mr. E
Icarian Member

 Posted: Sun Apr 07, 2002 3:52 am    Post subject: 27 But the thing is you are ASSUMING this theory. What if you do this, but when you separate them you end up getting all the red of most of the red in one group. And if there are thousands of them then you can not be sure.
SunStorm
Guest

 Posted: Sun Apr 07, 2002 5:04 am    Post subject: 28 PheonixMan Cha is entirely right. if you have millions of chips, and you seperate 129, no, you may not get any red chips in those 129. that means that all 129 are in the pile with millions. so if you got no red chips in pile of 129, and you flip them over, then you have one pile (all red) of 129. and since all the red in the original pile was in the stack of millions, then the stack of millions must have had 129 (the original number of red chips). now, look at your result. End up (as previously stated) with 129 red in pile of 129, and 129 red in pile of millions. it works if you have one red chip in the pile of 129 as well. think about it and do the math. You dont have to have each pile showing the same amount of green. congrats Cha.
Zealot
Daedalian Member

 Posted: Sun Apr 07, 2002 7:39 am    Post subject: 29 PhoenixMan was correct. He was not disputing the fact that there would be an equal number of red chips in both piles, he was correcting cha's statement that the number of green chips in each pile would be equal as well.
Minxy
Guest

 Posted: Mon Apr 08, 2002 2:57 am    Post subject: 30 Ok, Cha, obviously that is correct...but how did we get around the blindfold issue to know how many red we've pulled....???
Minxy
Guest

 Posted: Mon Apr 08, 2002 3:01 am    Post subject: 31 Scratch above message....ty mole...your very first reply helped!!!
Old Man
Guest

 Posted: Tue Apr 09, 2002 4:17 am    Post subject: 32 Simply sort into two equal piles and turn all of either pile over. The result will be equal numbers of same color in each pile. Simple!!!!!
alphaomegaQ
Guest

 Posted: Tue Apr 09, 2002 5:25 am    Post subject: 33 gather up all the bones,divide them into two equal piles.after dividing them into two piles flip each bone.is that wrong...how about pressing each bone tocken into the ground enough so all bones are standing on their edge.is that wrong...how about biting your hand and dyeing each bone from the fresh wound..is wrong?oh well,throw your head back caughting tribe-dude in the loin clothe then run like hell........
Minxy
Guest

 Posted: Tue Apr 09, 2002 3:56 pm    Post subject: 34 HAHAHAHA.......we are all posting reply after reply....sunstorm and cha, although your explanations are correct..does no one notice that mole's posting, the very first one on this page, has posted the correct answer?? Hats off to you mole!! =)
Mr. E
Icarian Member

 Posted: Tue Apr 09, 2002 9:11 pm    Post subject: 35 I don't understand what you mean.
Zealot
Daedalian Member

 Posted: Tue Apr 09, 2002 9:22 pm    Post subject: 36 Highlight mole's post... he used invisible text.
Daedalian Member

 Posted: Tue Apr 09, 2002 11:13 pm    Post subject: 37 Mole. What about Group 2 having 600 - x RED (The riddle states that there are hundreds of "discs") Then they are not equal
Daedalian Member

 Posted: Tue Apr 09, 2002 11:20 pm    Post subject: 38 Sorry MOLE You are correct
Mr. E
Icarian Member

 Posted: Thu Apr 11, 2002 3:56 pm    Post subject: 39 I am still havign problems what you mean. Sorry, you have to break it down for me.
cubestudent
3D Member

 Posted: Fri Apr 12, 2002 1:38 am    Post subject: 40 I'll go ahead and rephrase Mole's solution, then. there are N total tokens. (N is unimportant) 129 of them are currently red side up. mole says to separate out 129 of the tokens. call this group 1. the remaining tokens are group 2. group 1 will have x 'red-side-up' tokens in it. . and will have 129-x 'green-side-up' tokens. group 2 will have 129-x (total original red's minus those put into group 1) reds. when you flip all of group 1, it will have x 'green-side-up' tokens (formerly red). . and will have 129-x 'red-side-up' tokens (formerly green) so now group 1 and group 2 both have 129-x tokens with the red side up. fin.
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