Riddles & Brain Teasers

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Author	Topic: Riddles & Brain Teasers
The Immortal BANNED Member # 4	written Wednesday, October 26 2005 05:55 Profile Homepage #25 CO2 poisoning. Was für ein ein Besonderwagen fährt er? ;) -------------------- 人たちを燃えるために俺はかれらに火を上げるから死んだ Posts: 6936 \| Registered: Tuesday, September 18 2001 07:00
Khoth Post Navel Trauma ^_^ Member # 67	written Wednesday, October 26 2005 06:46 Profile Homepage #26 9.5 people is possible, but not like that. -------------------- Barcoorah: I even did it to a big dorset ram. desperance.net - Don't follow this link Posts: 1798 \| Registered: Thursday, October 4 2001 07:00
Student of Trinity Electric Sheep One Member # 3431	written Wednesday, October 26 2005 07:20 Profile #27 It was an electric car. -------------------- It is not enough to discover how things seem to seem. We must discover how things really seem. Posts: 3335 \| Registered: Thursday, September 4 2003 07:00
Or else o'erleap. Law Bringer Member # 335	written Wednesday, October 26 2005 08:06 Profile Homepage #28 Doesn't inhalation of exhaust fumes kill by anoxia from CO? In any case, while a corpse doesn't breathe can won't show any sign of CO inhalation, I'm not sure how "immediately" the police can test it. An electric car is a good answer, though. So is walking into the garage by a side door, not by opening the garage itself, and not falling unconscious or dying. The key to Khoth's problem is that the first prisoner has no information and is a 50% loss anyway. No other prisoner can convey information, because each answer is dependent on the known hat. That means that the first prisoner has to have a way to tell everyone else what their hats are. If the first prisoner counts the number of black hats he sees and says "black" if it's even and "white" if it's odd, each subsequent prisoner can count the number of black hats in front of him and knows how many went before. If he knows the other hats add up to an odd number of blacks and the first prisoner said even, his hat must be black. If the first said odd, his hat must be white. —Alorael, who has another question. If someone lays out a 10 by 10 grid of tiles with random numbers written on them, then lets you flip over as many tiles as you'd like in any order that you'd like, what is your chance of revealing the highest tile last (assuming you're trying to do so)? Why do these brain teasers on Spiderweb always end up mathematical? Posts: 14579 \| Registered: Saturday, December 1 2001 08:00
Zeviz Nuke and Pave Member # 24	written Wednesday, October 26 2005 09:12 Profile Homepage #29 For Khoth's puzzle, I still don't see how a person can effectively convey information to more than one other person, or both convey information and use information simultaneously. All schemes I've tried (telling majority colors for groups of various size, telling majority colors for sliding groups of 3, etc.) lead to worse results than just having every other person tell the color of the hat in front of him, when you work out the probabilities. Alorael, if the last guy just says majority color for the first 9, that might tell person 9 the color of his hat, but only if he sees equal number of black and white hats ahead of him. Otherwise, if he already sees 5 or more hats of one color ahead, his hat could be either part of a very large majority or a large minority. -------------------- Be careful with a word, as you would with a sword, For it too has the power to kill. However well placed word, unlike a well placed sword, Can also have the power to heal. Posts: 2649 \| Registered: Wednesday, October 3 2001 07:00
Khoth Post Navel Trauma ^_^ Member # 67	written Wednesday, October 26 2005 09:32 Profile Homepage #30 Alorael got it. Zeviz, it's not majority, it's parity. Another one: A group of five pirates has stolen 1000 gold pieces, and now they have to divide up the loot. They use the following method: The pirates are numbered from 1 to 5. Pirate 1 then suggests a loot distribution (eg "200gp each"), and they vote on whether to accept it. If the majority vote for it, that's what happens, otherwise pirate 1 is killed and pirate 2 suggests a division among the remaining four, which is then voted on, etc. If there is a tie, the result is taken to be "yes". All the pirates are very good at logic, and are totally greedy and bloodthirsty. They will vote in the way that gives them the most gold, and if it doesn't make a difference either way, they will vote to kill because they find that more fun. If called upon to suggest a distribution, they will pick the one that is best for them. How does the gold get divided? [ Wednesday, October 26, 2005 09:42: Message edited by: Khoth ] -------------------- Barcoorah: I even did it to a big dorset ram. desperance.net - Don't follow this link Posts: 1798 \| Registered: Thursday, October 4 2001 07:00
Drew By Committee Member # 4233	written Wednesday, October 26 2005 09:46 Profile #31 Pirates 4 and 5 would remain, and pirate 4 would end up with it all. EDIT: Also, an electric car isn't a good answer. At least in my experience with a hybrid electric car, the engine doesn't start running until the acceleration pedal is depressed. [ Wednesday, October 26, 2005 09:48: Message edited by: Drew ] Posts: 2242 \| Registered: Saturday, April 10 2004 07:00
Zeviz Nuke and Pave Member # 24	written Wednesday, October 26 2005 09:48 Profile Homepage #32 Looks like I didn't read Alorael's idea correctly. And I didn't even do the calculations for my ideas correctly either. Giving majority for a group of 3 works better than telling the guy ahead of you his color, if the guy ahead calls minority color whenever he sees 2 majority colors ahead. As for Khoth's new puzzle, I've seen it before, but in that version a tie meant "no", so the solution I remember needs a slight modification. EDIT: Drew, here is a hint: you are on the right track, now think how pirate 3 could use this to his advantage. [ Wednesday, October 26, 2005 09:55: Message edited by: Zeviz ] -------------------- Be careful with a word, as you would with a sword, For it too has the power to kill. However well placed word, unlike a well placed sword, Can also have the power to heal. Posts: 2649 \| Registered: Wednesday, October 3 2001 07:00
Drew By Committee Member # 4233	written Wednesday, October 26 2005 10:27 Profile #33 How about pirate 2 proposing to split the loot 998 for himself and 2 for pirate 5? 2 is more than pirate 5 will see otherwise, so he'd vote for it, which would create a draw, settling the deal. EDIT: Of course, if pirate four were aware of that, and pirate one was savvy enough, all pirate one would have to do is offer pirate five 3 gp, pirate four 1 gp, and keep the rest for himself. Because this amount is more than either pirate four or five will see otherwise per the rules, they will take it. Thus, all five pirates live, and pirates two and three will be screwed. [ Wednesday, October 26, 2005 11:25: Message edited by: Drew ] Posts: 2242 \| Registered: Saturday, April 10 2004 07:00
Zeviz Nuke and Pave Member # 24	written Wednesday, October 26 2005 12:28 Profile Homepage #34 You are almost there. Just think of whether pirate 1 can save even more money somehow. Alorael, for your riddle, do we know anything about the numbers written on the tiles, or are they 100 completely random integers, with no restriction on value or placement? -------------------- Be careful with a word, as you would with a sword, For it too has the power to kill. However well placed word, unlike a well placed sword, Can also have the power to heal. Posts: 2649 \| Registered: Wednesday, October 3 2001 07:00
wz. As Infiltrator Member # 3040	written Wednesday, October 26 2005 12:58 Profile #35 Pirate 1: 998 Pirate 2: 0 Pirate 3: 1 Pirate 4: 0 Pirate 5: 1 Assume it's down to Pirates 4 and 5. 4 can just keep all the money, and 5 won't get anything. So 3 needs to convince 5 to vote with him, so he'll offer one coin to pirate 5 and keep the rest himself. Pirate 2, knowing that 3 will screw over 4, would offer 4 one gold piece for his vote and keep the rest. Pirate 1 needs votes from Pirates 3 and 5, and gets the votes by offering them one piece each, which is one more than they'd get from 2's plan. [ Wednesday, October 26, 2005 13:01: Message edited by: wz. arsenic... ] -------------------- 5.0.1.0.0.0.0.1.0... Posts: 508 \| Registered: Thursday, May 29 2003 07:00
Or else o'erleap. Law Bringer Member # 335	written Wednesday, October 26 2005 14:18 Profile Homepage #36 If there were a pattern to the numbers, you could figure out which number would be highest in the pattern and keep flipping tiles until you found it. They're all random. In fact, I can't think of any reason why they have to be integers, or even rational. —Alorael, who still knows what to do after flipping over e, 2.998 x 10^8, ?, the Boltzmann constant, 42, -7 Planck time, and the square root of Avogadro's number. Posts: 14579 \| Registered: Saturday, December 1 2001 08:00
Ash Lael E Equals MC What!!!! Member # 5491	written Wednesday, October 26 2005 14:22 Profile Homepage #37 Given that if you truly pick a number at random it would be very difficult to write it down, I'd just see which tiles were outlandishly large. :P -------------------- Sex is easier than love. Posts: 1861 \| Registered: Friday, February 11 2005 08:00
Zeviz Nuke and Pave Member # 24	written Wednesday, October 26 2005 18:02 Profile Homepage #38 A random guess for Alorael's puzzle: Flip half of the tiles, then keep going until you get something that's larger than anything seen so far. This will guarantee that your biggest number is bigger than at least half the numbers on the tiles. (Unless the biggest number was in the first half of the tiles.) -------------------- Be careful with a word, as you would with a sword, For it too has the power to kill. However well placed word, unlike a well placed sword, Can also have the power to heal. Posts: 2649 \| Registered: Wednesday, October 3 2001 07:00
Or else o'erleap. Law Bringer Member # 335	written Wednesday, October 26 2005 18:18 Profile Homepage #39 That's the best I can come up with, although I'm not sure it's absolutely the best. —Alorael, who isn't even sure that he calculated a 25% chance of getting the highest tile that way correctly. Posts: 14579 \| Registered: Saturday, December 1 2001 08:00
Thuryl ...b10010b... Member # 869	written Wednesday, October 26 2005 18:42 Profile Homepage #40 I've heard that in order to maximise the chance that you'll get the highest tile last, the correct fraction of tiles to flip before settling on the highest tile you find after that is 37.5%. -------------------- My BoE Page Bandwagons are fun! Roots Hunted! Posts: 9973 \| Registered: Saturday, March 30 2002 08:00
Arancaytar Law Bringer Member # 2984	written Wednesday, October 26 2005 19:43 Profile Homepage #41 quote: Originally written by Student of Trinity: It was an electric car. Yup. :P (TM, I like the word "Besonderwagen". It should be a real word, because it sounds good.) [ Wednesday, October 26, 2005 19:46: Message edited by: Noreni ] -------------------- The Encyclopaedia Ermariana <-- Now a Wiki! "Polaris leers down from the black vault, winking hideously like an insane watching eye which strives to convey some strange message, yet recalls nothing save that it once had a message to convey." --- HP Lovecraft. "I single Aran out due to his nasty temperament, and his superior intellect." --- SupaNik Posts: 8752 \| Registered: Wednesday, May 14 2003 07:00
Spring Shaper Member # 5450	written Wednesday, October 26 2005 22:28 Profile Homepage #42 quote: Originally written by Thuryl: That's hardly even a puzzle. You can't add the amounts that are left in the account after each withdrawal and expect them to add to $50 -- that's just silly. It's only a carefully-crafted coincidence that the sum even happened to be close to $50 in this particular case, making you think there's some connection between the amounts when there isn't. (It's easy to see this by giving other examples. If you withdraw the $50 all at once instead of in little chunks, for example, the sum of amounts left after each withdrawal is obviously $0. If you withdraw it $1 at a time, the sum of amounts left after each withdrawal is, well, a lot more than $50.) It was in a Ripleys Believe It Or Not book, in my school library. You are right, though. If you took the money out in $1 lots, it would total well over $50. EDIT: $1275, to be precise. [ Wednesday, October 26, 2005 22:29: Message edited by: Spring ] -------------------- I'll put a Spring in your step. Polaris Posts: 2396 \| Registered: Saturday, January 29 2005 08:00
Khoth Post Navel Trauma ^_^ Member # 67	written Wednesday, October 26 2005 22:37 Profile Homepage #43 A group of monks are living in an isolated monastery. Some of the monks have red eyes. They are cursed, and if they find out that their eyes are red, they must kill themselves at midnight. However, there are no mirrors, and the monks have taken a vow of silence so they can't tell each other their eye colour, so the red-eyed monks have coexisted with the others for years. But one day, when all the monks are gathered for their lunch, a tourist wanders in, unaware of the curse and the vow of silence, and says "At least one of you has red eyes". What happens? -------------------- Barcoorah: I even did it to a big dorset ram. desperance.net - Don't follow this link Posts: 1798 \| Registered: Thursday, October 4 2001 07:00
Thuryl ...b10010b... Member # 869	written Wednesday, October 26 2005 23:45 Profile Homepage #44 The following answer assumes that the monks are all perfectly rational, all know that all the other monks are perfectly rational, all see everyone else's eyes every day, all believe the tourist's statement, all know that all the other monks believe the tourist's statement, all fully intend to kill themselves if they know they have red eyes, and all know that all the other monks fully intend to kill themselves if they know they have red eyes. It also assumes that none of the monks will ever kill themselves for any reason other than finding out that they have red eyes, and that all of the monks know this. Tell me if there are any assumptions that I missed. The red-eyed monks all commit suicide eventually, and all of them do it on the same night. Taking the first night after the tourist's announcement as night 1, the night on which they all kill themselves is the night equal to the number of red-eyed monks. Why? Well, if there's only one red-eyed monk, he can see that he must be the only red-eyed monk, since nobody else has red eyes. So he kills himself on night 1. If there are two red-eyed monks, then after the other one hasn't killed himself on the first night, each red-eyed monk knows that the one red-eyed monk he can see isn't the only red-eyed monk in the monastery (otherwise that monk would have killed himself), so he must be the other one. If there are three red-eyed monks, then after the other two haven't killed themselves on the second night, each red-eyed monk knows that the two he can see aren't the only red-eyed monks in the monastery (otherwise they'd both have killed themselves), so he must be the other one. ... and so on. [ Wednesday, October 26, 2005 23:56: Message edited by: Explode Thuryl Now ] -------------------- My BoE Page Bandwagons are fun! Roots Hunted! Posts: 9973 \| Registered: Saturday, March 30 2002 08:00
Student of Trinity Electric Sheep One Member # 3431	written Thursday, October 27 2005 00:58 Profile #45 Suppose in fact there are no red-eyed monks. All the monks kill themselves on the very first night because the tourist lied. He lied in order to loot the monastery. For this crime he is cursed with red eyes. He doesn't care. Alternatively, so many white-eyed monks are trying to kill themselves that at least one is prevented by a brother monk who knows he has white eyes. He kills the tourist to avenge his brothers. For this crime he is cursed with red eyes. He doesn't know. [ Thursday, October 27, 2005 01:01: Message edited by: Student of Trinity ] -------------------- It is not enough to discover how things seem to seem. We must discover how things really seem. Posts: 3335 \| Registered: Thursday, September 4 2003 07:00
Thuryl ...b10010b... Member # 869	written Thursday, October 27 2005 01:11 Profile Homepage #46 Okay, so it's also necessary to assume that the tourist's statement was in fact true. Whoops. :P -------------------- My BoE Page Bandwagons are fun! Roots Hunted! Posts: 9973 \| Registered: Saturday, March 30 2002 08:00
Spring Shaper Member # 5450	written Thursday, October 27 2005 01:27 Profile Homepage #47 quote: Originally written by Student of Trinity: Alternatively, so many white-eyed monks are trying to kill themselves that at least one is prevented by a brother monk who knows he has white eyes. He kills the tourist to avenge his brothers. For this crime he is cursed with red eyes. He doesn't know. The world is divided into red-eyed monks and white-eyed monks, now, is it? :P -------------------- I'll put a Spring in your step. Polaris Posts: 2396 \| Registered: Saturday, January 29 2005 08:00
Student of Trinity Electric Sheep One Member # 3431	written Thursday, October 27 2005 01:33 Profile #48 The puzzle as posed did stipulate that there were some red-eyed monks. I just think it's interesting to consider the possibility of none. It's an interesting feature of this puzzle, like the one about the prisoners, that it ignores realistic communication. The answer about 'Blaaaack' versus 'Black', for instance, would be perfectly good in reality; it's just not in the intended spirit of the puzzle. And here it would take monks who were awfully meticulous about an awfully broad interpretation of their vow of silence, never to indicate, by the faintest flicker of an eyelash, that some brothers were cursed. Hmmm. -------------------- It is not enough to discover how things seem to seem. We must discover how things really seem. Posts: 3335 \| Registered: Thursday, September 4 2003 07:00
Arancaytar Law Bringer Member # 2984	written Thursday, October 27 2005 01:43 Profile Homepage #49 A perfect example of total induction! If there is one monk with red eyes, he knows he must be the only one, so he kills himself on night one. Assume that if there are n monks with red eyes, they kill themselves on night n. It remains to be shown that from the assumption follows: If there are n+1 monks with red eyes, they kill themselves on night n+1. The monks know no more or less than we know. They behave perfectly logically, and know that all the others do so as well. Picture therefore an n+1th monk seeing the entire group of n red-eyed monks before him. He knows that n monks will kill themselves on night n. When they do not, he knows that the group is larger than n, and must include him. He will kill himself at the next night after this, which is n+1. Therefore, if the premise is true for n red-eyed monks, it is true for n+1 red-eyed monks. Since it is true for one red-eyed monk, it is true for any natural number of red-eyed monks. :) --- And if there are thousands of monks in the monastery, then the tourist will be long gone before his terrible statement has any effect. Then suddenly, the headlines will read "mass suicide in monastery!" And the tourist will think: "Hey, isn't that the quaint little place I went on vacation to ten years ago?" :D -------------------- The Encyclopaedia Ermariana <-- Now a Wiki! "Polaris leers down from the black vault, winking hideously like an insane watching eye which strives to convey some strange message, yet recalls nothing save that it once had a message to convey." --- HP Lovecraft. "I single Aran out due to his nasty temperament, and his superior intellect." --- SupaNik Posts: 8752 \| Registered: Wednesday, May 14 2003 07:00

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