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| Riddle Answering; Someone asks a riddle, the one who answeres correctly has to ask the next. | |
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| Tweet Topic Started: Aug 20 2010, 11:31 AM (12,849 Views) | |
| Blique | Sep 13 2010, 03:10 AM Post #101 |
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Jack-of-All-Trades and Such
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There are nine men, yet you say "Not necessarily 5 and 5"? Should there be nine or ten? @Shiki: In American stereotypes, both Scottish and Irish have red hair. =P The Irish are the green-wearing, shamrock-holding leprechauns and the Scottish wear plaid kilts and play bagpipes. Edited by Blique, Sep 13 2010, 03:15 AM.
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| amginE | Sep 13 2010, 09:14 AM Post #102 |
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bah, "not necessarily 4 and 5." |
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| kycse | Sep 14 2010, 02:05 AM Post #103 |
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I can save 7 (with luck more) of the 9, 2 have to be sacrificed. O < O < O < O < O < O < O < S < S O = saved ones S = sacrifice They discussed, that two people have to sacrifice themselves. With seven people, one type (bunny or cat) has less or equals 3 (7 to 0, 6 to 1, 5 to 2, 4 to 3, rest repeats). Now if they count the hats of the seven people with the lesser number, which is always less than 4. Thus you can tell them in 2 binary numbers. If there are 3, both answer with that hat type ( two times bunny or two times at), if there are two, only the second one answers cat or bunny. With one, only the first one answers cat or bunny and with a 7 to 0 ratio, they say two times the other hat type. Now the seven remaining ones can count that number down by seeing the hats before them and listening to the ones behind them. If the current number - the number of hats of the same type = 1, then say that type of hat and at the same time everyone else subtracts 1 from the current number. If it's 0, only the other type is remaining. Same for the calculation of the hats. If the current number is 3 and they see 3 hats, they have the other type. Example: C < C < B < B < C < B < C < B < C C = cat B = bunny Bunny is less than Cat. We have 3 Bunnies. (answering backwards) #0 answers Bunny #1 answers Bunny => All the other ones know there are 3 Bunnies. #2 answers Cat, because he sees 3 Bunnies and the vurrent number of bunnies is three. #3 answers Bunnie, because he sees 2 Bunnies and the number of current bunnies is thee. Thus the current number of bunnies becomes two. #4 answers Cat, because he sees 2 bunnies. #5 answers Bunnies, because he sees one bunny. Number of bunnies reduce by 1 #6 answers bunny, because he doesn't sees any bunny. Number of current bunnies is 0 Everyone else answers Cat, because there are no bunnies left. This way 8 People are saved in my example, because #1 got lucky and has the right hat on him. Generally they have a 1/4 chance to save 7, 1/2 chance to save 8 and 1/4 chance to save 9, resulting in an average of 8 people saved. (Math is fun) |
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| amginE | Sep 14 2010, 09:15 AM Post #104 |
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wow kycse, you really went overboard there...! i'm going to give you credit since you somehow found your own solution which was very close...but the correct answer and solution is as follows: you only need to sacrifice the last guy (who is impossible to save anyway since nobody can see his hat). how? the 9th guy takes a look at the 8 hats in front of him. if there's an even number of bunnies, he calls out bunny. if it's an odd amount, he calls out cat. let's say he sees 4 bunnies and 4 cats. he calls out bunny. the eighth guy sees 7 hats in front of him and knows that the 9th guy saw an even amount of bunnies. if he sees 4 bunnies as well, then it must follow that he has a cat hat. if he sees 3 bunnies, he must be the fourth bunny. So he calls out the correct one (let's say he's a cat) and is saved. the seventh guy knows out of the seven people remaining including himself, there's an even amount of bunnies (since #9 called bunny and because #8 was a cat). So if he sees an even amount of bunnies, he himself must be a cat. If he sees an odd number of bunnies, he must be the remaining bunny. this continues on until everyone (except the last guy who gets a 50/50 chance) is saved. the beauty of this answer is that you could have 100 people and it would still work with only one sacrificed, as long as they all have eaglevision and crazy good memory. another cool thing is that if someone in the middle messes up and dies, everyone else after him can still recover knowing that the dead guy was the opposite of what he said he was. the bad thing is that if either the last person or second to last person messes up, then everything goes kaput. because the 8th person will die, and everyone would be left wondering if it's person 8 or person 9 who messed up. (person 7 and everyone before him would then need to base their decisions on who they think messed up). even more interesting is the case when person 9 and 8 both mess up, in which 8 will live and 7 (who doesn't mess up) will die a very confused man. your turn ky =] |
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| kycse | Sep 14 2010, 10:58 AM Post #105 |
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Err. my answer was way off, still your turn... |
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| amginE | Sep 14 2010, 11:50 AM Post #106 |
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no way man. your answer was really unique and creative! i liked it a lot. and it was only 1 away from optimal. you definitely deserve full credit. so go ahead. |
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| kycse | Sep 14 2010, 12:40 PM Post #107 |
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(7 isn't equal to 8, so I don't think so. Besides it was just a answer based on binary numbers again) If you say so, but now I have to think of one... When traveling the world to the east, and going around once within 10 days and today is the 12th of the month, which day is it when I return home? |
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| amginE | Sep 14 2010, 02:26 PM Post #108 |
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when you say "going once within 10 days" does it mean if you push a stopwatch when you start and push it again when you return, it'll show 240 hours? if so, then it'll just be straight up 12+10 = 22nd, since 240 hours is 240 hours no matter where you are on the earth. if it means "you will see 10 sunrises during your journey" then it would be 21st, since by going against the direction of the sun, you will have seen one more sunrise during your journey relative to someone who stayed at home and didn't move. either that, or there's a trick somewhere that i didnt notice... |
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| kycse | Sep 14 2010, 02:35 PM Post #109 |
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No, you are right. it's the 21st then (I just noticed I used east instead of west...) I wanted to ask the inverse to the "80 days around the world" one, which would've been west and you would have to add a day instead of subtract. Anyways, your turn. |
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| amginE | Sep 14 2010, 03:47 PM Post #110 |
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A nice easy one...(?) You have a nice big birthday cake. You want to cut it up into as many pieces as possible, using only three straight cuts of a knife. How many pieces can you get at most? |
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| Blique | Sep 14 2010, 04:21 PM Post #111 |
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Jack-of-All-Trades and Such
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Eight? For the third cut, you line up the four pieces in a row. |
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| amginE | Sep 14 2010, 04:32 PM Post #112 |
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ah. thats right. some people would say 7 since they couldn't think in 3-d. basically, two normal cuts making fourths, then a third cut through the middle of the cake, splitting it into top and bottom, making 8. good job. |
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| Blique | Sep 14 2010, 05:39 PM Post #113 |
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Jack-of-All-Trades and Such
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Ooo, I wouldn't have thought of doing it that way... So is it my turn? |
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| amginE | Sep 14 2010, 07:08 PM Post #114 |
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yes |
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| Blique | Sep 15 2010, 11:11 PM Post #115 |
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Jack-of-All-Trades and Such
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A bookworm eats a straight path through an encyclopedia consisting of ten parts (books), all next to each other neatly on a shelf. Each part has 1000 pages. The bookworm starts on the front cover of the first part and ends on the back cover of the last part. How many pages did the bookworm eat through? |
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| kycse | Sep 16 2010, 01:39 AM Post #116 |
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Depending on the alignment of books: 8000, 9000 or 10000 pages. |
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| Blique | Sep 16 2010, 02:09 PM Post #117 |
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Jack-of-All-Trades and Such
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Assuming part one is on the left and part ten is on the right, 8000. Your turn, then~
Edited by Blique, Sep 16 2010, 02:09 PM.
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| kycse | Sep 17 2010, 03:19 PM Post #118 |
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Yea :> Well, it's well known, but I don't have any better idea right now, so: It cannot be seen, cannot be felt, Cannot be heard, cannot be smelt. It lies behind stars and under hills, And empty holes it fills. It comes first and follows after, Ends life, kills laughter. What is it? (Assuming someone knows this riddle, please tell them whether they are right or not, since I can't go online for the next few days.) |
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| guilover | Sep 17 2010, 11:22 PM Post #119 |
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i'm going to guess... space? lol. |
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| Shiki | Sep 18 2010, 12:31 AM Post #120 |
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My hovercraft is full of eels.
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It's not space. Very close though. Space can't really end life or kill laughter. -knows the answer- |
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| guilover | Sep 18 2010, 10:53 PM Post #121 |
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if you know it and don't say it, you're just trying to avoid coming up with a riddle for people to solve. is it darkness? lack of air? nothing? WHAT IS IT?!?!?!? |
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| Shiki | Sep 19 2010, 06:01 AM Post #122 |
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My hovercraft is full of eels.
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I don't want to go find a riddle. I'll have to spend a long time picking through google to find a good one. And that's no good. But if I know the answer, it feels a bit like cheating; riddles are supposed to be worked out, so it's best to leave it for the people who don't know the answer. Kycse said he wasn't going to be a here for a few days, so I'm just keeping the topic going xP But you just got it right just there. Darkness. |
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| guilover | Sep 19 2010, 05:14 PM Post #123 |
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but it's my turn now... so we don't need kysce. hmm, what to do... You throw away the outside and cook the inside. Then you eat the outside and throw away the inside. What did you eat? |
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| amginE | Sep 19 2010, 10:20 PM Post #124 |
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um, corn? |
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| Blique | Sep 20 2010, 01:40 PM Post #125 |
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Jack-of-All-Trades and Such
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Yup, corn. |
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| kycse | Sep 20 2010, 02:08 PM Post #126 |
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Back, yup it was Darkness and... @guilover: "so we don't need kysce.", I'm quite saddened by that xD |
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| amginE | Sep 20 2010, 09:05 PM Post #127 |
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here is one that will make you rack your brains. you are on a date with the magician. you couldn't decide amongst the filet mignon, the veal, and the chicken salad. your magician date offers to help by offering you ONE coin (because magicians are stingy like that) to help you decide randomly. the magician has all sorts of coins, fair and unfair, but didn't bring any that would land on its side (so only heads or tails). What kind of coin (fair or unfair, if unfair, what's the ratio of head/tails) should you take and what strategy should you use to minimize the average number of flips you would need to decide amongst the three dishes? Bonus if you also actually figure out the average number of flips taken using your strategy. |
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| kycse | Sep 21 2010, 02:39 AM Post #128 |
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Practitioner
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Oh dear, that's quite the math riddle. Binomial Distribution with unknown alpha and p. Warning! The following explanation is lengthy and probably kinda complicated! If "cheating" is allowed: 2 flips with a fair coin, the order is important: head, head --> nothing head, tail --> filet mignon tail, head --> veal tail, tail --> chicken salad. Another version could be: if you get head, head you have to repeat the whole thing... So we get 3/4 * 2 (flips) + 3 / 16 * (4 flips) + 3 / 64 * 6 (flips) + 3 / (4 * the one before) * (2 + flips before) and so on.... Short form: Sum from k = 0 to unlimited ( 3 / 4 power ( k + 1)) * (2 + 2k) (I'm too lazy to calculate, so: Solution of the Sum , this should do :>) Which equals to 2 2/3 flips. This solution is based on redoing the process to remove the unfavorable outcome. However I believe that isn't allowed, now is it? Well, with 2 coins getting a percentage of 1 third with the least flips is kind of interesting. It should be pretty much impossible with a fair coin, since it just has two stages of equal chance. And each flip multiplies the possibilities by 2, so it's 2 power n, n is the number of flips. And that is NEVER divideable by 3 (due to prime factorization). Doing this with one flip is impossible as well, since you can just get two states... So the times you have to flip the coins is at least 2 times. And the percentages of the three states have to be equal. Since my previous answer was 2 2/3, I have to do it this time with exactly 2. Therefore must two outcomes be the same as the other ones ==> outcome X + outcome Y = 1 third outcome Z: 1 third Since there are two times "outcome Z" (head, tails and tails, head) the second third is solved by Z. Now we have to find a percentage pair that resolves this: 0 < p, q < 1 (p and q are elements of the irrational numbers) p*p + q*q = 1/3 p*q = 1/3 Let's try this with rational numbers first: 1/3 is the same as 2 /6 and 3 / 9 = 4 / 12 = 5 / 15 and so on. This poses a problem however: p + q = 1, (since either p or q has to be an outcome), p = 1 - q. For rational numbers that means: the denumeratiors are the same for both numbers, and the numerators are: p's numerator: (denumerator - numerator q), which means we can only(!) create numbers whose square of the denumerator equals a number, divideable by three. Which would be: 3, 6 , 9... etc., since (as I mentioned earlier) numbers are only dividable by three if one factor of the prime factorization is 3. Therefore we can only make numbers like 9 (3*3), 36 (6*6) 81(9*9), which in turn poses a problem: 12 / 36 would be 1 / 3. However there are no numerators, that will create 12. ( 1 * 5 = 5, 4 * 2 = 8, 3 * 3 = 9) 27 / 81 would be the next, however there are again no numerators, that solve that equation. (I don't want to prove this right now, but believe me it's impossible to find numerators the solve the equations: 1 = p + q and 1 / 3 = p * q, at least not within the rational numbers.) Therefore a solution within the rational numbers is impossible! And I don't want to try the irrational numbers or complex numbers... (so I'll just ask wolfram alpha again... Solution of the system , sigh complex solution...) Therefore it's impossible to get a solution with 2 flips. And any other solution would have 3 or more flips. Therefore is my answer 2 2/3. Edited by kycse, Sep 21 2010, 07:23 AM.
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| amginE | Sep 21 2010, 09:13 AM Post #129 |
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hey ky, there's a slightly better answer than 2 2/3...but you need to think a little outside the box. hint: there must be a reason it's a magician and not just some random person, right? there's one unfair coin that will yield better results than a fair coin. and anything, anything is allowed, as long as each of the three choices yield a 1/3 chance overall. |
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| kycse | Sep 21 2010, 11:04 AM Post #130 |
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Okay, if anything is allowed, it's allowed that the coin switches it's percentages between the first and the second flip, right? Then I would assume: First flip: 1:2 ratio, second flip: 1:1 ratio (if you get the 2/3 side), if you flip 1/3 you won't flip again. Assuming head has 1/3 chance on the first flip and tail 2/3: So, if you flip head, you stop and choose filet mignom (well whatever would doo as well) If you flip tail, you flip again with 1/2 for tail and head. If you flip head you take the veal and if you flip tail you take the chicken salad. For the percentages: the chance to get the filet mignom is obviously 1/3. As for the other ones: you have a 2/3 chance to get into their arc. Then this chance is multiplied by 1/2, results in 2/6 each. and 2/6 is 1/3 (as stated in my previous post). Thus every choice has the same chance. As for the Average: 1/3 * 1 (flip) + 2 / 3 * 2 (flips) = 1 2/3 flips. At least, if that is allowed (well, based on your statement it is..) (amgine: as mathematican you think in formulas, (at least most of the time), there is usually no "outside the box", which my post should have made clear [yea a solution within the complex numbers, I calculated it just for fun sometime later today]) Edited by kycse, Sep 21 2010, 11:08 AM.
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| amginE | Sep 21 2010, 01:44 PM Post #131 |
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er...okay, i take that back. a coin that changes its chances is not |
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| kycse | Sep 21 2010, 03:39 PM Post #132 |
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Then he switches the coin, he's a magician after all. Or use the percentage of 1/2 +- i/squareroot(12)... Edited by kycse, Sep 21 2010, 03:58 PM.
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| amginE | Sep 21 2010, 03:57 PM Post #133 |
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no coin switching and percentages add up to 100%. |
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| kycse | Sep 21 2010, 04:17 PM Post #134 |
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Sigh, I give up... I can't solve it mathematically except within the complex numbers. For anyone who wants to try amgines riddle, those are the facts I think are at least right: 1. You have to flip the coin 2 times, no less, no more. 2. Appearently it's not a percentage of 1:1 3. Each result has to have the value of 1/3 4. The two percentages of head and tail have to be between 0 and 1 5. They are not rational. 6. The coin can't switch it's percentage Don't know whether they are all right, but they should be. |
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| guilover | Sep 21 2010, 04:48 PM Post #135 |
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answer: dump the magician, he's too tricky. hook up with the waiter, and ask him what you should order. go and live your life in happiness, never putting yourself in such an awkward situation again and proving that your math teachers actually served any sort of purpose. |
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| amginE | Sep 21 2010, 06:18 PM Post #136 |
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haha. i'm just going to post the answer. . . . you take a coin that lands 1/3 heads and 2/3 tails (or vice versa). first flip, if heads then dish A. else it's a choice between dish B and C. if you flip tails, you need to flip two more times. Possible results and decision: HH = 1/3*1/3 = 1/9 -> redo HT = 1/3*2/3 = 2/9 -> dish B TH = 2/3*1/3 = 2/9 -> dish B TT = 2/3*2/3 = 4/9 -> dish C This gives you an 8/9 chance of resolving between dish B and C on the third flip. If you're unlucky, then you will need to do this again, two more flips to decide between B and C. So the final list of flips 1 flip: 1/3 chance 3 flips: 2/3*8/9 chance 5 flips: 2/3*8/9*1/9 chance 7 flips: 2/3*8/9*(1/9)^2 chance and so on... The chances should add up to 1 as it should. The average # of flips turns out to be 2.5, slightly better than what you can get with a fair coin. |
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| guilover | Sep 22 2010, 12:54 AM Post #137 |
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so... who's next? |
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| amginE | Sep 22 2010, 09:25 AM Post #138 |
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ok. let me get a non-math one, since most you guys can't seem to stand numbers. which of these words do not belong and why? please don't go for the obvious answer. diaper nametag departer dogmatic deliver |
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| Blique | Sep 22 2010, 09:09 PM Post #139 |
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Jack-of-All-Trades and Such
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"deliver", because there aren't any letters that hang down?
Edited by Blique, Sep 22 2010, 09:10 PM.
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| Shiki | Sep 22 2010, 09:36 PM Post #140 |
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My hovercraft is full of eels.
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Dogmatic. The only word that doesn't make a word when spelt backwards. I almost freaked out, I couldn't decide if retraped was a word. Edit: I found the word for them is palindromes xD -really didn't know the proper name- Edited by Shiki, Sep 22 2010, 09:47 PM.
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| amginE | Sep 22 2010, 09:54 PM Post #141 |
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palindromes are actually words that are the same forwards and backwards, like mom or level. some people call these words, which spell sometheing different backwards, "semordnilap" (palindromes spelled backward), but i dont think it has an official name. anyway, you're correct, so it's your turn. |
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| Shiki | Sep 23 2010, 06:16 PM Post #142 |
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My hovercraft is full of eels.
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I couldn't remember the phrasing for the riddle I wanted to use. And not got the time to go hunting for it, I'll try have it for the next turn I get. So have a fail riddle. Only one color, but not one size, Stuck at the bottom, yet easily flies. Present in sun, but not in rain, Doing no harm, and feeling no pain. What is it. |
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| guilover | Sep 23 2010, 07:21 PM Post #143 |
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a shadow. right? |
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| Shiki | Sep 23 2010, 10:27 PM Post #144 |
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My hovercraft is full of eels.
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Yusssssssss. Your turn. |
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| guilover | Sep 25 2010, 11:49 PM Post #145 |
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me no wanna go... ok, delegation time. what is the 1st decimal place of pi? |
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| apoptoxin4869 | Sep 25 2010, 11:55 PM Post #146 |
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Ruler of the flying squirrels
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what is black and white and red all over? XD since you didnt want a riddle |
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| Blique | Sep 26 2010, 07:06 AM Post #147 |
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Jack-of-All-Trades and Such
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First decimal place? ...1? A penguin with a rash, a shy panda, or a newspaper. |
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| kycse | Sep 30 2010, 02:49 PM Post #148 |
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a black and white and red <insert nearly anything here>. Mouldy black pudding |
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| guilover | Sep 30 2010, 05:07 PM Post #149 |
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well, pretty much anyone who wants to take this can. |
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| apoptoxin4869 | Oct 2 2010, 12:56 AM Post #150 |
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Ruler of the flying squirrels
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LOL agreed XD i didnt really think of an answer so i think kycse gets this round XD |
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