Trahern Posted May 28, 2008 Report Share Posted May 28, 2008 No, it's not what I have in mind, but it's very similar. Try something a bit more "elemental ". Hmm... there are no E's? @What am I? A mosquito! Link to comment Share on other sites More sharing options...
Sacosphilz Posted May 28, 2008 Report Share Posted May 28, 2008 Hmm...there are no E's? @What am I? A mosquito! You got both right. That paragraph is unusual because it is lacking the most frequently used letter in English, E. Link to comment Share on other sites More sharing options...
Grido Posted May 28, 2008 Report Share Posted May 28, 2008 my ans was for the uncrossed riddley btw EDIT oh looked at diff para oops Link to comment Share on other sites More sharing options...
Sacosphilz Posted May 29, 2008 Report Share Posted May 29, 2008 @my maths puzzle: I don't know if anyone's still working on it , but here's a hint: All possible solutions to the solvable settings have a few things in common. If you can solve one or more of them, you can use a similar method to solve the rest. My friends at the other forum manage to solve it already. One of them summed up the nature of the solutions quite nicely. Exact quote: assumptions ใน operator สามอันที่ต้องเลือกมี / แน่นอน (ไม่งั้นคงคิดออกไปนานแล้ว) operator ที่ทำอันสุดท้ายต้องเป็น *,/ (เพื่อให้เศษส่วนที่เกิดขึ้นระหว่างทางกลับเป็นจำนวนเต็ม) อีก operator นึงเป็น +,- (สังเกตจากคำตอบของโจทย์ข้อที่คิดได้แล้ว) Hmm.... what? What do you mean, you can't read it? Link to comment Share on other sites More sharing options...
Grido Posted May 29, 2008 Report Share Posted May 29, 2008 it's all (not) Greek to me Link to comment Share on other sites More sharing options...
Trahern Posted May 29, 2008 Report Share Posted May 29, 2008 1,7,8,9 -> 36 8/(1-(7/9)) = 36 3,3,8,8 -> 24 8/(3-(8/3)) = 24 1,4,5,6 -> 24 6/((5/4)-1) = 24 I'll edit as I come up with more... 1,2,8,9 -> 76 1,4,7,8 -> 58 1,4,8,9 -> 74 1,6,7,8 -> 24 2,4,6,8 -> 51 3,4,7,8 -> 10 3,6,6,9 -> 22 Link to comment Share on other sites More sharing options...
Sacosphilz Posted May 30, 2008 Report Share Posted May 30, 2008 @Trahern: The first two solutions you just posted are among the few that most people managed to solve last. Interesting. Good luck with the other 7, they should be easier. Link to comment Share on other sites More sharing options...
Rendril Posted May 30, 2008 Report Share Posted May 30, 2008 Here is an easy one for now: There are 3 boxes each with 2 marbles in them. In one box there are 2 white marbles, in another there are 2 black marbles and in another there is one white and one black marble. The boxes all have labels on them indicating what marbles they contain. Somebody comes and rearranges the labels so that no box has the correct label anymore. You are able to select one box and take one marble out. How will you determine which label belongs to which box. No, the boxes are not see-through etc, no trick just a simple riddle. Link to comment Share on other sites More sharing options...
Sacosphilz Posted May 30, 2008 Report Share Posted May 30, 2008 I hereby ban myself from answering Rendril's (first) riddle. Link to comment Share on other sites More sharing options...
Rendril Posted May 30, 2008 Report Share Posted May 30, 2008 Perhaps this one will tickle your fancy more? Also a simple riddle, this time more math orientated: An ordinary deck of playing cards contains 52 cards divided into 4 suits, diamonds, hearts, clubs, spades. Each suit contains 13 cards of the following denominations from smallest value to greatest: 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king), and A (ace). The game of poker is played with an ordinary deck of playing cards in which a player will have a five-card hand. There is a certain five-card holding called two pairs. It contains 2 cards of one denomination, 2 cards of a second denomination, and a fifth card of a third denomination. How many five-card poker hands contain two pairs? If a five-card hand is dealt at random from an ordinary deck of playing cards, what is the probability that the hand contains two pairs? Link to comment Share on other sites More sharing options...
Trahern Posted May 30, 2008 Report Share Posted May 30, 2008 I hereby ban myself from answering Rendril's (first) riddle. Is that just to give others a chance? Maybe I should too... though I thought that was what spoiler tags were for. I wouldn't have jumped in so quickly with mosquito before; it was mostly a guess - but I suppose that's why they always tell you to go with your first answer. Link to comment Share on other sites More sharing options...
Metal Bunny Posted May 30, 2008 Author Report Share Posted May 30, 2008 Lol What gets wetter the more it dries? Link to comment Share on other sites More sharing options...
Rendril Posted May 30, 2008 Report Share Posted May 30, 2008 LolWhat gets wetter the more it dries? I see two possible answers: 1) A towel, by drying some it will get wetter. or 2) A block of ice, by drying (getting hotter) it will melt and hence get wetter. Sorry if I don't use the spoiler bar correctly, I'm new to this. Link to comment Share on other sites More sharing options...
Sacosphilz Posted May 30, 2008 Report Share Posted May 30, 2008 @Rendril: Sorry, it's not that I don't like your riddle. It's just that once a riddle is solved, even with spoiler tags, people will stop bothering with it (or it has always been the case so far in this thread). Since that riddle was easy for me, that post was a way of showing off without spoiling the answer. @Two-pair poker hands: Not sure if I got it right, but here goes anyway: Let C(A,N) = A!/(N!(A-N)!) Number of five-card poker hands containing "two pairs" = (number of ways to possible 1st and 2nd denominations) * (number of possible suits of the two pairs) * (number of possible third denomination) * (number of possible suits of the third denomination) = C(13,2) * C(4,2)^2 * (11) * (4) = 78 * 36 * 11 * 4 = 30888 (I just added *4 after getting the first wrong 'answer', and I forgot to recalculate) = 123552 The number of possible 5-card hands dealt from a standard deck of 52 cards = C(52,5) So the probability that one such 5-card hand contains "two pairs" = ( C(13,2) * C(4,2)^2 * (11) * (4) ) / C(52,5) = 30888/2598960 = 99/8330 ~ 0.011884753901560624249699879951981 (From previous mistake) = 123552/2598960 = 198/4165 ~ 0.047539015606242496998799519807923 Link to comment Share on other sites More sharing options...
Rendril Posted May 30, 2008 Report Share Posted May 30, 2008 @Sacosphilz I know what you meant, I didn't think you didn't like it, I just thought I'd give you another one to keep you busy. As for your solution, it is the correct method by the wrong answer. 78 * 36 * 11 * 4 does not equal 30888. Otherwise, with the correct follow through you got it. Link to comment Share on other sites More sharing options...
Sacosphilz Posted May 30, 2008 Report Share Posted May 30, 2008 Ack, how careless can I be? I'll edit the answer into my previous post. Link to comment Share on other sites More sharing options...
Trahern Posted May 30, 2008 Report Share Posted May 30, 2008 The first two solutions you just posted are among the few that most people managed to solve last. Interesting. Good luck with the other 7, they should be easier. That's about par for me... I'll think of the situation that no one else will, but have a tendency to completely miss the blatantly obvious. The 3,3,8,8 combination I had been challenged with before though, so it was more remembering how I solved it the last time; the other just fit the pattern nicely. Link to comment Share on other sites More sharing options...
Sacosphilz Posted May 30, 2008 Report Share Posted May 30, 2008 That's about par for me... I'll think of the situation that no one else will, but have a tendency to completely miss the blatantly obvious. The 3,3,8,8 combination I had been challenged with before though, so it was more remembering how I solved it the last time; the other just fit the pattern nicely. In my case, the one I had been challenged with before was the 1,4,5,6 combination (and I couldn't solve it by myself and had to resort to some programming , and then I was inspired to come up with 3,3,8,8 on my own. I never knew it was going around somewhere else before. Great minds think alike, huh? @Your current 3 solutions: Two weeks ago, chewett sent me a PM asking for confirmation if 9 out of those 10 combinations can truly be solved, because he could only solve 6 of them at that time and had 4 remaining (1 fake). The three combinations you just solved, Trahern, were all in those remaining 4. Link to comment Share on other sites More sharing options...
Eldrad Posted May 31, 2008 Report Share Posted May 31, 2008 @Your current 3 solutions:Two weeks ago, chewett sent me a PM asking for confirmation if 9 out of those 10 combinations can truly be solved, because he could only solve 6 of them at that time and had 4 remaining (1 fake). The three combinations you just solved, Trahern, were all in those remaining 4. Sooo... I just spent the last 10 min not realising there was a fake. I got 9 out of 10 of them... and ummm kept trying. (1/2+9)*8 = 76 6/(5/4-1) = 24 (1/4+7)*8 = 58 (1/4+9)*8 = 74 1678->24 (trap) 8/(1-7/9) = 36 (2/4+8)*6 = 51 8/(3-8/3) = 24 (3-7/4)*8 = 10 (6/9+3)*6 = 22 Oh also on the crown puzzle: There are 4 optimal solutions bellow's an example of another one. First person answers white if they see two white, and black if they see two black. Second person black if they see a black and a white (in that order) or white if they see a white and a black. Third person answers black if they see a black and a white (in that order) and white if they see a white and a black. Basically anything which has only one person answering when they're right and has all three people answering when they're wrong will give a 75% success rate. Link to comment Share on other sites More sharing options...
Eldrad Posted May 31, 2008 Report Share Posted May 31, 2008 MB I think you made a mistake on the 'I have no Brother's or Sisters. But This Man's Father is my Father's Son.' puzzle. If the speaker is the answer then one would be able to replace "this man's" with "my"... "My Father is my Father's Son." Which clearly doesn't work. It does work if "this man" refers to the son of the speaker. Link to comment Share on other sites More sharing options...
Eldrad Posted May 31, 2008 Report Share Posted May 31, 2008 And here's another (some what famous one). There are 100 pirates who have 100 gold pieces from a recent plunder. The pirates are ordered from the highest ranking to the lowest ranking. At any point the highest ranking pirate (who is alive) is the Captain. The Captain gets to decide how to divide up the plunder any way he/she likes. However if after the treasure is divided up more than 50% (including the captain and strictly more not greater than or equal to) decide to rebel then they will rise up and kill the captain making the next in line re-divide the plunder. Now these of course happen to be a rare breed of perfect rational actor pirates and each know that the other pirates are similarly perfectly rational. They have the following motivations. First they don't want to be killed, second they want as much money as they can get (without getting killed), and third if it means they get the same amount of money they'd rather kill the captain just to see some blood. Give the above how does the captain distribute the booty and why? Link to comment Share on other sites More sharing options...
Metal Bunny Posted May 31, 2008 Author Report Share Posted May 31, 2008 Lol Eldrad, don't double post and I never specified who I meant with 'this man', it's one of those brainteasing riddles that are not really worth solving because their answers leave questions Link to comment Share on other sites More sharing options...
Eldrad Posted May 31, 2008 Report Share Posted May 31, 2008 Sorry, I don't normally double post but editing doesn't work correctly with the new forum skin (I should switch back but haven't yet). Link to comment Share on other sites More sharing options...
Sacosphilz Posted May 31, 2008 Report Share Posted May 31, 2008 @Eldrad: On 9/10 math puzzle: You left out one solution. On 3 crowns riddle: Trahern's answer made me think that there might be such possibilities, but I never fully thought them out. Good job on finding and defining the optimal alternatives. On 100 pirates: I've seen that one before, but I've completely forgotten the answer, so I'll be working on it. By the way, does 50% mean 50% of the other pirates (excluding the captain), or is the captain included in the total too? Link to comment Share on other sites More sharing options...
Eldrad Posted May 31, 2008 Report Share Posted May 31, 2008 Including the captain (I knew I was going to forget something). EDIT: Put in the math bit I dropped out from before (missed it when I copied stuff) Link to comment Share on other sites More sharing options...
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