EPTPuzzleing Question! (HARD) - Page 58
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BigBalls
United States5354 Posts
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BigBalls
United States5354 Posts
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Jim
Sweden1965 Posts
On November 24 2004 16:02 BigBalls wrote: booyah jimbo And I thought I was fluent in English.. :/ | ||
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BigBalls
United States5354 Posts
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MoltkeWarding
5195 Posts
You only turn over 1 card. Therefore the "last card you turn over" as worded in the question will always be the max. Now where's my prize? | ||
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BigBalls
United States5354 Posts
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bio.Joe
Czech Republic21 Posts
the solution is 7. why? imagine there is only 1 man who commited adultery. then his wife would know it the first day, because she knows that no other man commited adultery. imagine there are 2 men who commited adultery. then both their wives know about 1 case of adultery, so they both wait 1 day, if one of them announce it. it doesnt happen and so the 2nd day they both know (if they are clever that there must be another man. and because each of them knows only 1 man who commited adultery, so each of them can easily figure out that the second one is her husband.and in the given case, nothing happened untill 7th day, so it clear that there must be 7 men who commited adultery, because every of those smart women easily figured out, then when she know 6 men, then she must wait until 6th day and if nothing happens, then the last one must be her husband. hope u all get it. if not just post message me, i will try to explain it better | ||
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Jim
Sweden1965 Posts
On November 24 2004 16:05 BigBalls wrote: As a follow up question, there is an algorithm that allows you to win nearly 37% of the time. Can you come up with this one? My intuition says my algorithm works 37,5% of the time. (I base it on this probably faulty assumption; you find a higher card in the second half on average after half of the seconds half's cards) edit: I dont like it but had to say something. | ||
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BigBalls
United States5354 Posts
Its actually n/e. Dont ask me why this one works, but if you flip over the first 1/e cards, you will maximize your chances of winning. With your algorithm, the winning percentage is about 1/3. But it is clearly >25% always | ||
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wAt-74
United States137 Posts
3 guys go stay at a hotel and the desk clerk charges them $30.00 for the room. The 3 guys decide to split the cost in 3 so that all 3 guys pay $10.00 each(10.00x3=30.00). Later the manager tells the deskclerk that he over charged the men and the actual cost is 25.00. So the clerk gives 5 dollars to a bellboy to give back to the men. The bellboy decides to pocket some of the money and takes 2.00 and gives each men 1.00. Now each men have payed 9.00 each for the room so $9.00 x 3 men = $ 27.00 plus the 2 dollars the bellboy took so $27.00 + $2.00 = $29.00!?!? So what happened to the extra one dollar? | ||
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gg2w
Canada118 Posts
personally i prefer the solution where we all just admit 3x9 = 28 and it all works out | ||
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wAt-74
United States137 Posts
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BigBalls
United States5354 Posts
You have a balance (a way of measuring if things are equal) and 12 identically sized silver balls. However, one of the balls is a different weight than the others. Using 3 or less weighings, find the different weighted ball AND whether it is lighter or heavier. Have fun. | ||
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wAt-74
United States137 Posts
On November 22 2004 11:59 wAt-74 wrote: EDIT2: i changed my mine a couple of times on this problem, so i'm not too sure on the answer Hey guys about the boy and girl puzzle the answer IS 2/3. I'm double majoring and one of my majors is probability & statistics and this is a problem that you get in a beginning class. Lemme try to break it down a bit. you are given the fact that you know one of the childs genders, which is a girl. Now you want to find the probability that the other child is either a boy. You can break down this problem into a simple equation: P(the other child is a boy | given that at least one child is a girl) = P(having a girl and boy)/ P(having at least one girl) < -- there is a long proof for this equation but i dont think you would want to hear it. Trust me its true, all statisticians now it, its a basic equation and you can find the proof in any basic statistics book. so what is the probability of having a girl and boy? well your choices are... GG BB GB BG so P(having a girl and boy) = 2/4 // BG GB whats the probability of having at least one girl? P(having at LEAST one girl) = 3/4 // GG BG GB you plug it into the equation and you get (2/4)/(3/4) = 2/3 (if you cant do the division, use a calculator :p) Some people here think BG GB are the same when they are not. Both come in different order and that is very important to accept it. Here is a visual: EDIT: forget the visual i cant get it to line up properly... new visual:1st you have G-->G or G-->B or B-->G or B-->B As you can see they are different, i hope tat helps Oh btw haha GG is still the same as GG(same goes for BB) If you still dont believe me ask a statistician or stats professor about this, its pretty simple he/she should confirm this. AND if you still dont believe me then go to college and spend thousands of dollars so you can solve this. AND BTW: this is a great thread and keep them coming!  hey guys i hate to bring back an old topic but i hated the fact that the answer was kind of unresolved, well in my book it was. Initially i thought it was 2/3rs then i jump to 1/2 now i think its 2/3rds again. I showed this to one of my stats professor and told him the problem and he told me that he has heard this problem before. He told me that the key was the wording in the problem and that it drastically changed the answer to the problem. But all in all the answer is 2/3rds He even gave me a link to a website that discusses this same problem: http://mathforum.org/dr.math/faq/faq.boygirl.choose.html So now we finally have an answer to this thing and put it to rest | ||
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seeyoulater
970 Posts
last 2 sentences were sarcasm | ||
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MoltkeWarding
5195 Posts
On November 24 2004 19:53 seeyoulater wrote: The fact is, one child does not effect the probability of another child. Flip 2 quarters on the ground. Look at 1. The other 1 has 2/3 odds of coming out the opposite. So despite popular belief, a quarter flip is not 50/50. last 2 sentences were sarcasm Edit: Read sarcasm part  | ||
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MoltkeWarding
5195 Posts
On November 24 2004 19:19 wAt-74 wrote: hey guys i hate to bring back an old topic but i hated the fact that the answer was kind of unresolved, well in my book it was. Initially i thought it was 2/3rs then i jump to 1/2 now i think its 2/3rds again. I showed this to one of my stats professor and told him the problem and he told me that he has heard this problem before. He told me that the key was the wording in the problem and that it drastically changed the answer to the problem. But all in all the answer is 2/3rds He even gave me a link to a website that discusses this same problem: http://mathforum.org/dr.math/faq/faq.boygirl.choose.html So now we finally have an answer to this thing and put it to rest This is so screwy. BB BG GB GG only works when assuming the set of two children are random In this case only 1 child is random. I have a sibling, older or younger doesn't matter off on vacation. What is his/her gender? 50% Girl, 50% boy. It doesn't matter whether I was born first or he/her was born first. Let's say the girl's "sibling" was similarly on vacation. No difference whatsoever in the odds. | ||
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KarlSberg~
731 Posts
On November 24 2004 13:30 BigBalls wrote: Well, v*t = d, so likewise v/d = t the two trains meet at time (speed1+speed2)/D the fly travels at a constant speed for a certain time. So, his velocity*his time traveled is his total distance flown. his velocity is speed f, his time is (speed1+speed2)/D, so the total distance of the fly is [SpeedF*(speed1+speed2)]/D i saw this one right away Wrong as something got inverted in what you wrote but yeah, you got the idea. | ||
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LTT
Shakuras1095 Posts
Edit: I actually found the fortran program I made a long time ago when I wanted a way of checking my algorithm :D | ||
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KarlSberg~
731 Posts
***************** SPOILER ************************** First you just flip half of the cards. Then you flip the rest until you get a card which is higher than any card in the 1st half. It is very hard to write this down clearly without a complicated pageing, but this strategy gives more than 25% (at least 1/4+1/16 from what I calculated) if the number of cards is infinite (an infinite which could be split into 2 of course :p) so I suppose it works for any number of cards. EDIT: hmm took so much time to write this I didn't notice Jim gave the same | ||
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