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mikegoo
Daedalian Member

 Posted: Sat Mar 16, 2002 10:03 pm    Post subject: 1 I did not write this, but do like it. Picture a ginormous wire strung out to form a circle. On this big wire circle there are some birds. How many birds? you ask. A LOT of birds (to use nice precise mathematical language). So what are all these birds doing? you ask. Each bird is watching its closest neighbor. So the question is: What percentage of birds are being unwatched? edited to fix bad phrasing/vagueness of problem (apologies...my first attempt at posting a puzzle) 1) birds are randomly spaced and don't touch (no one brought that up yet) 2) a bird is never equidistant from 2 birds (which actually arises from the first additional condition as the probability of a bird randomly being placed exactly half way between 2 other birds is 1/oo which equals zero) [This message has been edited by mikegoo (edited 03-16-2002 11:17 PM).]
Lepton
1:41+ Arse Scratcher

 Posted: Sat Mar 16, 2002 10:41 pm    Post subject: 2 25%
impossibleroot
Hi-Keeba!

 Posted: Sat Mar 16, 2002 11:47 pm    Post subject: 3 Okay, but if they're all equally spaced and each looks, say, left, won't that be 100% watched?
Ghost Post
Icarian Member

 Posted: Sun Mar 17, 2002 12:08 am    Post subject: 4 The least number of unwatched birds is none if the birds are all sitting in pairs. The most number of unwatched birds is half if they are sitting in groups of four with the two centre birds closer to each other than they are to the birds at either end. [stating the obvious]The number of unwatched birds where neither the number of birds or the distances between them is known must fall between these two[/stating the obvious]. At best you would be able to work out a probability. Given the possible combinations, 25% is as good a guess as I'd be able to come up with.
Coyote

 Posted: Sun Mar 17, 2002 12:15 am    Post subject: 5 Or, if they're equally spaced, the number of birds is divisible by four, and they're bunched into groups of four like this: *> *> <* <* The number of unwatched birds could be as high as 50%. Either the problem isn't specific enough to have a definate answer or there's some sneaky semantical trick involved, I guess.
Coyote

 Posted: Sun Mar 17, 2002 12:16 am    Post subject: 6 Dangit, typed tooo slow again!
Ghost Post
Icarian Member

 Posted: Sun Mar 17, 2002 12:21 am    Post subject: 7 At least you picked up that there might be a sneaky semantic trick
ctrlaltdel
Member of the Daedalians

 Posted: Sun Mar 17, 2002 12:49 am    Post subject: 8 sometimes its just up to the birds... lets say we have a bird at 12, 3, 6, and 9 oclock. each watching its closest neighbor poses a little split personality problem for them but since their brains are incapable of such psychological dilemmas, they each pick a random neighbor to watch. which may result in 0, 1, or 2 unwatched birds...
quercitron
Don't trust Robinson

 Posted: Sun Mar 17, 2002 2:45 am    Post subject: 9 There are no birds. q
impossibleroot
Hi-Keeba!

 Posted: Sun Mar 17, 2002 4:15 am    Post subject: 10 "a LOT" of birds implies a limit, don't you think?
mikegoo
Daedalian Member

 Posted: Sun Mar 17, 2002 4:18 am    Post subject: 11 Hope I cleared up all the vagueities.
Lepton
1:41+ Arse Scratcher

 Posted: Sun Mar 17, 2002 5:35 pm    Post subject: 12 Lets say that A LOT = oo birds looking left = oo/2 birds looking right = oo/2 half of the birds looking left have a bird on their left looking right. oo/4 half of the birds looking left have a bird on their right looking left. oo/4 half of the birds looking left have a bird on both sides looking towards them. oo/16 half of the birds looking right have a bird on their right looking left. oo/4 half of the birds looking right have a bird on their left looking right. oo/4 half of the birds looking right have a bird on both sides looking towards them. oo/16 9/16 birds looking right and 9/16 birds looking left are being looked at. I can use dedection to guess that 7 out of every 16 birds are not being looked at. personally, I still prefer 25%, though... [This message has been edited by Lepton (edited 03-17-2002 12:36 PM).]
Lepton
1:41+ Arse Scratcher

 Posted: Sun Mar 17, 2002 5:40 pm    Post subject: 13 I dislike the idea of limiting this problem. Seems too straightforward for complicated stuff...
ralphmerridew
Daedalian Member

 Posted: Sun Mar 17, 2002 6:01 pm    Post subject: 14 I think the correct question should be "What is the expected percentage of the birds that are not watched?" (For example, the problem "John flips a fair coin 99 times. What is the number of times it came up heads?" can be anywhere from 0 to 99. However, "What is the expected number of times it came up heads?" has a definite answer: 49.5.)
groza528
No Place Like Home

 Posted: Sun Mar 17, 2002 6:03 pm    Post subject: 15 I would never expect a coin to show heads 49.5 times
ChienFou

 Posted: Mon Mar 18, 2002 1:21 am    Post subject: 16 I bet that the answer uses e. I'll hazard 100/e per cent.
Chuck
Daedalian Member

 Posted: Mon Mar 18, 2002 6:35 am    Post subject: 17 8%, or 9% maybe?
Death Mage
Raving Lunatic

 Posted: Mon Mar 18, 2002 6:38 am    Post subject: 18 It's a trick question. All the birds are dead, so none are really being watched.
afreet
Icarian Member

 Posted: Mon Mar 18, 2002 4:55 pm    Post subject: 19 All the birds are being watched. By us, who are trying to figure out which ones are being watched.
mikegoo
Daedalian Member

 Posted: Mon Mar 18, 2002 6:03 pm    Post subject: 20 Ummmm....the "correct" answer has been posted with no explanation (it wasn't ChienFou). I say "correct" because it is the answer I think is correct, but it may be open to debate.
Quailman
His Postmajesty

 Posted: Mon Mar 18, 2002 6:42 pm    Post subject: 21 Every bird is looking at another bird (the nearest neighbor). That bird is either looking away or right back. It seems that the ratio would be 50-50 if they are randomly spaced. So half the birds are being watched by the bird on their left, and half are being watched from the right, so they're all being watched? No, that's not right, because half are being watched from one direction, one quarter from none, and one quarter from both. That's my guess anyway.
Lepton
1:41+ Arse Scratcher

 Posted: Wed Mar 20, 2002 4:21 am    Post subject: 22 That was my original thought as well, but people kept arguing about it and I made up another one.
mikegoo
Daedalian Member

 Posted: Wed Mar 20, 2002 4:48 am    Post subject: 23 Since its been posted and explained. I believe the 25% (expected value) is accurate...and here is my reasoning (please feel free to poke holes in it). Taking 5 random birds in a row (ABCDE) and looking at the probability that bird C is not being watched p1(B not watching C) = .5 since there is an equal probability of B watching A or C. Same thing goes for D not watching C i.e. p2 = .5. Now the probability of C not being watched by either bird is p1*p2=.5*.5=.25 or 25% chance that a random bird is not being watched. Since all the birds are randomly placed this should hold true for each and every bird which means 25% of the birds won't be watched (again, expected value for a large sample). Anyone disagree or see any flaws (or can do a better explanation of why)?
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