2014 people and 2015 hats
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by I-G-D
Hey, everyone.
Since I’m bored, I’ve decided to write an article about a cool logical problem that I know. I might have mentioned it to some of you, but I don’t remember revealing the solution to anyone but a few people, so I think it will be a cool idea to share it with the others, via article.
So, this is the problem:
You have a team of 2014 people, and they are arranged in a column, with #1 facing everyone and #2014 facing no one. There are 2015 hats, numbered from 1 to 2015. The game goes like this: one of the hats gets hidden, and the remaining 2014 hats are put on the 2014 people, one hat per person. However, nobody can see the number on their own hat.
Each person can see the numbers on the hats of ALL the people sat in front of them. This means that person #1 sees everyone’s hat number except his own, person #2 sees everyone’s hat number except his own and person #1’s, etc. The people don’t know which hat is hidden.
After the hats are handed out, everyone must try to guess their hat number by saying one number from 1 to 2015. The person at the back of the column starts first (#1), then goes the person in front of him (#2), and so on. The only condition is that no two people say the same number.
Clarification: After someone says a number, the people that go after him know which number he has said, but they don't know if he was right or wrong.
For each person that guesses his hat number correctly, the team earns one point.
The 2014 people know how the game functions before the hats are handed out, so they can talk to each other and try to form a strategy to maximize the number of points they earn.
The question is: what is the optimal strategy? What’s the best way for them to figure out their hat numbers? What’s the smallest number of points that the team can theoretically earn if they use that optimal strategy?
Post your opinion and/or solution about this problem via comment… Also, it would be awesome if you could subscribe to my newspaper, I need like 250 more subs for my next Media Mogul medal, so the subs mean a lot to me.
I hope you enjoy solving this. By the way, the problem was not created by me, its creator is some Russian dude whose name I don’t know…
I-G-D
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2013 bi moglo ako se organizuju
Kako?
da ti kazem pa i ti da znas 😃
Znam ja, ja sam ga resio, haha. 😃
Prilicno sam siguran da postoji jedna strategija koja recima moze da se objasni.
mind = hurt
to best strategy is to ask siri
no more mind blown
Just a question to clarify something. Do they know what the other people have guessed? eg. Does #2 know what #1 has said? And if he does, does he know whether it's right or wrong?
#2 knows what #1 has said, but he doesn't know if it's right or wrong.
pa onda teoretski ja bih rekao da u najgorem slucaju mogu da osvoje 0 poena, jer uvek imas mogucnosti da pogodi ili promasi, pa ako svi ispromasuju eto nule, a neku posebnu strategiju ja ovde ne vidim, nastavljam pratiti da cujem resenje 😉
Nije tacno. Postoji nacin da se obezbedi da uz odgovarajucu strategiju izvuku bas mnogo poena. 🙂
Tesko je naci strategiju, ali postoji.
why 2014 ? xD
'cause it's the year 2014. 😃
Prvi u nizu zna sve brojeve osim svog i skrivenog broja tako da ima 50% sanse da pogoti tacan broj,dalje ako svi ostali znaju koji je broj onaj prethodni izabrao i sve brojeve ispred sebe sve zavisi od toga da li misle da je onaj prethodni pogodio broj,dakle ako svi misle da je prethodni pogodio sve zavisi od toga da li je prvi rekao tacan broj ako jeste to je 100% poena a, ako nije to je nula.
Kako? Prvi kaze broj, i ako drugi misli da je taj broj tacan, on ima da bira izmedju neka dva broja... Odakle on da zna koji je na njemu, a koji na dodatnom sesiru? On ne zna koja dva broja je prvi mogao da izabere, samo cuje broj sto je prvi izabrao, ne zna koji je onaj drugi broj.
recimo da brojevi idu redom prvi ce videti brojeve od 2 do 2014 znaci bira 1 ili 2015, ako je recimo uzeo 2015 sledeci u nizu vidi brojeve od 3 do 2014 i ako odbaci 2015 ostaje mu 2 ili 1 i tako do krajnjeg u nizu
Ako se to desava, postoji sansa da se uzme 0 poena. Fora je smisliti strategiju sto bolju, tako da je minimalan moguci broj poena sto veci.
recimo onda da 50% njih misli da je onaj prethodni pogresio a 50% da nije tako se dobija bar 50% sigurnih poena,ukoliko nisam zesce omanuo 😃
Zeznuto je da pratis to onda. To su verovatnoce i verovatnoce... 🙂
Reci cu ti odmah da je moguce napraviti strategiju tako da SIGURNO dobijaju broj poena koji je mnogo vise od pola.
ovo je nesto sto mi je prvo palo na pamet i verovatno da nije najbolje resenje ali voleo bih da vidim i tvoju verziju edituj kasnije clanak sa tvojim resnjem ukoliko se ne javi neko sa boljim rezonovanjem 🙂
Hocu.
Ovo resenje je teoretski najbolje moguce. 😛
Ne postoji bolje, dokazao sam.
Нека се договоре да први каже број онога задњег, други предзадњег и тако редом.
Када дођу до пола- ова друга половина ће апсолутно знати које шешире носе.
Ево, ти рече има систем који сигурно доноси број поена већи од пола, мени ово овако на брзину "паде на памет"...
Nije poenta da na kraju oni znaju ko je koji sesir, nego da oni sto oni KAZU bude to sto je njihov sesir.
Pa jarane- ta druga polovina će tačno REĆI koji je njihov šešir. Jar ako si ti zadnji a ja prvi- ja ću reći koji je tvoj šešir i ti ćeš to zapamtiti kada dođe red na tebe da kažeš i rećićeš taj broj. Pa tako redom- drugi/predzadnji itd...
Uslov je da ne mogu da kazu jedan broj dva puta... Tipa, ako jedan kaze taj broj, niko sledeci ne sme da ga ponovi.
Taj uslov si sada dodao XD XD XD
To ne stoji u tekstu- a gde je pečat? XD
Haha. 😃
Bilo bi lako doci do pola da nema tog uslova...
Fora je da likovi kazu 2014 razlicitih brojeva.
A inace, resenje je mnogo vise od pola, ali pola verujem da moze da se nasteluje.
Jeste brt, ali nisi nigde gore napisao to kao uslov, da 1 broj može da se kaže samo jednom 🙂
Sad me jbš da razmišljam brteeeeee XD
"The only condition is that no two people say the same number."
Tu pise.
Razmisljaj, nije lako ovo. 😃
Fak, evo čitam- postavljeno je to kao uslov. Jbg, na brzinu sam iščitao...
Brt, ne mogu sada. Al baš me zainteresovalo. Ujutru ću uz kaficu. Nemoj editovati članak, ako neko u komentarima i provali.
Pa ti pišem ujutru ako se dosetim.
I mogao bi bar jednom nedeljno da postaviš neki ovakav članak 😉
Пази, ако постоји могућност да поред изговора броја изговоре још нешто (што није број), могли би то урадити на следећи начин:
Узмимо да свако слово азбуке симболише једну цифру:
А=1, Б=2...И=0, рецимо да број 2016 узму као основу и да то буде слово Ј, и рецимо да минус означе словом К -број један би за себе могао рећи један од два преостала броја и (ако бр. 2 нпр носи шешир бр. 7) да изговори, нпр- " 1 ЈКБИИЗ [2016 (J) - (K) 2(Б) 0(И) 0(И) 9(З)" чиме би бр.2 знао да носи шешир бр. 7.
Он би изговорио "7" и по већ задатој формули број онога испред себе...
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where to find so much people but good I like it 🙂
JEBOTE MORAM SAD DA MUDRUJEM! HAHAHAHAHAHAHA
E ti bebeli ako ne provališ- niko neće XD
KO ZADNJI IKSDEUJE - NAJSLADJE IKSDEUJE HAHAHAHAHA
One way would be for the first person to guess 1 number less the person #2, and having spoken before hand person #2 would know his number was 1 more than the guess and then correctly guess his number.
Going forward all the odds guess one number less than the evens so they all can guess correctly.
In the event that the number that is one number less has already been said (i.e. Person #6 is 80 so person #5 guessed 79, Person #10 is 81 and therefore 80 is not an option) then the odd person simply guess the next available number that is less that the correct number. As each person know what has been guessed the Even guesser can simply simply guess the next available number above the one the odd person guessed.
If the even person has the number 1, then the odd person will simply say 2015.
This will mean at least 50% correct answers, There is a chance that the odd will get a correct answer or two in there as well.
That was with a couple minutes of thinking, If I thought more perhaps I could come up with something else, but If I think longer then it will move into my drinking time, then I will be drunk and have no answer.
Nevermind, this won't work, too many even people's numbers will be said by odd's in the clue guesses.
damnit.
It's possible to get a lot more than half.
If I was in that line and I hat to listen 2014 people say a number, there's no way I'd memorize more than 7 numbers... Therefore, the solution is realistically impossible. 😃
I'm close, each person only has 2 numbers to guess. I'm going to work from there while I sleep.
Too much to me.
I like puzzles, but only when I got the time and I mean veeeeeeeeeeeery much time
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http://www.erepublik.com/en/article/2014-people-and-2015-hats-analysis-hints--2419390/1/20