Devedander

Very good but if I am following properly it doesn't quite work... First guy calls out the guy in front of his hat color. He may well be wrong about his own, but that's ok, one guy can get it wrong. Ok, next guy knows his color now, sweet, he says whatever it is. Now we are on guy 3... he knows how many black and white are in front of him, but nothing about his own hat... Uh oh... Let's look at B W W B B W Last guy says white. He is wrong. Oh well. Second guy knows there are 2 whites and 2 blacks ahead, but has to say white to answer his own color. Third guy now sees 2 blacks and 1 white ahead... but what can he do with this info? But you are on the right track.

Hate me because I am beautiful but do not hate me for my logic puzzles If it helps at all, you are correct that the infinite hats means no probability exists and that indeed the first guy to answer plays a very pivotal roll in the whole ordeal.

No, it means that you cannot in anyway pass information about anything besides the color you think your hat is ie if you use an accent to answer thus providing the guy in front of you some info about his hat in the process you are passing more info than the color of your hat. You will see that it is specifically an example that they may not use accents to denote extra information. To make it more clear I will edit the original question.

I do not see the similarity... in fact I do not think that means what you think it means...

@ Ken and @jdkno I think that's covered here:

Good try though! Don't give up!

Again, no googling or searching an answer. If you are sure you have the right answer, maybe say so or even drop a subtle hint but try not to ruin it for others (PM me if you want to check if your answer is right). So the ninjas from the previous problems are irked at the 100 pirates ability (with Pugs assistance) to defeat their captivity and decide to spite them by putting them in yet another devious trap. This time all the pirates are moved to an underground prison. This prison is massive and each pirate is to be put in a cell alone completely out of earshot and site of every other pirate. Randomly a guard will come by and randomly pick one pirate to take to "the room". In this room are 2 switches on the wall (standard light switches) which don't control anything, they can be flipped on or off and will stay that way if untouched. This is the only thing that the pirates are allowed to interact with and when taken to "the room" each pirate will be given 1 minute alone and must flip 1 and only 1 switch 1 time (ie if it's on it can then be flipped off and vice versa). Then the pirate will be taken back to his cell. The ninjas inform the pirates that at any time a pirate may tell the ninja that they have all (pirates) been to the room at least once. If he is right, they are all let free, if they are wrong they are all killed. The pirates are told this info and given a day to prepare. After this day they will be moved to their cells at which point they will have no futher communication. No pirate can communicate with any other pirate (other than flipping switches in "the room") and the pirates will not see the ninja guard comming to get his pirate a day or him escorting said pirate to "the room" (the ninjas have teleporters). Remember the guard is doing this randomly, he may bring the same pirate to the room 200 days in a row if he so feels. What is the strategy the pirates come up with to be able to answer with certainty that every pirate has been to the room in the fewest number of days? (Note: there is no "X days answer like 200 days because of the whole random guard thing, so the answer should be a strategy and why the strategy will let them answer at the earliest they can be certain that every pirate has been to the room at least once. Assume smart pirates with perfect memory who will not die of natural causes throughout this problem.) EDIT: I think I added too much info to the problem... the problem should not include days... the ninja can take any pirates at any time so no one can know they were the first one in the room just because it's the first day. I apologize as I think both Pug and Edd have it right as it was originally worded. I have updated it so I think it's accurate now.

Good try, but let's say the last three pirate hats are B W B Last pirate says "white" so the guy in front of him knows, guy 2 says "White" knowing his hat is white... but now what about guy 3? He knows nothing to help with his hat... He could simply guess opposite of the guy behind him but then what about B W W

Pirates and ninjas have been moved to their own thread http://www.rwg.cc/members/index.php?showtopic=21583

This was moved out of the triangle thread: Googling searching for the answer is cheating! Brain power only 100 pirates are captured by a clan of ninjas. The ninjas tell the pirates that being honrable they will give the pirates a way to be let free, they will be blindfolded and a hat put on their heads (white or black - there are infinite white and black hats to draw from). They will then be lined up on a hill facing down the hill such that each pirate can see every pirate and hat in front of him and blindfolds removed. Then starting from the back of the line, the last pirate will be asked his hat color, he must whisper his answer to the ninja guard so that no one else can hear. The guard will call out his answer in a plain fashion so that everyone can hear. They will progress up the line until all 100 pirates have answered from back to front what their hat color is. If more than 1 gets his wrong, they are all killed. If 1 or 0 gets his wrong, they are all let go. The pirates are given a day to plan a strategy for this situation. What strategy will get them released? Assume smart pirates, math skills and other cognotive skills sufficient to carry out a complex strategy. These are harvard pirates. There is to be no information traded between pirates after blindfolds are put on other than the color of their hats - ie they cannot cough, shout or answer in a strange accent to give info, essentially this is a logic solution, not a funny business solution EDIT The description has been changed slightly to make it clear and impossible what kind of information each pirate can pass to the other pirates. The only information that can be considered is what the pirate thinks his own hat color is. This does not change the problem or the solution at all, just clarifies. Bonus points: Solve this for 101 pirates and/or solve for any number of pirates.

Spoilsport!

No harm no foul... but to keep this from being "pugs personal brain teaser" post if you know this one, just say so until others get a chance 100 pirates are captured by a clan of ninjas. The ninjas tell the pirates that being honrable they will give the pirates a way to be let free, they will be blindfolded and a hat put on their heads (white or black - there are infinite white and black hats to draw from). They will then be lined up on a hill facing down the hill such that each pirate can see every pirate and hat in front of him and blindfolds removed. Then starting from the back of the line, the last pirate will be asked his hat color. Every pirate can hear what every pirate before him says. They will progress up the line until all 100 pirates have answered from back to front what their hat color is. If more than 1 gets his wrong, they are all killed. If 1 or 0 gets his wrong, they are all let go. The pirates are given a day to plan a strategy for this situation. What strategy will get them released? Assume smart pirates, math skills and other cognotive skills sufficient to carry out a complex strategy. These are harvard pirates. There is to be no information traded between pirates after blindfolds are put on - ie they cannot cough, shout or answer in a strange accent to give info, essentially this is a logic solution, not a funny business solution

Besides simply saying who, it's important to detail how (as there are only 4 choices as to who) the strategy works out. I think you know the answer Pug so let's see if someone else gets it

It's a setup... a more tricky hat question to follow

Pug you party pooper!! Dun be a pug! 4 soldiers are captured by the enemy and told they will be given a chance to be let free. The captors have 2 black hats and 2 white hats. The men will be blindfolded and a hat placed on each mans head, then 3 will be lead into one room and 1 into another. The 3 in one room will be lined up facing a wall such that each man can see the hats of the men in front of him, but not his own - crude illustration: ROOM 1 ROOM 2 | O> | | <O <O <O | The < represent the guys nose so you can see which way he is facing when lined up. The guy by himself can see nobodys hat and the other guys can only see the hats of men in front of them (so one of them can see 2 hats, one sees 1 hat and one sees no hats). After the blindfolds any man may shout out the color of his own hat. But if he is wrong they are all killed. If he is right they are all released. They have 10 seconds for someone to respond or they are all killed. What strategy can be used to garauntee they are released?

Cheaters BTW Ken unless they do things different down under, I don't think you meant to say vertical... Here's another one: 3 Guys are on a tirp and need a room at a hotel, they ask how much and are told $30 a night. So they each pony up $10 and go to the room. A few minutes later the frontdeskman realizes it's only $25 a night and sends the bellhop with $5 change to the mens room. Upon arrival the men thank the bellhop but can't split $5 between 3 people so they each take $1 and give the bellhop a $2 tip. So since each paid $10 but got back $1 they each spent $9. $9 * 3 people is $27. They tipped the bellboy $2. $27 + $2 is $29 Where is the last $1?

Wonder if there is a discount if you buy 2... and does customs seizure replacement still hold up?

Just look at it from the side while installed, if you can see through the link, it's hollow, solid links don't have any visible holes when they are attached.

Seriously... scan that document up and post it... As for customs seeing this post... unless there is some identifying link or they want to do a lot of detective work to prove it, I don't think a post on a forum can really tie anyone to any kind of guilt...

Ok so here is a nerdy one for you... These two triangles are made out of the same parts, just moved around, but the bottom one is missing a square of coverage... how did this happen?

Hmm... that picture... then somene mentions AsianTS... I am not sure what you thought you might come up with googling AsianTS that you might want to see...

You a big AsianTS fan there TT?

From the grid shot, something is off... if you follow the line that runs down the left side fo the top end link down, it does not come near the left side of the bottom endline... something odd there... and OP as was said already, there is no way you got 31 jewels swiss movement in there... if you can get teh case off the back for a picture that would be good... it may say swiss 31 jewels but at the price there is no way... also dealers here easily get manual paul newman for $200-250... My main concern buying from your guy is either he knows and is lying to you/ripping you off or he doesn't know in which case... well why buy from him? Basically if it is a swiss 7750 then he gave you a good deal, but someone who gives that kind of deal isn't going to lie and tell you it's a 31 jewel... if it's an asian 7750 then you paid too much and whatever the reason you shouldn't be going back. Nice watch but you could do better here I think...

It's possible if they are asian auto and not ETA/Chrono movements... elsewise it sounds pretty unlikely unless he is part of the mafia or something...

Heck, leave the flash on... it's not that bad...