bonanova

Oh wait! I completely misunderstood the question. I thought you were asking if the dice you proposed could fit the problem. Hmmm... Now I need to think OP edited for clarity.

Excellent!

Sorry - I missed the idea. Correct. That method will guarantee 99% and give a cointoss chance to the 100th. And also correct - you can do better than that, even if the others do not know the fate of the 1st prisoner - or of any of them.

To clarify about the list: if there were four prisoners named Al, Bob, Chuck and Dave, for example, the warden might go to Al's cell and give him a black hat, go to Bob's cell and give him a black hat, go to Dave's cell and give him a black hat, go to Chuck's cell and give him a white hat. He would then go to his office and type up a list. It could be in any order. He might choose to alphabetize it, but he doesn't have to. The list might look like this. Al - black Bob - black Chuck - white Dave - black Al would see: Bob - black Chuck - white Dave - black Bob would see: Al - black Chuck - white Dave - black Chuck would see: Al - black Bob - black Dave - black Dave would see: Al - black Bob - black Chuck - white

I guess it was hardest because I started with a square and looked for ways to morph it. All the others had equilateral triangles or points on a main line of symmetry. The trapezoid - 4/5ths of a pentagon - did not. Thanks for the post, and interesting question for others to answer. - bn

I love your riddles ... they're much more intuitive than analytical. Actually that's why I hate them ...

Did that work like I thought it would? You get too many distinct lengths with your method. The idea was to have all the lengths equal to one of only two different values. There is a trapezoid that works, however. See it here.

Question 1: Prove that P is maximised when all the elements of S are equal in value and rational.

out of the range of hearing of the other prisoners means that the others will not hear what another prisoner says. But even if they could hear the answer, your answer is too low.

All I can say so far is that you guys are smarter at getting around a restriction on communication than I am at prohibiting it. So I've restated in a way that negates all previous answers [sorry, but ... no communicating means no communicating] and makes it clear what you CAN do. Here goes. Tomorrow, each prisoner will remain in his cell. He cannot see and he cannot hear any other prisoner. The warden will visit each cell and place a hat on the prisoner. The hat will be black or white. The prisoner will not know his hat's color and he may not look at it, or determine it by any other manner. When all 100 prisoners will have received their hats, the warden will go to his office and type a list of all prisoners and the hat colors. He will then distribute copies of the list to each prisoner - with that prisoner's name and color totally and irretrievably erased. The warden will then visit each cell and asks the color of that prisoner's hat. If the answer is correct [i.e. black or white, appropriately] the prisoner will be set free; he will be executed if not. Tonight the prisoners ask you for guidance in forming a strategy. How do you advise, and how many do you anticipate you can save from execution? I've edit the OP to point to this post.

From the values of the numbers I thought it might be, but after thinking more,

Great job imran. Unfortunately there's late word that we still have a problem. That sly old Warden Smith... He may be compassionate, but he's also underhanded enough to plant a bug in the strategy room. The whole exercise is still on for tomorrow, but now he's changed the rules slightly again. Realizing that he inadvertently set up a old puzzle, each prisoner now will be lead into an interrogation room, out of the range of hearing of the other prisoners. There he will tell the warden what he believes the color of his hat is, and there his fate will be decided. OP has been edited to reflect this change in plan.

What's written on the paper is final. Everyone writes on his paper before the first paper is read.

Warden Smith has heard about Warden Jones's plan to deal with his prisoners. Read about that here. Smith decides on a similar plan for his prisoners. Being a creative man, he changes the rules slightly. He informs his prisoners that tomorrow is their day of reckoning. They will be led into a courtyard and the warden will place a black or a white hat on each or their heads. No prisoner will see the color of his own hat, but each will see the color of all 99 of the other hats. Since Smith is a compassionate man; he decides to give the prisoners 15 minutes, during which they may ponder their fate, do mental arithmetic, decide on their lucky color ... whatever. During this time they must remain inside the courtyard. They will not be permitted to speak to each other, nod their heads, wink, clear their throats, tap their toes, make any sound at all; you get the idea. At the end of 15 minutes, they will be [edit: last minute change in plan] taken individually into an interrogation room, out of the hearing of the other prisoners, and there be asked the color of their hat. If a prisoner answers correctly he will be set free; incorrectly, and he will be executed on the spot. The prisoners are meeting tonight to form a strategy. They have called you to their meeting as an expert logician to assist them. If you give them the best possible advice, how many of the prisoners can you guarantee will be freed?

Warden Jones has informed 100 prisoners that tomorrow is their day of reckoning. For a clearer statement of this puzzle, please click here. They will be led into a courtyard and the warden will place a black or a white hat on each or their heads. No prisoner will see the color of his own hat, but each will see the color of all 99 of the other hats. The prisoners will be given a pencil and a piece of paper. They will each write W or B on the paper. The warden will go to prisoners at random and read what has been written. If it matches the color of the prisoner's hat [W-white or B-black] the prisoner will be set free. If it does not match, the prisoner will be executed on the spot. Tomorrow the prisoners will not be permitted to communicate with each other in any way. They will be permitted to look at any of the other hats, and to write a W or B on their paper and nothing else. The prisoners are meeting tonight to form a strategy. They have called you to their meeting as an expert logician to assist them. How many prisoners can you guarantee will be set free?

Here's my latest research on that matter.

Sounds fascinating. Maybe you could post a sound bite..

Hi ash, You may have missed the idea of the question in the OP. A pair of marbles is a "couple" only if there is one Red and one Blue marble. Your answer [and method] makes sense if they can be any color, but that's a trivial puzzle. This one requires a little more thought. See what you can come up with.... - bn

What are some words that do work?

Correct. The answer depends on R and B. For starters, try R=6 and B=10. Then try for a formula for arbitrary R and B such that R+B is even.