[Nmoitmzr]

Originally posted by Krill
Is the person allowed to look at all of the other hats so that he knows the distribution of the colours? If each individual person can do this, each person simply notes which colour is the most common and everyone says that. After the hats have been given out the people don;t have to communicate. You have no information about the distribution.

[Rurcextedutty]

theoretically my way COULD work... i played DnD for 15 years... im just not good with numbers...

[neniajany]

YAY!

[toopyimport]

I'm going to poach some from my CS homework. This was an easy one (we had to prove it, but the proof is trivial once you know the answer):

After a long day of 251 homework, you and your friend decide to treat yourself to a circular 12-cut
pepperoni pizza. When the pizza gets to your place, Vocelli’s makes dividing the pizza hard on you, by
putting a different number of pieces of pepperoni on each slice. You and your friend decide to divide
the pizza in the following way. First, you choose and eat any slice from the pizza. Then, you both
alternate turns by taking and eating a slice from the pizza, but only one of the slices that borders the
gap left by the removed slices.

Is there a strategy you can use to ensure that you will have eaten at least as many pieces of pepperoni
as your friend, once the pizza is fully consumed?

[genna]

251 is the course number.

[MYMcvBgl]

Oh come on people, the answer is trivial.

[GinaIsWild]

I think math freaks should be rounded up and shot .

[espabamar]

for the pizza, take the piece iwth the most pepperoni. your friend will probably do the same. then from the available pieces take the one with the most amount of pepperoni.

or you could eat with a vegetarian.

[ronaldasten]

Originally posted by b etor
for the pizza, take the piece iwth the most pepperoni. your friend will probably do the same. then from the available pieces take the one with the most amount of pepperoni. That doesn't guarantee you'll have at least as many pepperonis.

[Bvghbopz]

Originally posted by Kuciwalker
Obviously they schedule themselves. Line up, every minute the next one gives a guess. This guess is based on some function of the others. I'm trying to think of one that guarantees a correct guess by the last one in line. The only information each has is that all previous guesses were false. They do not hear each other's guesses, or cannot base themselves on that.
IF they could, they could jsut agree that number one will guess number two's color and vice versa, and whoever guesses right...
read my post above.

[FoetAgerhot46]

They do not hear each other's guesses, or cannot base themselves on that.

I didn't say they could hear each other. The 1st person makes a guess (silently), and if they're freed, then no one else need make one. After 1 minute the 2nd person knows that the first person's guess was wrong, which gives him a little extra information. By the last person someone will have gotten enough information to guess correctly.

Of course, this still works if they all guess simultaneously, but I found it easier to think of it with them guessing in sequence. You'll notice I already posted a solution.

[CtEkM8Vq]

If the other person's guess is (partly) a function of your hat, and you know that the guess was not the same as the color of their hat (which you can see), then you know a little bit more about your own color. Assuming the function was properly constructed, of course.

[Anypeny]

Originally posted by Kuciwalker
F(ith person) must take on a different value for each different color of j's hat. Therefore, if j knows F(i) != some number, j has just eliminated a possible color for his hat. Can you show me a system where you can guarantee more than "one person will guess his hat correctly"?
If not, then there is no extra information.

[imictiorume]

Well, there's a little. By the last person, if no one has gotten it right, he knows his guess is correct.

[Uojeyak]

Originally posted by Kuciwalker
Well, there's a little. By the last person, if no one has gotten it right, he knows his guess is correct. In the solution to the original problem with "no extra information", he also knows that if nobody else gets it right, his guess is correct.

I guess it all depends if the game stops when somebody is right.
You could argue that in your version, if they keep playing after somebody gets it right, then they can all get it right.