In my last post I asked the following question (I've shortened it here but its the same really.)
An infinite number of people, labelled 1,2,3,... have hats on their head, RED or BLUE. They must all shout at the same time a color. They want only a finite number of people to not guess their own hat color correctly. Can they do this?
YES they can manage to get all but a finite number of hats wrong. Here is what they do in their strategy meeting:
1) Define an equivalence class on infinite strings of R's and B's: x==y iff x and y differ on only a finite number of places. It is easy to show that this is an Equiv relation. We think of the string as telling us the hat colors of all the people.
2) An Equiv Rel induces a partition. Choose, for each part of the partition, a representative. They all know all of the representatives.
NOW, once the hats are put on here is how each person reacts:
Gee, I see all of those hats out there! Since I see all but my hat I KNOW which partition the hat coloring is in. I look at the representative of that partition. I guess the hat color that I have in that representative.
Note that ALL of the people will use the SAME rep, and that rep differs from reality in only a finite number of hats. Therefore the number of incorrect answers is finite.
POINT- when I show this to my class they often don't like it. Some of course do not understand it, but even those who do sometimes say that's not practical or the people would need infinite brains! These are both true, but I like it anyway. HOW ABOUT YOU- does the solution satisfy?
NEW PROBLEM (are any problems really new?) I heard this from Peter Winkler who heard it from Sergui Hart who does not claim to have invented it.
Alice and Bob play a game (My darling complains that when a math puzzle uses the word game its usually not a fun game. This problem will not be a counterexample.)
There are an infinite number of boxes labelled 1,2,3,....
Alice puts into each box a natural number.
(CLARIFICATION based on a comment- Alice an put any number she wants in any box.
She wants to put 18 in both box 199 and box 3999 she can do that. NO restriction on what
Alice puts in the boxes except that every box has SOME natural number. ALSO- she need not
use all the naturals- if she wants to put a 1 in every box, she can.)
Bob opens all but a finite number of boxes.
For one of the boxes Bob does NOT open he guesses what the number in it is.
They then open that box. If Bob is right, he wins. If not then Alice wins.
Would you rather be Alice or Bob?
Feel free to post thoughts in the comments, though if you want to solve it without help you may want to avoid the comments. I'll post the answer next week.