GAME THEORY

[flyingfaeries]

Q1.
3 people are in a prison together. All wear either a red or a blue hat, but they do not know the color of their hat (in fact all wear red hats). No communication allowed.

The guard tells them that he will ring a bell every hour and if somebody knows that he is wearing a red hat, he or she can leave. (ETA: the guard also tells them that anyone

Comments

[anonymous]

It does not make a difference and all should leave on the first hour as there are no explicit punitive measures mentioned against blue hats who leave. =)

[flyingfaeries]

smart ass! :)

[madhairday]

THATS WHAT I SAID hahahahaha

[flyingfaeries]

hahaha i added another sentence for all you smartalecs.

[dsng]

Yes - everyone leaves after 3 hours.

[flyingfaeries]

why?!

[dsng]

It's all based on assuming the other guys are just as capable of deduction.

We assume all can see each other. Let's call them A, B, and C, although the logic applies to all equally. From A's point of view, he can see that B and C have red hats. He knows if B or C saw two blue hats, B or C would leave within the first hour (because the assumption is that there is at least one red - that's what the guard's new announcement changes). First hour passes. No one leaves.

Now everyone also knows there are either 2 or 3 red hats on (because the first hour has passed and the possibility that there is only 1 red hat has been eliminated). So A knows that if B or C saw a blue hat on him, they would leave at the 2nd hour, because they would both know that they have the two red hats. But no one leaves after the 2nd hour.

So now everyone knows there are 3 red hats. Which means that each one has a red hat. So they all leave after 3 hours.

QED.

[jemauvais]

Hahaha!  That's quite smart!

[anonymous]

are they standing in a line?

[anonymous]

assumption: if they are standing in a line and each person is only able to see the person(s) in front, i.e., A and see B & C; B can see only C, C cannot see anybody.

after 1 hour - nobody leaves (obviously)
after 2 hours - since A did not leave after 1 hour, B will know that either him or C is wearing a red hat, or both are wearing red hats. since he sees that C is wearing a red hat, he will be unable to deduce whether he himself is wearing a red hat. nobody leaves.
after 3 hours - since B did not leave, C will know that he is wearing a red hat and will leave.

[flyingfaeries]

i think you're right!

[dsng]

Oops - thought I spotted an error, but it was my mistake in logic. But still, I think the "all in one line" assumption is a very strange one to make!

[flyingfaeries]

I've heard a version of this one before (the all standing in a line one). but i think your answer makes more sense!

[anonymous]

hmmnn dsng: your logical line doesn't seem to run. All parties in the game already know there are at least two red hats before the first hour. The problem is that none of A,B or C will be able to ascertain what hat they are wearing nor will they be able to ascertain if the other parties see two red hats or one blue and one red. This cannot be solved no matter how many times the bell rings, hence no one will leave because the risk / reward ratio for the individual that his might be the blue hat is disproportionate.

Anon: i know this one. hahaha. I don't believe that it is a valid assumption to make as then, it would not be anything akin to game theory.

i might be wrong. =)

YSA

[dsng]

You _are_ wrong :) There is information being communicated in the first hour by everyone not leaving, and it is a fallacy to assume that because it overlaps with information that seems apparent that it is identical.