Nicely constructed! I will be "V". I haven't looked yet.

I think I get this. Person X can see everyone's hats except his own, person IX can see everyone's except his and X's, person VIII can see everyone's except his, IX's and X's and so on until person I, who can't see anyone's hat.

Well, a crude strategy would be for X to say the colour of IX's hat, then IX can say that colour and be safe. VIII would then say VII's colour and VII would be safe. Then VI would save V, IV would save III and II would save I. I'm almost certain that there has to be something better than that, as that one would guarantee the safety of only 5 people.

Well, I think this is one of those reverse-logic thingummys, where the person who knows least has the most chance of survival, so on that basis I'll be person I.

*looks at hide tag*
Ooh, helpful.

Last edited by mathsyperson (2005-12-12 10:50:27)

I'll be III. I think I understand, but I'm not sure of a plan yet.

I'm meant to look at my hidden text now, right?

Oh, Mathsy have already gave a strategy!
But there's better. Otherwise the game would be too easy.

So we have time to plan. Let's get the silly questions out of the way first.

1) Can't we just tell each other our hat's colors, since we have time to plan?

2) What is stopping the people in front from looking down the line in the other direction?

3) The person in back can see all of the hats. Can't he just be spokesperson and tell the colors?

4) I assume the natives are watching and wouldn't be happy if we just took our own hat off to look at the color.

5) Can we use nonverbal communication? For example, starting at the back of the line, we tap shoulders. A tap on the left indicates a black hat, and a tap on the right a white. Each person would pass on the sequence of taps until the person in front receives nine taps. Everyone then knows their hat (except he in the back).

6) Look up into the sun, only hold your hand up so it blocks the sun's sphere from your view. If you notice light reflected onto your arm/hand, you have a white hat; if not, it's black.

7) jU appears to have become an orangutan, like the Librarian in Terry Pratchett books.

If 9 of us are right we all go free.

If "X" (krassi) tells "IX", and "IX" tells "VIII", on down the line, we all go free.

Or not?

oh thanks being an orangutan is real fun

In another version of this I know, if you say anything else apart from 'black' or 'white', you are executed. So we better check that before krassi_holmz says 'BWWBBWBWW'!

Is the chance of having black/white equally likely? If we knew that there was going to be exactly 5 white and 5 black, we could guess correctly, but I'm sure it won't be that easy.

mathsyperson wrote:

krazzi_holmes says BWWBBWBWW and then he would get executed...

I thought that would be letting the team down!! Do you mind, krassi?

krassi_holmz wrote:

They are telling is that if 9 of us to guess the color of their hat we'll be free.

A picky question: does this mean we'll all be free (all 10 people), or the just the people who guessed correctly (not the person who got it wrong)?

Well done, Mathsy!
You're very close to the absolute strategy!
And the strategy when we know exactly how many are the white hats is very close!
Here's a hunt: The idea of binary code is right, but the nineth person didn't needs to know the exact number of white hats.
That waser VERY BIG HINT!

OK. You have to reduce your strategy only a little to give 9.5 survivours.

But it is. I just figured it out. The people don't need to know the exact amount of white hats, because they would already know how many there are with a range of 1.

e.g. IX sees 6 white hats, so he knows that there are either 6 white hats if his is black, or 7 white hats if his is white.

So, all X needs to do is to indicate whether the total amount of hats is odd or even, and everyone else will be able to figure it out.

Let's say that the sequence, from I to X, is BWWWBBWWBW.

Everyone will have previously agreed that X saying 'white' means that can see an even number if hats and if he says 'black', then he can see an odd number. He can see 5 white hats, so he says 'black'. He's wearing a white hat, so he died. Aaw.

But, IX now knows that the total number of white hats is odd. He can see 5 white hats, so he knows that his must be black.

"As IX's was black, the total number of hats is still odd," thinks VIII, "but I can see an even number, so mine must be white!"

VII would then know that the total of white hats was even and as he could see an odd number, he would know that his was white too.

This would continue all the way to I, who would know that there was an even number of white hats left and so his was black.

Survival Rate: 9.5