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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

PLEASE READ THIS AND THEN ACCORDING TO THE RULES CLICK ONE OF THE BUTTONS. DON'T CLICK ANOTHER, PLEASE.

If you have some suggestions and comments please post it.

10 forumers from mathsisfun went to some kind of voyage. But the ship sunk and they are now at small tropical island in the ocean. But the natives didn't like them. So they are arranging them in a row, one after another. They are putting hats on each head. I am last, so I can see that the hats are black or white. They are telling is that if 9 of us to guess the color of their hat we'll be free. First the last (me) tells his hat's color, then the nineth and so on to the first. The natives left us time to make a plan.

If someone wants to play he must say which of the nine is and everybody must think out a plan how to save themselves.

Here is what every person see:

When you examine a plan I'll tell that my hat is black or white according to the plan.

Please do not read other persons

*Last edited by krassi_holmz (2005-12-10 21:25:49)*

IPBLE: Increasing Performance By Lowering Expectations.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,604

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

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Yo-ho-ho! Someone have understood me! Right, mathsisfun! Now you and I have to wait for someone else to get in and we're ready to make a plan. Now you can look at the V person hide, but not in others.

I'll wait and see what will happen.

IPBLE: Increasing Performance By Lowering Expectations.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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 (2005-12-11 11:50:27)*

Why did the vector cross the road?

It wanted to be normal.

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

I want IV!

El que pega primero pega dos veces.

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**justlookingforthemoment****Moderator**- Registered: 2005-05-26
- Posts: 2,161

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?

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

We have 4 players already. And we can start thinking a plan. A hunt- satifice me.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Oh, Mathsy have already gave a strategy!

But there's better. Otherwise the game would be too easy.

IPBLE: Increasing Performance By Lowering Expectations.

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**jU****Real Member**- Registered: 2005-08-17
- Posts: 1,923

ook ook ook ook ook ook ook

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

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.

El que pega primero pega dos veces.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

1) We only get given our hats when we are doing the trial. We don't wear them while we are planning.

2) Everyone is buried up to their neck.

3) No. I don't know why, but no.

4) I refer you to point 2.

5) I refer you to point 4.

6) I refer you to point 5.

7) I agree, apart from that she didn't become one, she just is one.

Disclaimer: I know none of these things for a fact (apart from 7), and you will need to wait for krazzi_holmz to see whether they are true or not.

Why did the vector cross the road?

It wanted to be normal.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,604

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?

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

So, if IX has a white hat, then X would say 'white' to tell IX that they have a white hat. But then, IX would have to say 'white' to survive. What happens if VIII is wearing a black hat?

Why did the vector cross the road?

It wanted to be normal.

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**jU****Real Member**- Registered: 2005-08-17
- Posts: 1,923

oh thanks being an orangutan is real fun

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,604

mathsyperson wrote:

So, if IX has a white hat, then X would say 'white' to tell IX that they have a white hat. But then, IX would have to say 'white' to survive. What happens if VIII is wearing a black hat?

So, if you can only say your own hat colour to survive, and 9 of us have to guess right, then one person must start the deduction process and everyone else has to get it right ... right?

So ... hmmm ... "X" could say a word like "BWWBBWBWW", and everyone else would know their hat!

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**justlookingforthemoment****Moderator**- Registered: 2005-05-26
- Posts: 2,161

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.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

That would technically still work. krazzi_holmes says BWWBBWBWW and then he would get executed and everyone else would go free. 9 corrects = we win.

You're right about the 5 of each hats. krazzi_holmes would count 4 of one hat and 5 of another and so deduce his own hat, then IX would do a similar thing by looking at the 8 hats he can see and using what krazzi_holmes had said, and so on through the line, making it a 100% survival rate. But that relies on the 5 hats thing being true.

Why did the vector cross the road?

It wanted to be normal.

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**justlookingforthemoment****Moderator**- Registered: 2005-05-26
- Posts: 2,161

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)?

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I've had a thought. Using the same reasoning that I used with the 5 black, 5 white hats, the same thing could happen for any number of white hats, as long as the number was known.

So, the first four people could make their responses into a binary code to tell everyone how many white hats there are. The first person would have to do the units, as he's the only one that knows the exact amount. The second person could then do the 2's, as he knows the amount within a tolerance of 1 and so on.

The others could track how many of each hat had already gone by remembering the people's responses and whether they died or not. This method would guarantee the survival of 6 people. It's not brilliant, but still a better strategy than before.

Why did the vector cross the road?

It wanted to be normal.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

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!

IPBLE: Increasing Performance By Lowering Expectations.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Oh, I get it now. The binary code is only for the benefit of the people in front, so it doesn't need to go up to 10. X, IX and VIII can all make a binary code to indicate any number of hats between 0 and 7 and as there are only 7 people after that, then that will guarantee their safety.

This strategy would give an average of 8.5 survivors.

Why did the vector cross the road?

It wanted to be normal.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

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

IPBLE: Increasing Performance By Lowering Expectations.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Hmm. That means that we need to guarantee 9 survivors. It's impossible to guarantee the safety of X, because he doesn't know anything, so he needs to give a response that everyone else can use to figure out their hat colour.

In other words, he needs to give one of two responses and the people need to use that to figure out which situation they are in out of a possible 512. It doesn't seem possible.

Why did the vector cross the road?

It wanted to be normal.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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**

Why did the vector cross the road?

It wanted to be normal.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Great! Congratulations! Unbelieveble!

Well done, Mathsy!

This is the strategy!

Now, the end:

I'm saying "white"

IPBLE: Increasing Performance By Lowering Expectations.

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