Pay It Forward - new game

Originally Posted by cata_mint

Its cat T because she knows A is blue,
If T were blue S would know,
As there are 2 of each colour that means S would be red
S would have said the colour of its hat
But since S is quiet that means S doesn't know the colour of its hat
Which means S can see one red hat and one blue
Since A is blue, T knows she is red.

Yeah?

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Yay! You win!

Why must you win so quickly!

Prize on its way!

Haha! sorry i won so quickly!

Ok here's mine, its kinda similar to Going Nova's, but not in answer.



10 prisoners are due to be executed. The prison master decides to be lenient and let some of them free in a game of chance.

He decides he will bury the prisoners up to their necks in the ground on a slope, and randomly give each prisoner a red or blue hat.

Prisoner 1 will be able to see prisoners 2-10. Prisoner 2 will see prisoners 3-10 etc. Prisoner 10 will see no one.

The prisoners are allowed to talk to each other before they are buried but have no say in what order they are buried or what colour hat they wear.

Once they have been buried the prison master will ask them what colour their hat is. The prison master starts at 1 and goes downhill. If the prisoners get it wrong their head is chopped off. If they say anything other than red or blue, their head is chopped off. If they speak out of turn their head is chopped off.

If they can say what colour their hat is they are free to go.

What is the most effective strategy to save as many prisoners as possible?

Btw, the order of hats as shown in picture is just an example. The correct answer works just as well if all hats are red, or all are blue or any combination in between.

Also, if its too badly explained then ask for clarification. If its too hard I can give a clue or 2

Good luck!

Oh my....I'm going to take a stab in the dark.

Wouldn't the most efficient way be for each prisoner to tell the prisoner in front of him his hat color? Eg. 9 tells 10 8 tells 9. That way only one prisoner is lost and that would be prisoner 1

They aren't allowed to talk to each other once they have their hats on and have been buried. And can only say one word (red or blue) when asked.
Sorry if it wasn't clear.

They're allowed to talk to each other before though, right? Maybe they can work up some sort of code. Blink right eye for red, left eye for blue...

something like that. Then the odds are at least 9/10...

They can only see the backs of each others heads, so blinking wouldn't be noticed.

Clue: Before they are buried and hatted, one prisoner says to the others "I have a plan to save us, but it is not guaranteed to save no1"

Ok, so they talk to each other ahead of time... and # 1 can see everyone's hats, but his own... but he cant them them anything, all he can say is red or blue....

If #1 says the color of the person in front of him, then #2 would know what color hat he had on.... but then how would #3 know? They could tell #4 their hat color but this is only saving half of them...

still thinking...

Another clue: In the example image I gave of the 10 people on the slope wearing the hats #1 is saved. On average he is saved 50% of the time, and in that picture he gets lucky.

I'm such an idiot. I've just realised that there are always 2 answers to this problem, and they are mirror images of each other. In the answer I've always known he is saved, but in the equally correct mirror image solution he is not.

Note- Mirrors are not actually used in the solution of this problem

Originally Posted by cata_mint

Another clue: In the example image I gave of the 10 people on the slope wearing the hats #1 is saved. On average he is saved 50% of the time, and in that picture he gets lucky.

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So he must say red.... hmmm... that blows my idea out of the water. Still, they must work out some sort of system beforehand. hmmmm

Well, I imagine #1 is saved 50% of the time because there is always a 50/50 chance. But other than that I'm not very good at this.

Originally Posted by kluchetta

But other than that I'm not very good at this.

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I'm right there with ya!

Clue: #1 answer tells #2 - #10 something very important about the distribution of their hats. Yes there is a code.

I'd forgotten how hard this one was! I've heard it a couple of times, so I know the answer off by heart

There is some even odd mathematical code............I think................I really want to know the answer to this one, was hoping someone had it by now!

Originally Posted by cheylink

There is some even odd mathematical code............I think................I really want to know the answer to this one, was hoping someone had it by now!

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You're really close!! And if no one gets closer then I'll give you the gift.

Originally Posted by cata_mint

You're really close!! And if no one gets closer then I'll give you the gift.

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based on an even count code agreed on earlier, no. 1 states he is red seeing 6 blue hats and 3 red. From that point, the others count down from his lead..........if he is killed, they know he was blue and odd count applies, if spared, red and even count applies.
Maybe................

None of them figure it out and they all die.

I am dieing to know!

The prisoners use a binary code where each blue hat = 0 and each red hat = 1. The prisoner in the back of the line adds up all the values and if the sum is even he says "blue" (blue being =0 and therefore even) and if the sum is odd he says "red". This prisoner has a 50/50 chance of having the hat color that he said, but each subsequent prisoner can calculate his own color by adding up the hats in front (and behind after hearing the answers [excluding the prisoner in the back]) and comparing it to the initial answer given by the prisoner in the back of the line. The total number of red hats has to be an even or odd number matching the initial even or odd answer given by the prisoner in back.

I cheated, so don't give me the points, LOL

Oh my! Even after hearing the answer, I'm still not sure I get it



Trophy or not - Kim, it's now your turn What's your question?!

Originally Posted by kluchetta

The prisoners use a binary code where each blue hat = 0 and each red hat = 1. The prisoner in the back of the line adds up all the values and if the sum is even he says "blue" (blue being =0 and therefore even) and if the sum is odd he says "red". This prisoner has a 50/50 chance of having the hat color that he said, but each subsequent prisoner can calculate his own color by adding up the hats in front (and behind after hearing the answers [excluding the prisoner in the back]) and comparing it to the initial answer given by the prisoner in the back of the line. The total number of red hats has to be an even or odd number matching the initial even or odd answer given by the prisoner in back.

I cheated, so don't give me the points, LOL

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That sounds like a technical version of my answer.................Hmmmmmmm