Logic Puzzles

[PsYcHoZeRoSk8eR]

They don't answer...

They all answer the same (example: all answer black)

They all say that they cannot see anything

[ashleychauntel]

Well... there is no definite number of people that have to be saved... so...

The last guy, that can see the heads almost has to die. When asked he can say the color of the hat in front of him. He has a 50/50 chance of sharing the same color, black or white.

Then, when the next guy has to say his color he will know what his is. HOWEVER, the flaw is that if they don't share the cap color someone else can die - so, if the colors are different they can cough before they announce there own color so the person in front of them knows what color to say.

I think that works... maybe, well, for 9 of them anyway.

[BLaCkHawKDowN615]

i think there's something in this one that we're missing. it seems too simple.

[Rolenquin]

Originally Posted by BLaCkHawKDowN615

have the 10th guy tell everyone else their colors and let him guess based on which colors have been picked the most. this would definately work if the masks were 5 white 5 black but idk if that is the case.

Edit: or the guy in the back could just rub his hat on the guy in front of him and hope some fibers come off

You're only allowed so say one of two words, Black or White. Assume the prisoners are far enough away from each other that they cannot touch.

Originally Posted by Doug31

Assuming there will always be 5 of each of black hats and white hats, the guy in the back says the color that there are only 4 of in front of him, then the next guy says the color that there are only 4 of still in front of him and counting what the previous guy said, and so on, until they get to the front. Or, can they only see the hat of the person in front of them?

You're on the right track but there is a solution that will work for any combination of colors of 10 hats. The prisoners can see all the hats in front of them but none behind.

Originally Posted by PsYcHoZeRoSk8eR

They don't answer...

They all answer the same (example: all answer black)

They all say that they cannot see anything

They can see what's in front of them as they are wearing hats. Their eyes are not covered

Originally Posted by ashleychauntel

Well... there is no definite number of people that have to be saved... so...

The last guy, that can see the heads almost has to die. When asked he can say the color of the hat in front of him. He has a 50/50 chance of sharing the same color, black or white.

Then, when the next guy has to say his color he will know what his is. HOWEVER, the flaw is that if they don't share the cap color someone else can die - so, if the colors are different they can cough before they announce there own color so the person in front of them knows what color to say.

I think that works... maybe, well, for 9 of them anyway.

There is a way to save 9 of the 10 prisoners definatly in any combinaton of hat colors

[aperson]

Logic puzzle:

Axioms:
Ax 1: [A -> (B -> A)]
Ax 2: (A -> B) -> {[A -> (B -> C)] -> (A -> C)}
Ax 3: (~B -> ~A) -> (A -> B)

Prove:
~A -> (A -> B)
then prove
~~A -> A
then prove
(A -> B) -> (~B -> ~A)


where -> is the standard implication structure

[Rolenquin]

For the answer to mine since no one has posted the 100% correct one yet... since the last prisoner will see 9 people in front of him, there has to be an odd number of hats and an even number of hats. If he calls out the odd number of hats first lets say white, he has a 50/50 chance of living. The person in front of him now knows that there should be an odd number of white hats in front of him IF he is not a white hat. If he is a white hat then there will be an even number. This process can go on all the way to the front of the line.

[Shashakiro]

For Kit-'s answer: So wtf does an inmate do when he opens box 34, it has ID 53 in it, and then he opens box 53 and it has ID 34 in it?

[Rolenquin]

then he found his own nametag Shash I thought about that too but it works itself out...