[bonanova]

Warden Smith has heard about Warden Jones's plan

to deal with his prisoners. Read about that here.

Smith decides on a similar plan for his prisoners.

Being a creative man, he changes the rules slightly.

He informs his prisoners that tomorrow is their

day of reckoning.

They will be led into a courtyard and the warden will

place a black or a white hat on each or their heads.

No prisoner will see the color of his own hat, but

each will see the color of all 99 of the other hats.

Since Smith is a compassionate man; he decides to

give the prisoners 15 minutes, during which they

may ponder their fate, do mental arithmetic, decide

on their lucky color ... whatever. During this time

they must remain inside the courtyard.

They will not be permitted to speak to each other,

nod their heads, wink, clear their throats, tap their

toes, make any sound at all; you get the idea.

At the end of 15 minutes, they will be [edit: last

minute change in plan] taken individually into an

interrogation room, out of the hearing of the other

prisoners, and there be asked the color of

their hat. If a prisoner answers correctly he will

be set free; incorrectly, and he will be executed on

the spot.

The prisoners are meeting tonight to form a strategy.

They have called you to their meeting as an expert

logician to assist them.

If you give them the best possible advice, how many

of the prisoners can you guarantee will be freed?

[Guest]

99 can be saved. They can decide that the first one will tell red if he sees odd red hats and black if he sees even red hats

[bonanova]

99 can be saved. They can decide that the first one will tell red if he sees odd red hats and black if he sees even red hats

Great job imran.

Unfortunately there's late word that we still have a problem.

That sly old Warden Smith...

He may be compassionate, but he's also underhanded enough to plant a bug in the strategy room.

The whole exercise is still on for tomorrow, but now he's changed the rules slightly again.

Realizing that he inadvertently set up a old puzzle, each prisoner now will be lead into

an interrogation room, out of the range of hearing of the other prisoners. There he will

tell the warden what he believes the color of his hat is, and there his fate will be decided.

OP has been edited to reflect this change in plan.

[Guest]

Great job imran.

Unfortunately there's late word that we still have a problem.

That sly old Warden Smith...

He may be compassionate, but he's also underhanded enough to plant a bug in the strategy room.

The whole exercise is still on for tomorrow, but now he's changed the rules slightly again.

Realizing that he inadvertently set up a old puzzle, each prisoner now will be lead into

an interrogation room, out of the range of hearing of the other prisoners. There he will [edit for emphasis]

tell the warden what he believes the color of his hat is, and there his fate will be decided.

OP has been edited to reflect this change in plan.

I am assuming that other prisoners will know the fate of the first prisoner.

still 99. He will still say white for odd white and black for even white. Now since others know the color of his hat. They can guess what he said based on his fate.

[bonanova]

I am assuming that other prisoners will know the fate of the first prisoner.

still 99. He will still say white for odd white and black for even white. Now since others know the color of his hat. They can guess what he said based on his fate.

out of the range of hearing of the other prisoners

means that the others will not hear what another prisoner says.

But even if they could hear the answer, your answer is too low.

[Guest]

out of the range of hearing of the other prisoners

means that the others will not hear what another prisoner says.

But even if they could hear the answer, your answer is too low.

they don't need to hear his answer, just know his fate (and honesty). According to imran's plan the first prisoner will commit to a color based on the even/oddness of what he sees. All the other prisoners know the color of his hat so they know what he said by his fate. Of course if a guarunteed (Priisoners ARE honest!!!) 99% survival with a 50% chance of 100% is too low....

[Guest]

If all else fails

[bonanova]

they don't need to hear his answer, just know his fate (and honesty). According to imran's plan the first prisoner will commit to a color based on the even/oddness of what he sees. All the other prisoners know the color of his hat so they know what he said by his fate. Of course if a guarunteed (Priisoners ARE honest!!!) 99% survival with a 50% chance of 100% is too low....

Sorry - I missed the idea.

Correct. That method will guarantee 99% and give a cointoss chance to the 100^th.

And also correct - you can do better than that, even if the others do not know the fate of the 1st prisoner - or of any of them.

[Guest]

I'm not sure if this is allowed, but I think that it would ensure a 100% survival rate.

When they're in the courtyard, the prisoners could split into 2 groups. Those that can see an even number of black hats and those that can't.

Your hat color, will be the same as the hat color of everyone else in your group. If no-one else is in your group and you're in the even hat group then your hat is black. If no-one else is in your group and you're in the odd hat group then your hat is white.

[Guest]

get in groups of some size. If you see white, turn your back, if you see black, face the person. Then they will know what color they have.

[Guest]

I would tell them to line up in a row, one at a time. When you enter the row, go in the middle, with a black hat on your right and a white hat on your left. The last person would need to guess their hat color, so you have a 0.5% fatality rate.

[bonanova]

I'm not sure if this is allowed, but I think that it would ensure a 100% survival rate.

When they're in the courtyard, the prisoners could split into 2 groups. Those that can see an even number of black hats and those that can't.

Your hat color, will be the same as the hat color of everyone else in your group. If no-one else is in your group and you're in the even hat group then your hat is black. If no-one else is in your group and you're in the odd hat group then your hat is white.

Interesting, and you're on the right track.

But how does this work?

Say I see two groups: one group has black hats and one group has white hats.

Say I see an even number of black hats.

How do I know which group to join?

p.s. I can't walk over and ask someone, "Is this the group that sees an even number of black hats?"

[bonanova]

get in groups of some size. If you see white, turn your back, if you see black, face the person. Then they will know what color they have.

Warden notices the back-turning and infers the prisoners are communicating.

Everyone is executed.

[bonanova]

I would tell them to line up in a row, one at a time. When you enter the row, go in the middle, with a black hat on your right and a white hat on your left. The last person would need to guess their hat color, so you have a 0.5% fatality rate.

Great Idea!

And best of all, no communicating!

But we're back to ensuring 99.5 prisoners are safe - actually guaranteeing 99.

I claim we can get them all home safely, 100% guaranteed.

[Guest]

Do it twice?

[bonanova]

Do it twice?

Yes.

Actually, only one more person -either of the end people - has to go to the middle, to remove the uncertainty of the 100^th prisoner.

Great job.

OK, someone is going to say this is a form of communication.

All I can say is ... uh maybe.

But It's my puzzle, and I get to make the call ...

[Guest]

Interesting, and you're on the right track.

But how does this work?

Say I see two groups: one group has black hats and one group has white hats.

Say I see an even number of black hats.

How do I know which group to join?

p.s. I can't walk over and ask someone, "Is this the group that sees an even number of black hats?"

Sorry, I didn't explain myself particularly well. The idea is that if you are a prisoner, you look at all your fellow prisoners hats. If you see an even number of black hats, you join group 1. otherwise, you join group 2. The positioning of the groups will have been predetermined the night before ( e.g north side of courtyard is group 1, south side is group 2). Once everyone has separated into the two groups, you can look at the hats of the people in your group. You will all have the same colour hat.

[Guest]

Yes.

Actually, only one more person -either of the end people - has to go to the middle, to remove the uncertainty of the 100^th prisoner.

Great job.

OK, someone is going to say this is a form of communication.

All I can say is ... uh maybe.

But It's my puzzle, and I get to make the call ...

I don't quite get this. Surely there's a 0.5% chance that the first person will die on the first attempt?

[Guest]

I would say that was whar the meeting was about!

Twas a good one Bonanova GM

[bonanova]

I don't quite get this. Surely there's a 0.5% chance that the first person will die on the first attempt?

The answer is described by jcalonego in post #11, but it needed a slight tweak.

See posts 15 and 16 for the tweak.

If it's unclear, it can be explained in more detail.

Hint: posts 11, 15 and 16 describe a way to tell everyone his hat color.

And it's all done during the 15 minutes before any of them are asked their color.

Does that make sense?

[Guest]

The answer is described by jcalonego in post #11, but it needed a slight tweak.

See posts 15 and 16 for the tweak.

If it's unclear, it can be explained in more detail.

Hint: posts 11, 15 and 16 describe a way to tell everyone his hat color.

And it's all done during the 15 minutes before any of them are asked their color.

Does that make sense?

Thanks bonanova. I missed that post. That method works quite nicely - but I still prefer my solution , although counting prisoners would get more and more impractical as the numbers increased/time decreased.