Riddle lovers try solve this

A bit of a long one. A group of 10 men are captured by aliens and given a test to save themselves and humanity. They are lined up in a row facing the same direction and they are given either red or blue hats. Starting from the back the aliens will ask what colour there hat is, get it wrong and they die. They were told before hand what will happen and had time to decide what to do. They are not allowed any codes or to say anything but red or blue. There is no tricks of sight they can only see people's heads in front of them. The bad news is you need to guarantee to save 9 lives to save humanity.

So, if I get this right, they can decide upon a tactic first, allowing them to say whatever they want amongst eachother, knowing what is going to happen. Afterwards they are only allowed to say one word (red or blue). I would suggest the following: [spoiler] - The first one to speak (the last one in the row) is going to say the colour of the hat in front of him. (50% chance he dies, too bad for him). - The second one now knows the colour of his hat, so he will say that colour. However, if the person in front of him has the same colour, he will say it very slowely, like "Bluuuuuueeeeee". Otherwise he will say it really fast, like "Blu" - Now the third one knows the colour of his hat as well, and can repeat the trick, etc. [/spoiler] 50% chance no one dies, 50% the first one dies - all others are saved - humanity is saved.

Creative but not the best answer, I said no codes so there's a good chance the aliens being a super smart race would pick up on this and count it as a code and were all dead. The best answer has no change of tone just "red" or "blue".

I have played supreme commander, believe me, aliens aren't that smart xD But I will have to think some more now :S.

Just a question out of curiosity, am I right in assuming you cannot see the colour of your own hat, but only the colours of all ppl that are yet to name their colour? Edit, you misspelled "their hat". You do mean that every human has to name the colour of his own hat. Edit, I am also assuming the total number of blue and red hats is unknown. If they know in advance there will be 4 blue ones, and 6 red ones its easy (and everyone can be saved guaranteed). (Just to rule out some mean cases, riddles then to have those...)

[spoiler]The person in the back looks in front of him and calls "blue" when he sees an even amount of blue hats in front of him and "red" when he sees an odd amount of blue hats in front of him. This will give him 50%-50% chance to be right. However the one in front of him can look at all his hats in front of him. If he sees an odd number of blue hats but the person behind him just told him there was an even amount in front of the last one, he knows that he has to be wearing a blue hat himself. Analogously the other three cases follow. For any prisoner in the row, everything he has to do now is count how many blue hats were behind that person and look at the ones before him. In total he knows if this number has to be odd or even and therefore can deduct his own colour. In the end you have 50% on 9 survivors and 50% on 10 survivors, so 9 is guaranteed and therefore I just saved humankind... *goes back to bed*[/spoiler]

Hmmmz, I think the parity bit tric will work here. [spoiler] The first person sais red if there are an odd number of red hats before him, and blue if there are an even number of red hats before him. Assume he sais red. Now the next person either sees an odd number of red hats (so his one is blue), or an even hat (so his one is red). Based on his answer the next one knows his hat colour as well, and so on. [/spoiler] Edit, dang, Cireon was a bit earlier.

That's right, fair play there's some brains on this site

[quote=lionheart - Apr 27, 07:00 AM]That's right, fair play there's some brains on this site[/quote] You have been talking to two of the programmers here. You should be happy we have some brains XD