Riddle Minute

[Merry Christmas, DC]

when I draw animals, they have feet...

[Arloest]

Well then you suck at drawing animals.

[Muninn]

no, I'm actually curious which answer is correct. the 16-feet one or the 10990-feet one?

Hyperion is in fact right.

[Arloest]

Damn it Jack now I'm not sure which one is correct either.

[Merry Christmas, DC]

are you just being witty, or do you actually not realize that hyperion gave both of those answers

[Hyperion]

He's being witty.

[Muninn]

So at this point I'm going to stop trying to make every new post in this thread of mine repeat how 10990 is the right answer because anyone who's still saying they're unsure is doing it on purpose and I'm bored of that and a new riddle would be nice.

[Merry Christmas, DC]

I seriously couldn't tell which answer you were talking about. THANK you.


okay I have one. (this is another math riddle)

why is 6 afraid of 7?

[Segovia]

Because 7 raped 6's mother and threaten to kill him if he said anything.

[Doc Sigma]

Because 0x315

[nickspoon]

10 straight-jacketed prisoners are on death row. Tomorrow they will be arranged in single file, all facing one direction. The guy at the front of the line (who cannot see anything in front of him) will be called the 1st guy, and the guy in the back of the line (who can see the heads of the other nine people) will be called the 10th guy. An executioner will then put a hat on everyone's head; the hat will either be black or white, totally random. Prisoners cannot see the colour of their own hat. The executioner then goes to the 10th guy and asks him what colour hat he is wearing; the prisoner can respond with either "black" or "white". If what he says matches the colour of the hat he's wearing, he will live. Otherwise, he dies. The executioner then proceeds to the 9th guy, and asks the same question, then asks the 8th guy, and so on. This continues until all of the prisoners have been queried.

Assuming that the prisoners are able to arrange a plan beforehand, how do they save as many lives as possible?

[Doc Sigma]

Just an initial attempt... more like thinking outloud...

#10 calls the color of #1's hat, #9 calls the color of #2's hat, etc. That way, 1 through 5 are guaranteed to live, and 6 through 10 each have a 50/50 chance of living.

There's probably an even better way, though, right?

EDIT: Question... is there guaranteed to be at least one black hat and at least one white hat, or is it possible that all 10 hats are the same color? If the former, I think I have a better idea. Specifically, I think I have an idea where 9 of the prisoners are guaranteed to survive and one of them has a 50/50 shot.

EDIT 2: Actually, I think the stipulation I just imposed doesn't matter... there could be no guarantee of at least one of each color hat, and the plan would still work (9 definitely survive, one is a coin toss). But I'm still working it out. If someone beats me to it, more power to you!

[nickspoon]

It doesn't matter how many there are of each colour hat. And it is always possible to save at least nine of the prisoners.

[Merry Christmas, DC]

do they agree to "pause for a considerable amount of time" if the color they call is the same color as the next person?

[nickspoon]

That would be one system, but not a particularly good one. It can be done without any communication besides the words "black" and "white".