That's exactly where I got up to last night! I can't think how to put that into a per-round equation though... to be honest it's been 18 years since I really put anything into equation form (y'know, because maths is pointless and you don't need it in the real world ).
this strategy works for 5 people too. i dont think it works for 6 though. the distribution is red or blue if im prisoner 1 and see that 2 - r 3 - r 4 - b 5 - r and i see 4 doesnt see 4 of the same colour (4 reds) then i must be blue or else he would have guess his own.
I don't have the slightest idea how to put that into an equation, either, but I think it can be solved without that... Ah, that's true. I didn't think much about each number of prisoners. I missed this
Wait though - that would still be 50/50 for that person. If you see two colours you have a 50/50 chance, if you see all one colour you know 100% you are the opposite colour.
But after the first 10 s passed and nobody answered, you know that nobody sees 100% of the same colored hats, because they would have answered. And if you see all the same and one different, that means you see the least number of "minority" colors and therefore you have one, too.
Ahhh! Clever! I find it hard to think about co-operating with the other prisoners... I'm too busy fashioning a shiv from my toothbrush... I'll show that sadistic guard!
again the gamblers fallacy. there only needs to be one red hat, there is no minority or majority of hats necessary apart from one hat of each colour at least. there is no other relation between hats. the odds of getting a red or blue hat is still 50/50. and one slip up is still 50/50 EDIT: just understood what you meant. will convene with the other prisoners. if we dont reply back, then it hasnt worked
I know, but if they decide on the rounds, the first round will take care of the possibility that there all except for one have the same color. If they don't answer in the 10 s they decided on, then they know there's more than one hat of a different color (that's what I meant by "minority" color). If they still don't answer, it means there are more and eventually that there's an equal number of colors (if there is). But then everyone should be able to answer, because they all see the same ratio and know that nobody sees more hats of the same color than they themselves do, otherwise the others would have answered before. - I'm sorry, I sometimes have trouble explaining myself in English Could this be the answer to the puzzle, then? Or is there another one?
So we got to about 50 prisoners and overpowered the guards but the warden escaped. Here's how we did it with Nachi's method: Spoiler So we decided that if once the "game" started; - if you saw all hats were the same colour then you guess your hat was the opposite colour (we've won!) if not then stay silent - after 10 seconds if you see there is one hat that is not the same colour (1 hat minority) as the others then you guess that your hat is also part of the "minority"(we've won!) if not then stay silent - after another 10 seconds if you see there is two hats that are not the same colour (2 hat minority) as the others then you guess that your hat is also part of the "minority"(we've won!) and so on. So each prisoner assumes they are in the minority but waits 10X seconds (X being the number of minority hats they see) before guessing tha they are in the minority to ensure that everyone in the true minority has the opportunity to guess correctly. i'll use the example of 7 and 8 prisoners. So for seven prisoners, the colour of hats (red=r, blue=b) looked like this: 1 - r 2 - r 3 - r 4 - b 5 - b 6 - b 7 - b If i was prisoner 1, 2 or 3 i would wait to 20 seconds because i see 2 hats (red) in the minority and then guess that i am also in the minority (red). if i was prisoner 4, 5, 6 or 7 then i would wait to 30 seconds before guessing because i see 3 hats in the minority (red or blue) and guess that i was in the minority. but the people in the true minority (1, 2 and 3) have already guessed at 20 seconds, freeing us. It works with 8 people the same way: 1 - r 2 - r 3 - r 4 - r 5 - b 6 - b 7 - b 8 - b If i was prisoner 1, 2, 3 or 4 i would wait to 30 seconds because i see 3 hats (red) in the minority and then guess that i am also in the minority (red). if i was prisoner 5, 6, 7 and 8 then i would wait to 40 seconds before guessing because i see 4 hats in the minority (red) and guess that i was in the minority. but the people in the true minority (1 to 4) have already guessed at 30 seconds, freeing us.
Right so we're trying to escape the prison now but we've encountered a problem from our scouts ahead. They say there is a maze to the exit. Each corridor except the last corridor turns to two corridors, one to the lefthand way and one to the right way. How do we get to the exit? (This one will make you cry inside from its lameness)
assuming symmetrical maze, they meet in the middle*, so one aspect is probably simply moving away from the starting point. without knowing where the exit is, however... * straight ahead, l/r fork. left, f/b fork (left takes you back, right away). right, same but inverted. go forwards, l/r fork, right from initial left-right meets with left from initial right-left.
wait what. "Each corridor except the last corridor turns to two corridors, one to the lefthand way and one to the right way." go right. if last corridor, exit, else fork. go right. if last corridor, exit, else fork. go right. now you're either back at the start or at a dead end.
Hahaha, he got you! Think of a really easy answer and read the puzzle very carefully And the maze isn't necessarily symmetrical, is it?
Put one hand on one wall, keep turning in the same direction if presented with a corner so that you keep your hand on the same wall, in the finish you must turn back on yourself and if you keep your hand on the same wall in due course you must surely find your way to the exit?
