Puzzle

div0 · 23 Jun 2008 at 10:15

iraiguana said:
to me you would not have enough parameters to solve the puzzle as its purely random in which case they would have a 50/50 chance as a guess.

The odds for all to survive would be low, but you could not say they would not all guess correctly.

Nah, there's no strategy to save them ALL, but there is a strategy they could use to be certain of saving SOME of them

For example you could save at least 50% of them quite easily.

If they agree that the guy at the back of the line says the colour of the hat on the guy at the front of the line. Then he MAY die, or he MAY be saved, depending what colour his hat is - but he has told the guy at the front of the line what colour to guess when it is his turn. So the guy at the front of the line will definitely be saved.

If the guy 2nd from the back, says that his hat is the colour of the guy who is 2nd from the front, then again he may die, or he may survive, depending on luck, but he has told the guy 2nd from the front how to survive.

Rinse and repeat 5 times and the back 5 people may die, but the front 5 will definitely survive.

Not sure if this is the optimum solution or not, but it will save at least 5 of them for sure (and if they get 'lucky' then all 10 of them will survive).

Siliconslave · 23 Jun 2008 at 10:38

arr...

all bar one (maybe) the guy at the back says white if there are an odd number of white hats, or black if there are an even number of white hats. He has a 50% chance of being right, but the guy in front of him (9) can work out by the number of white hats he can see what colour he has and so on going forward

div0 · 23 Jun 2008 at 10:43

Sounds good to me

davethe11 · 23 Jun 2008 at 10:46

ah yeah, I misread the puzzle and thought each prisoner could only see the prisoner right in front of him rather than all the prisoners in front of him, even so the minimum number that could be saved surely is zero because the puzzle says nothing that there are equal numbers of black and white banners so it's possible that they all guess wrong (?)

[edit] think its five actually, the first guy says the colour of the guy in front (which may not be his colour so he has a 50/50 chance) so the guy in front knows what colour he has, he says it and lives but the next guy (each odd guy) has a 50/50 chance as he has to say the guys colour in front

thesubstitute · 23 Jun 2008 at 12:33

Siliconslave said:
arr...

all bar one (maybe) the guy at the back says white if there are an odd number of white hats, or black if there are an even number of white hats. He has a 50% chance of being right, but the guy in front of him (9) can work out by the number of white hats he can see what colour he has and so on going forward

Correct, but to elaborate:

The 10th person simply counts the number of either odd or even hats (odd or even needs to be agreed the night before) he sees, and calls it. He has a 50% chance of being right and surviving.

Let's say he calls odd hats, and calls white.

The 9th person then counts the number of white hats in front of him, if there're odd numbers of white hats, then he's obviously black; otherwise he's white.

The 8th person, based on the last person's call, does the same. Let's say 9th person calls white. 8th person counts odd number of white hats in front, then he's obviously got white because his own white + 9th person's white keeps the white hats odd.

And so on up the line.

So the minimum people saved would be 9.

Want anymore puzzles?

Siliconslave · 23 Jun 2008 at 12:39

sure

Dj_Jestar · 23 Jun 2008 at 13:00

thesubstitute said:
You're onto something with the train puzzle, but it's simplier than that. Just to simplify it a bit - let's assume that both trains stops at at instant and leaves immediately, but you'd be able to catch it if you're present when it arrives.

I was going to post that it's the same train, and the station is the end of the line.

thesubstitute · 23 Jun 2008 at 13:08

Let's try another hat one which is a bit easier:

This time, 3 intelligent prisoners are to be executed, but just before the execution they're given a chance to save themselves by the infamous hat game.

They're lined up as such that the 3rd person can see both of the previous people's hats, 2nd person can only see the 1st person's hat, and 1st person can't see any hats.

They're all shown 5 hats - 3 blacks and 2 whites.

Only one of them needs to call out their own hat's colour and they'll all go free.

Can they save themselves in all possible scenerios?

Dj_Jestar · 23 Jun 2008 at 13:13

Yes, in a similar way to the 3 princes - it's a case of logic. The person in the middle will know their colour before the others, as the 1st can't see jack and the 3rd will remain quiet.

tl;dr = yes, 2 will always know their colour.

thesubstitute · 23 Jun 2008 at 13:17

Dj_Jestar said:
Yes, in a similar way to the 3 princes - it's a case of logic. The person in the middle will know their colour before the others, as the 1st can't see jack and the 3rd will remain quiet.

tl;dr = yes, 2 will always know their colour.

Not necessarily. If their arrangement is:

Black, Black, Black, how can the second person know what colour his hat is?

iraiguana · 23 Jun 2008 at 13:23

thesubstitute said:
Correct, but to elaborate:

The 10th person simply counts the number of either odd or even hats (odd or even needs to be agreed the night before) he sees, and calls it. He has a 50% chance of being right and surviving.

Let's say he calls odd hats, and calls white.

The 9th person then counts the number of white hats in front of him, if there're odd numbers of white hats, then he's obviously black; otherwise he's white.

The 8th person, based on the last person's call, does the same. Let's say 9th person calls white. 8th person counts odd number of white hats in front, then he's obviously got white because his own white + 9th person's white keeps the white hats odd.

And so on up the line.

So the minimum people saved would be 9.

Want anymore puzzles?

I disagree, as you have assumed that they have put in an even number of black and white hats. If they did not do this then your plan falls down.

I would say if you could tap the guy in front on the left or right to indicate which hat they had then you could save at least 9/10 and the first one has a 50/50 chance.

//edit

div0 I stand corrected I did not read that the plan would inform the others by the direction of his guess a bias to the rest. As I assumed it was random you could have say 7 hats at the end being black etc

div0 · 23 Jun 2008 at 13:37

iraiguana said:
I disagree, as you have assumed that they have put in an even number of black and white hats. If they did not do this then your plan falls down.

No it doesn't.

The plan is as follows...

The guy at the back agrees to 'guess' WHITE if he can see an odd number of WHITE hats.

So, say the guy at the back can see 5 black and 4 white. He doesn't know his own hat's colour (obviously).

He sticks to the plan and guesses his hat is BLACK. He guesses BLACK because he sees an even number of WHITE hats in front of him (4).

He may be right (or he may be wrong) that his own hat is black - but he's now told all the guys in front of him that (between the 9 of them, in front of him) there is an even number of WHITE hats.

So if the next guy counts an odd number of white hats (in front of him), then his must be white (the guy at the back saw the odd number, plus his WHITE hat, to make even).

The next guy does the same, remembering that the guy behind was one of the white hats that the guy at the back counted.

They can all then work out their hat colour.

Only the guy at the very back is the one who may, or may not live - depending on whether his hat colour happens to fit in with his 'guess'.

kaiowas · 23 Jun 2008 at 13:48

thesubstitute said:
Let's try another hat one which is a bit easier:

This time, 3 intelligent prisoners are to be executed, but just before the execution they're given a chance to save themselves by the infamous hat game.

They're lined up as such that the 3rd person can see both of the previous people's hats, 2nd person can only see the 1st person's hat, and 1st person can't see any hats.

They're all shown 5 hats - 3 blacks and 2 whites.

Only one of them needs to call out their own hat's colour and they'll all go free.

Can they save themselves in all possible scenerios?

If the guy at the back can see 2 white hats he'll know he is wearing black and will shout out. If he can see any black hats he won't know what he is wearing and will stay quiet.

The middle guy will then know that either he or the person at the front (or both) is wearing a black hat. Therefore if he can see a white hat he'll know he is wearing black and will shout out. Otherwise he will say nothing.

The continued silence from those behind him would then tell the guy at the front that he is wearing a black hat and he will shout out.

mame · 23 Jun 2008 at 14:20

well if we're digging all these up again might aswell have the lightbulb one:

There are three light switches up in the attic of an old house... They control three light bulbs down in the basement. No dimmers or other tricks, it's one switch to one bulb. The problem is that you don't know which switch is connected to which bulb. You can only make one trip down to the basement after you have messed with the switches. The only thing you know is that all 3 bulbs are OFF to start with. How do you find out which switch controls which bulb?

thesubstitute · 23 Jun 2008 at 14:48

mame said:
well if we're digging all these up again might aswell have the lightbulb one:

There are three light switches up in the attic of an old house... They control three light bulbs down in the basement. No dimmers or other tricks, it's one switch to one bulb. The problem is that you don't know which switch is connected to which bulb. You can only make one trip down to the basement after you have messed with the switches. The only thing you know is that all 3 bulbs are OFF to start with. How do you find out which switch controls which bulb?

I like this one, nice bit lateral thinking puzzle. I shan't spoil it

div0 · 23 Jun 2008 at 14:58

Spoiler?: (think I've heard this before?)

Turn on 2 of the lights and leave them on for a while. Turn one of them off again, then go downstairs.

The bulb that is ON, is the switch that you've left on.

The bulb that is OFF and HOT, is then switch you turned on and off.

The bulb that is OFF and COLD, is the switch you didn't use.

Highlight above to see...

tripitaka · 23 Jun 2008 at 15:00

EDIT:
Deleted to avoid spoiler

thesubstitute · 23 Jun 2008 at 15:25

Another simple one:

You have two fuse wires and you know that the fuse wires are known to burn at various speeds along its own length. However you do know for certain that each fuse will burn out in an hour exact.

With no other equipment besides a lighter, how do you measure 45 minutes?

Siliconslave · 23 Jun 2008 at 15:38

fold one in half, fold one in 4 and burn one then the other, although i'm not convinced that doubling a bit over would totally work as it might well burn faster due to the increased heat...

thesubstitute · 23 Jun 2008 at 15:46

Siliconslave said:
fold one in half, fold one in 4 and burn one then the other, although i'm not convinced that doubling a bit over would totally work as it might well burn faster due to the increased heat...

Folding wouldn't actually work too well, since each of the 1/4 segments would still be burning at various speeds, some will burn out faster and others slower, which means they won't burn out at the right time.

You are thinking along the right lines though.