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Once I've been given the pencil, I use it to stab the warden through the eye, every prisoner runs for it, some get shot down but the majority of us survive.

Jean Eugene Roberts wrote:the warden.

I take it no-one here has had real experience in a prison. Pah. Warden...singluar! Not likely!

This one is a lil' tricky in my opinion, so if anyone wants me to either post the answer, or PM it to them, let me know and .......... well you never know .

Lenoir wrote:

Jean Eugene Roberts wrote:the warden.

I take it no-one here has had real experience in a prison. Pah. Warden...singluar! Not likely!

Nooo! Prison break lied to me!

Moonbeam, I for one vote for you revealing the answer, I don't think anyone's going to have luck with this one.

I still stick by my original answer...no rules should be allowed to be changed

*spits out tongue*

Jean Eugene Roberts wrote: Moonbeam, I for one vote for you revealing the answer, I don't think anyone's going to have luck with this one.

I'll leave it a day or 2 (prob 'til after I get back from Blackpool on Sunday), then I'll post the answer if it's still not been solved.

It is possible to guarantee to save 19 prisoners, with a 50/50 chance of saving the 20th - but how do they do this

The rules stay as they are Beardy

Jean Eugene Roberts wrote:Moonbeam, I for one vote for you revealing the answer, I don't think anyone's going to have luck with this one.

........... and here it is:

The answer is parity bits. Let red = 1 and blue = 0. The prisoner at the rear of the line counts up the number of red hats he sees. If the result is odd he says "red", otherwise "blue". (This is called "even parity". You could do it the other way around; the decision is arbitrary.) If his hat happens to be the opposite colour he will be killed on the spot: a 50% probability if the hats are given out randomly, and the only risk any of them has to take as long as nobody screws up.

Now all the prisoners now know the parity of the remainder of the line - call this the "parity bit". If any prisoner says "red", the others know that the line after that position has opposite parity, and mentally flip the parity bit. This continues all the way down the line, with each prisoner knowing the parity of the remaining part of the line and toggling this value every time someone says "red". When it's his turn, he counts the number of red hats. If his own hat is red then the parity bit and the count will disagree. Therefore, if the parity of the count is the same as the parity bit he says "blue", otherwise "red".

An example. Let's say prisoner #20 says "red", and is freed (or not). Everyone knows that there are an odd number of red hats distributed among the other 19 prisoners. Now prisoner #19 counts the red hats in front of him and comes up with 5. Since he knows the line from himself forward has odd parity and so does the part of the line in front of him, he must have a blue hat; otherwise the count would have been even. (If the parity bit was even and he counted 4 red hats, he would reach the same conclusion.) He says "blue" and is freed. Prisoner #18 counts and sees 4 red hats - an even number, but the parity bit is odd. He knows that his hat must be red, and after he says "red" everyone else knows that the parity bit is now even. And so on.

Right....

There is a ladder that is 125 feet high. It is pointing straight up (for some reason this is a magical ladder that never falls over, even without support )

It is pointing straight up towards the sky (so it isn't lying flat) in the middle of a field (so it isn't leaning against anything).

There is a man. He is naked. No parachute or anything (I'm trying to cover all the angles here ).

He climbs up said ladder, before shouting to the world "I can't live like this!" and jumps all the way back to the solid hard not-soft-in-any-way ground.

He stood up, realised that due to his survival he maybe does have something to live for, and walks off without a scratch, broken bone, or in any way damaged.

How does he survive?

He only climbed up one step...?

Got it in one! I'll have to think of a harder one next time.......

Two guys are at a restaurant. They order tea. The waitress brings them identical drinks. The first guy finished his drink and leaves. The second guy reads a magazine while drinking finishes twenty minutes after the first guy then dies. They both finished their drinks and the drinks were exactly the same.

How was the second guy killed?

He got shot in the back of the head by a lone sniper with a .50cal Sniper Rifle.

Hart attack?

sleightlycrazy wrote:Two guys are at a restaurant. They order tea. The waitress brings them identical drinks. The first guy finished his drink and leaves. The second guy reads a magazine while drinking then dies. They both finished their drinks and the drinks were exactly the same.

How was the second guy killed?

He died of old age many years later. The tea is an irrelevence.

sleightlycrazy wrote:The first guy finished his drink and leaves.

Perhaps the "leaves" are in fact the tea leaves. By eating/drinking these leaves (some sort of antidote) as well as the tea the first guy is not poisoned by the 'evil tea'. However guy two fails to drink the leaves and thus dies, as the tea is evil... pure evil.