[bonanova]

Warden Jones has informed 100 prisoners that tomorrow

is their day of reckoning.

For a clearer statement of this puzzle, please click here.

They will be led into a courtyard and the warden will

place a black or a white hat on each or their heads.

No prisoner will see the color of his own hat, but

each will see the color of all 99 of the other hats.

The prisoners will be given a pencil and a piece of paper.

They will each write W or B on the paper. The warden will

go to prisoners at random and read what has been written.

If it matches the color of the prisoner's hat [W-white or

B-black] the prisoner will be set free. If it does not match,

the prisoner will be executed on the spot.

Tomorrow the prisoners will not be permitted to

communicate with each other in any way. They will

be permitted to look at any of the other hats, and

to write a W or B on their paper and nothing else.

The prisoners are meeting tonight to form a strategy.

They have called you to their meeting as an expert

logician to assist them.

How many prisoners can you guarantee will be set free?

[Guest]

OK here it is ...

There is a strategy that will guarantee 50% survival whatever the warden does.

What is it?

This solution is really quite clever.

And having said that of course, I have to disclaim ownership of it.

Prisoner 1 knows color of hat of prisoner 2. When the warden asks him color of his hat, he tells the color of hat of P2. P2 knows the color of P1's hat. Depending upon the outcome of P1 (whether he is set free or executed) he will correctly be able to tell the color of his hat. Similarly P3 & P4, P5 & P6....

[Guest]

To clarify about the list:

if there were four prisoners named Al, Bob, Chuck and Dave, for example,

the warden might go to Al's cell and give him a black hat, go to Bob's cell and give him a black hat, go to Dave's cell and give him a black hat, go to Chuck's cell and give him a white hat.

He would then go to his office and type up a list. It could be in any order. He might choose to alphabetize it, but he doesn't have to.

The list might look like this.

Al - black

Bob - black

Chuck - white

Dave - black

Al would see:

Bob - black

Chuck - white

Dave - black

Bob would see:

Al - black

Chuck - white

Dave - black

Chuck would see:

Al - black

Bob - black

Dave - black

Dave would see:

Al - black

Bob - black

Chuck - white

Ok, unlike the other this is true probables....

4 possible circumstances -

Even hat split - 50-50

near even split - 51/49 or 52/48 NS1 and ns2

Non-even split.

if even split as a prisoner you see 50/49 and would choose the lower one - all live (choosing higher all die therefore bad idea)

if near even split ns1 you see either 51/48 or 50/49 Choosing lowest 49 live (49 see 51/48, 51 see 50/49)

if near even split ns2 you see either 52/47 or 51/48 choosing lowest 48 live (48 see 52/47)

53/47 means 53 / 46 or 52/47

So logically if you see more than 53 of one color - choose that color otherwise, choose opposite

at 55+ number of max hats live

at 54 all die

at 53/47, 46 live

at 52/48, 47 live

at 51/49, 48 live

at 50/50 all live

so you have a chance to lose all no matter what advice you give (unless it is ignore the list and guess)

I would then advise as follows.

If you see 55 or more of a color on the list, choose that color, and lets hope the warden did not hear us choose that number.

Otherwise, choose the lower number

worse case all die

second worse, 44 live

45 live 46 live 47 live 48 live 49 live 55 live 56 etc... all the way to all living

If you apply human logic, you choose a number at random between 52 and 59 excluding 55 and you have a better chance than 55.....