<blockquote>quote:</font><hr>Originally posted by Donut:


I think that only saves 50% Lioness. If you tell the man in front of you the colour of his hat he has to say the colour of his own hat to save himself and he can't help the man in front of him. So 50% can be saved and 50% of the rest hav a 50/50 chance - so that's approximately 75% saved.<hr></blockquote>

That's why I said that'll only work for the first guy...I think coughing is the best idea so far...

Okay, I have it. He looks at the shiny, polished boots of the despot from the corner of his eye so he can see his reflection...

Or, he asks the despot if he's colourblind or something, because he needs to ask. Then when the despot says he's not, the prisoner dares him to prove it by telling him the colour of his hat. ;)

Or he can just agree with each prisnoer to say the following for every 'other' prisoner, "Black, White, Black, White, Black, White, Black, White".

75% of prisoners may be saved.

And these things vary, so there are many more solutions in my head hehe.

<font color="darkblue"> I can do this easily if we take the number of people down to 3.
They are in a Line as Follows-
<Person 1 <Person 2 <Person 3

We will say that person 3 has a black hat on. He can see the hats of both person 1 and person 2. If he sees 2 white hats, he knows that his must be black because there is at least 1 black hat in the lineup. If however, he sees 1 Black and 1 White hat, he says nothing.
If person 3 doesnt say anything, Person 2 knows that his hat must be different to person 1's, so he knows the colour of his hat.

Then, out of the remaining 2, 1 further can be saved. If person 3 calls out that he has a Black hat, Person 2 knows his must be the same as person 1s ie. White.

If person 2 figured his out, person 3 cant do a thing.


So the maximum which can be saved is 2/3. And this riddle is not mathematical, but rather scientific. It is a control experiment- the results are recorder twice, the seond time after changing 1 variable. </font>

Yes, but every prisoner can only see the hat on the one person directly in front of him. So person number three can only see the hat on person number two. Not the one on person number one, not the one on person number four, not his own, only that of number two.

Plus, you don't know how the hats are divided. For all we know there's one black hat on person 1 and all the other 99 have whites. Or, it could be the other way around. Or, more likely, somehting inbetween.

I still like my cricket solution best ;) [img]tongue.gif[/img]