For those who like logic puzzles, here is one I found on BrainDen.

You are one of 20 prisoners on death row with the execution date set for tomorrow. Your king is a ruthless man who likes to toy with his people's miseries. He comes to your cell today and tells you:
“I’m gonna give you prisoners a chance to go free tomorrow. You will all stand in a row (queue) before the executioner and we will put a hat on your head, either a red or a black one. Of course you will not be able to see the color of your own hat; you will only be able to see the prisoners in front of you with their hats on; you will not be allowed to look back or communicate together in any way (talking, touching.....).

The prisoner in the back will be able to see the 19 prisoners in front of him. The one in front of him will be able to see 18…

Starting with the last person in the row, the one who can see everybody in front of him, he will be asked a simple question: WHAT IS THE COLOR OF YOUR HAT?

He will be only allowed to answer “BLACK” or “RED”. If he says anything else you will ALL be executed immediately.

If he guesses the right color of the hat on his head he is set free, otherwise he is put to death. And we move on to the one in front of him and ask him the same question and so on…

Well, good luck tomorrow, HA HA HA HA HA HA!”

Now since you all can communicate freely during the night, can you find a way to guarantee the freedom of some prisoners tomorrow? How many?

Change hats at night!

The guy way in the way back can swap hats with the guy in front of him.. then the 2nd to last guy can switch with the 3rd to the last guy ... etc.
That way they all know the color of their new hat.

That saves 19 of them for sure but the guy at the very front is at a 50/50

Change hats at night? They don't get the hat until the morning. If they change hats then, they will all be executed. The solution has one similarity, though.

The solution: 

First guy is a coin toss - let's wish him good luck.
His job is to establish the parity of black hats visible to him.
He says "Black" if he sees an odd number of black hats; "Red" otherwise.
By paying attention to what has been said, each prisoner will know his hat's color.

Example:
Second to speak hears "Black" and sees an even number of black hats. 
He knows his hat is black [odd changed to even - must be his is black] and says "black".

Third guy has heard "black" and "black" and sees an even number of black hats.
He knows his hat is red [even stayed even - his hat can't be black] and says "red".

And so on, to the front of the line.

General algorithm:
The first time you hear "black", say to yourself "odd".
Each time your hear "black" after that, change the parity: "even", "odd", ... etc.
When it's your turn, if the black hats you see match the running parity, you're Red; Black otherwise.
Call out your color.

thats a stiffy... i'll need to put my thinking cap on... now , shall i wear the red one... or maybe...

What if everyone can see everyone else's hats? there are 7 people and to win, nobody can guess wrong. At least one guess must be right (passes allowed).

However, everyone must submit their answer on a piece of paper, so no even odd idea.

Solution- message me.