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I actually like this one:

There are five pirates in a ship and they have found 100 coins.

The biggest pirate offers a way to divide the coins. If at least half of them agree on the division, it will be done. If not, they will drop him into sea and the next biggest pirate will make an offer.

What is the number of maximum coins that the first one (the biggest) can get?

Other assumptions:

The pirates first like to live, then money, then killing each other, but most of all they believe in democracy!

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its easy but I like it – Mahdi Jul 11 '14 at 14:07
You make the impression that you have an answer allready. If so then why do you ask this? – drhab Jul 11 '14 at 14:10
@drhab yes I know the answer but this question was so nice that I decided to share it with other ones! – Mahdi Jul 11 '14 at 14:12
The general case has been posted and solved here on puzzling – Ross Millikan Jul 11 '14 at 16:02
FYI: A comment to this question lists numerous variants of the puzzle on Math.SE. – Blue Jul 11 '14 at 16:26
up vote 9 down vote accepted

If we have only pirates $D,E$, then $D$ may suggest that he gets all the coins, a suggestion supported by the required 50% (himself).

Consequently, with three pirates $C,D,E$, pirate $E$ will support any suggestion that gives him more than $0$ coins, $D$ will object to any suggestion (including the suggestion to give him all as this results in some killing fun). Therefore $C$ can suggest to keep $99$ and give $1$ to $E$, which will be supported by himself and $E$.

Consequently with four pirates $B,C,D,E$, pirate $D$ will support any suggestion giving him more than nothing. This allows $B$ to suggest $99$ for himself and $1$ for $D$. Note that if he suggested to keep all, then $C$ and $E$ would object while $D$ might not bother: His choice does not influence his survival or his income - but objecting gives him the satisfaction of killing $B$!

Consequently with five pirates $A,B,C,D,E$, pirate $C$ will support any suggestion giving him anything at all; same for $E$. Therefore $A$ suggests to keep $98$ coins and give $1$ coin each to $C$ and $E$.

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According to the official timekeeper, my identical answer was posted one second before yours. I suggest we upvote each other :-) – TonyK Jul 11 '14 at 14:20

If it gets as far as $P_4$, then $P_4$ will suggest a division of $(0,0,0,100,0)$, and successfully vote in favour of it. So $P_5$ can't allow this.

Hence if it gets as far as $P_3$, then $P_3$ will suggest $(0,0,99,0,1)$, which wil be approved by $P_3$ and $P_5$. So $P_4$ can't allow this.

Hence if it gets as far as $P_2$, then $P_2$ will suggest $(0,99,0,1,0)$, which will be approved by $P_2$ and $P_4$.

Hence if $P_1$ suggests $(98,0,1,0,1)$, then it will be approved by $P_1,P_3,$ and $P_5$.

If $P_1$ tries $(99,0,1,0,0,)$ or $(99,0,0,0,1)$, thinking that $P_3$ or $P_5$ have nothing to lose by voting in favour of $0$ coins (they will get $0$ anyway), then the third tie-break rule applies (they like killing each other).

Hence the biggest pirate gets $98$ coins.

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