Here is a nice little logic puzzle for you all to play, it’s called the Pirate Loot problem and it goes something like this:

Imagine you are the oldest in a band of five pirates, where no two pirates in the group are of the same age. It is your responsibility as the oldest pirate to decide how to divide the group’s booty, which comprises one-hundred pieces of gold. Once you announce how you will allocate the gold, all the pirates in the group (including you) vote either “yes” or “no” on your decision. If at least half of the votes are “yes”, the pirates divide the loot as you directed, after which everyone carries on about his business. But, if more than half of the votes are “no”, you are then killed, and the task of deciding the allocation of gold lies with the next oldest pirate in the group.

Knowing that all the other pirates are just as smart, and just as logical as you, and furthermore, knowing that they all want the largest amount of gold they can get for themselves, how do you divide the gold so that you get the maximum amount of gold pieces possible, and guarantee that you will receive enough “yes” votes to stay alive?

For those of you who would like to work it out for yourselves please look away now – here be spoilers (and indeed Pirates).

The best way to solve the problem is to chunk it. Deal with smaller bits of it and build it up. So if you think of the 2 pirates situation then in that situation the older pirate can keep 100 gold pieces. And since there would be two pirates before there was one pirate (if they kept killing the oldest) the youngest pirate will always vote for you if you give him one gold piece, all it takes is giving him slightly more than he would otherwise get, to get the vote.

In the situation that there were three pirates that’s exactly what the oldest pirate would do, he is bound to vote for himself. So he can just give the youngest pirate 1 gold piece and get his vote and keep 99.

If there are four pirates then you can’t bribe the next youngest guy (in fact you’d never be able to bribe him however many younger people there are). And so you could choose to either bribe the youngest or the second youngest to get your vote.

So then in the five pirates problem (which after all was the one that I set) you can bribe the youngest, which gives you one vote at the cost of one gold coin. But who else can you bribe cheaply? Well you can bribe the second youngest with one coin because we know that in the three pirates problem he won’t get anything.*

So the answer is that if you’re the oldest pirate then you get to keep 98 of the coins and only have to dole out 2 coins to get the votes you need.

The thing I really like about this puzzle is that it is so close to real life. If we look at the economy around us then we see that the powerful can easily bribe the weak to agree with them. In fact the powerful can often bribe the weakest and ignore their middle strength rivals and still retain power and wealth. Just look at the situation in Iraq for example a variety of small countries are in the “coalition of the willing” circumventing the power of much larger rivals like France and Germany.

Basically this game really shows the mechanics of how the rich stay rich in the real world. Remember 2% of the world’s population hold 50% of the worlds wealth, it seems like that would be impossible, but this game shows exactly how this comes about.

*Some accounts of this problem will say that as part of the solution you have to bribe the middle pirate with one coin rather than the second youngest. The reasoning for this flows from imagining the problem the other way around. In their minds you give the gold to the second youngest to bribe them not the youngest. And then they go on to say that in the five pirate problem you would be unable to bribe the second youngest with one gold coin because he will demand two coins for his vote rather than just one. It doesn’t matter which of the two you bribe in the four pirate problem which means that in the five pirate problem there isn’t a guarantee for the middle pirate that he’ll get a coin if the oldest pirate is killed. This way he guarantees his coin so he would vote yes for one piece of gold.

This is a very strange problem. How would Kasper Hauser solve it, I wonder? My first idea was ‘well give everyone 20’, but equality obviously has nothing to do with capitalism.Once I had figured out how the whole thing works my second idea was ‘what if the youngest and the second eldest are in league together?’ But then all you do is bribe the middle and the second youngest. Anyway, it’s like all logical problems in that they work perfectly only within the boundaries the ‘questioner’ sets for them: it revolves around people wanting money, and the more of it the better. What if the other pirates don’t wany any money, but just want to kill you? They’ll refuse all your bribes and vote you out.The easiest solution is killing the other four pirates and keeping everything for yourself. By the way, the question I was thinking of asking you to solve was: How does the Traffic Alert function on a car radio work?