## Dividing up the loot…

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Five rational and extremely greedy pirates, ranked from most senior to most junior, have 76 gold coins to divide amongst themselves. They have agreed to use the following method to divide them, and we know that they will all abide by the final result:

The most senior pirate will propose a split, and all the pirates will vote on it. If 50% or more vote for the proposal, then they split the coins as proposed, otherwise the most senior pirate has to walk the plank and the next most senior pirate will propose a split to begin the system again.

However, apart from this agreement, the pirates do not trust each other and they will not make or honour any other agreement. The pirates are also a murderous lot, and would rather another pirate walked the plank so long as they got the same number of coins themselves. Knowing all this and that all of the other pirates know it too, what should the most senior pirate propose so that he gets as much treasure as possible?

27, 25, 24, 0, 0

Akshat RathiMay 27, 02012 at 22:07

He can do even better… Trick is that they are all super rational and greedy…

Christo FogelbergMay 27, 02012 at 22:53

28;24;24;0;0

jennyMay 28, 02012 at 09:31

74,0,1,0,1

Edward MooresMay 28, 02012 at 07:57

74,0,1,0,1

Rob NewnesMay 28, 02012 at 08:50

I’ll post a trickier version of this puzzle shortly, but in the meantime, in case anyone’s wondering how Rob and Ed might have got to their answers…

The trick to this puzzle is understanding what will happen in the later games, and (given the pirates’ greediness and rationality) what this means for the earlier games with more pirates in them.

Let’s label the pirates 5, 4, 3, 2, 1, from most senior to most junior. The latest/smallest possible game is two pirates. In this game, 2 would allocate 76 gold coins to himself, vote for himself, and get the gold. I.e., with two pirates the allocation would be {76, 0}.

Knowing this and knowing that 1 knows this as well, pirate 3 could offer 1 a single gold coin for his vote in the three-pirate game, and the allocation would be {75, 0, 1}.

In the four-pirate game, 2 knows that he will never have a chance to play the two-pirate game and that he gets 0 gold in the three-pirate game. Therefore if 4 offers him any gold he will be better off and will vote for it. 4 knows this as well, so the allocation would be {75, 0, 1, 0}.

Finally, in the five-pirate game, 5 knows that he has to persuade two other pirates to vote for him. Therefore, by the same logic, the allocation (and answer) is {74, 0, 1, 0, 1}.

Addendum: {74, 0, 1, 1, 0} and {74, 0, 0, 1, 1} might appear to be solutions as well. However pirate 2 also gets 1 gold in the 4 pirate game and all else equal for himself he wants pirate 5 to walk the plank. Therefore pirate 2 would not vote for either of these proposals.

Christo FogelbergJune 2, 02012 at 14:17

so pleased you put up “why” – thank you!

jenhnzJune 2, 02012 at 22:06

[…] a comment » Last week, we met 5 pirates with 76 gold coins. Here’s an interesting tweak on the same […]

Dividing up the loot… again! « Qua locusJune 3, 02012 at 22:58