# Dead Men Solve No Puzzles

Okay, enough generalities for a while. I heard a delightful puzzle that has a surprising answer. The origin is unknown to me. If anyone has a reference, I will gladly give full credit.

The hypothetical scene is somewhere in the Caribbean at a pirate hideaway. A nasty crew of five of the bloodthirstiest pirates ever to sail the sea has just scored on a raid and netted 100 Doubloons. The question now is how do they divide this ill-gotten gain?

Dividing it equally, 20 Doubloons per pirate is out of the question because the captain, a very nasty sort, wants more. To prevent post-raid arguments, the pirates have established the convention that the captain proposes before all of them an appropriate distribution. Then they vote. Everyone, including the captain, gets a single vote.

If a majority agrees (argh-ree? Hmm… is there an emoticon that uses a question mark for a pirate’s hook?), then the distribution is made (at least half of the votes must be aye to pass — there could be an even number of pirates voting — read on). Getting a positive vote is vitally important to the captain because if he loses, the majority obviously knows they have power and will mutiny, killing the captain.

Pirates being what they are, even those who support the initial vote will also turn on the captain and help to send him to Davy Jones. Then the remaining pirates pick a new captain and the process starts over. The pirates have worked together for years and all know each other’s relative strength. That is, each one is pretty sure of what fights he would win and which he would lose. This means that if the captain is killed, the number two guy is unanimously elected to replace him.

On the other hand, even if some among the crew are highly dissatisfied with the vote (assuming it passed), they will not mutiny because they do not have a majority of their colleagues to support them and no rational pirate will mutiny without having a secure majority.

Since this system has been in place for several raids and has worked, we assume the captain is reasonably shrewd about how he makes the distribution.

Also, we can suppose that this haul is particularly attractive to the captain because he wants to take his share and retire to Miami. So he has an incentive to keep as much of the haul as possible without regard to the ongoing enterprise.

What does he propose that will net him the most gold and not result in walking the plank?

I will post the analysis in a few days. If you work it out first, you will be surprised.

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• John Swallow

Maybe a bit of a moot point, mate. “Pyrates being what they are…” are somewhat democratic in a sense so that chaos does not ever take a hand in life at sea – most Captains and crew agreed to a set o’ guidelines (aka “the Code” or “the Articles”) that were strict enough to keep order, but fair enough to do the same. To wit’…excerpts from two such “articles”:

I. Every Man Shall obey civil Command; the Captain shall have one full

Share and a half of all Prizes; the Master, Carpenter, Boatswain and

Gunner shall have one Share and quarter.

VIII. If any Man shall lose a Joint in time of an Engagement, shall have

400 Pieces of Eight; if a Limb, 800.

[Capt. John Philips]

IX. No man to talk of breaking up their way of living, till each had shared £1,000. If in

order to this, any man should lose a limb, or become a cripple in their service, he was

to have 800 dollars, out of the public stock, and for lesser hurts, proportionately.

X. The captain and quartermaster to receive two shares of prize: the master,

boatswain, and gunner, one share and a half, and other officers one and a quarter.

[Capt. Bartholemew Roberts (aka Black Bart)]

Capt. John Swallow, Buccaneer

• Chris Moore

I think we have to work this backward, and also assume that all the pirates are perfectly rational, not just the current captain.

If only two pirates remain, the captain (D) takes all the money for himself, since he only needs his own vote and doesn’t care what the other pirate (E) thinks.  D=100 E=0

If three pirates remain, the captain (C) has to convince just one of the other two pirates to vote with him. Since E will get no loot at all if they kill C, he’ll be happy to accept just one coin, and so the captain proposes C=99 D=0 E=1 and E agrees.

If four pirates remain, the captain (B) again only has to get one of the other three on-side.  Since D stands to get nothing if B dies, D will accept any non-zero reward.  The captain proposes B=99 C=0 D=1 E=0 and gets D’s vote.

So when all five pirates are alive, the captain (A) needs to get two of the other four votes.  He can do this by giving a coin to C and E, since they’ll both get nothing if he dies.  He proposes A=98 B=0 C=1 D=0 E=1 and gets C and E to vote with him.

• Barry Etheridge

I’ m not sure that this approach takes account of the fact that there is a set order of succession because the pirates are not equal in strength..

A & B = 100%A, I agree as B is too weak to do anything about it (although he might be looking for another job!)

But A, B, C is a different proposition, Whatever B is offered, it is always logical to vote against it because a majority will result in his becoming captain and having the whole treasure. If C is offered anything then he knows B is strong enough to take it off him (probably with extreme prejudice) so there’s no incentive to vote for that proposal. If he is offered nothing, on the other hand, there is no incentive to vote against the proposal because nothing is what he will always end up with however many captains are bumped off.

Following this logic forward it must be that the 2nd in succession will always vote against the captain no matter how many pirates there are but it is never in the interest of those below him to vote for anything that gives them any reward for fear of reprisals from above. The captain, in other words, always takes 100% no matter how many pirates in the crew.

• Chris Moore

> If C is offered anything then he knows B is strong enough to take it off him

I don’t think B would take anything off C if the coins were distributed after voting.  Both A and C want C to have the coin, so B is in the minority.  The pirates go with whatever the majority vote for.

Seems a little odd to add new rules to the puzzle after it’s been stated, and use them to come up with a different solution.

• Rasmus

I get the theory but it seems to only apply in theory. If – in the case of 3 pirates – the captain offers a 99-0-1 split, wouldn’t the 2 agree to kill the captain for a 50-50 split? Same deal applies whenever more than 2 pirates are in the equation, I guess?!?
Therefore the captains solution – given that he’s retireing to Miami and doesn’t care about his crew at all – is to steal the loot while the others are asleep, and keep the 100 gold pieces for himself

• Chris Moore

In the case of 3 pirates, if they kill the captain it isn’t a 50-50 split.  The stronger of the remaining two pirates takes everything because he only needs his own vote.

Saying that they can agree to a 50-50 split and go against the ancient pirate code is changing the puzzle.

It’s possible the stronger of the last 2 pirate will promise a 50-50 split to try to get the weakest pirate to vote with him, but they both know what will really happen – he will renege on his deal and take all 100 coins, because he’s stronger, and he’s a pirate.

• Chris Moore

In the case of 3 pirates, if they kill the captain it isn’t a 50-50 split.  The stronger of the remaining two pirates takes everything because he only needs his own vote.

Saying that they can agree to a 50-50 split and go against the ancient pirate code is changing the puzzle.

It’s possible the stronger of the last 2 pirate will promise a 50-50 split to try to get the weakest pirate to vote with him, but they both know what will really happen – he will renege on his deal and take all 100 coins, because he’s stronger, and he’s a pirate.

• Chris Moore

In the case of 3 pirates, if they kill the captain it isn’t a 50-50 split.  The stronger of the remaining two pirates takes everything because he only needs his own vote.

Saying that they can agree to a 50-50 split and go against the ancient pirate code is changing the puzzle.

It’s possible the stronger of the last 2 pirate will promise a 50-50 split to try to get the weakest pirate to vote with him, but they both know what will really happen – he will renege on his deal and take all 100 coins, because he’s stronger, and he’s a pirate.

• Manolo

Game Theory!