# Puzzle with pirates

That one I'm pretty low on ideas of how to approach it.

Five pirates of different ages have a treasure of 50 gold coins. On their ship, they decide to split the coins using this scheme: The oldest pirate proposes how to share the coins, all of them will then vote for or against it. If more than a half of pirates vote for it, then the coins will be shared that way. Otherwise, the pirate proposing the scheme will be thrown overboard, and the process is repeated with the pirates that remain. Assuming that all 5 pirates are intelligent, rational, greedy and do not wish to die, how much will the 2nd youngest pirate get?

• 97
• 98
• 0
• 20
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97 and 98 are not very likely with a treasure of 50 coins in total. Anyway, the oldest pirate can propose to give $1$ coin to the youngest and "middle" pirate, and keep the remaining 48 coins. – user133281 May 6 '14 at 9:41
Well, yes - I'm quite surprised with the first two options, but perhaps this is just done as all other tasks in the problem list have 4 options. Why would the middle and the youngest pirates voting for $(48,0,1,0,1)$ though? – S.D. May 6 '14 at 9:43
The middle pirate would vote for $(49,0,1)$, so the 2nd youngest pirate will vote for $(49,0,1,0)$ with $4$ pirates left. Hence the youngest and middle pirates get nothing when the oldest pirate dies, so it suffices to give them $1$ coin. – user133281 May 6 '14 at 10:00

An important detail in the problem is missing: you need to specify how a pirate will vote if he gets the same amount of money either way, and does not die either way. I will assume that in this case the pirate is malicious and will automatically vote against.

Now, we start with the case of $1$ pirate, and slowly increase the number of pirates.

• If there is ever one pirate remaining, she will just take all the 50 coins for herself and vote for her own proposal.

• If there are two pirates remaining, the younger pirate knows that she can get $50$ coins herself if the older pirate's proposal fails. Thus she votes against the proposal no matter what. The older pirate dies, and the younger pirate keeps $50$.

• If there are three pirates remaining, the second-to-oldest pirate does not want to die, and he knows he will die if the proposal does not go through. Thus he votes for the proposal no matter what. Given this, the oldest pirate takes all the money ($50$) for herself, and gets $2$ out of $3$ votes (including her own), and the proposal passes.

Here is where it gets interesting.

• If there are four pirates remaining, the oldest needs to win $3$ votes for his proposal to pass (including his own vote). The second-to-oldest will vote against no matter what, because she knows if the proposal fails she will get all of the gold. So the oldest cannot win her vote. But the two youngest pirates know they will get nothing if the proposal fails. Thus the oldest pirate need only bribe them with $1$ gold each, and they vote for him. So, the oldest pirate proposes $48$ for himself and $1$ for each of the two youngest, and the proposal passes.

• Finally, if there are five pirates remaining, the oldest also needs to win $3$ votes. In order to win the second-to-oldest's vote, she needs to pay him $49$ gold. In order to win the third-to-oldest's vote, she needs to pay her $1$. And in order to win the two youngest's votes, she needs to pay them $2$ each. So she pays $2$ to one of the two youngest, and $1$ to the third-oldest, and keeps $47$ for herself.

The 2nd to youngest pirate will either get $0$ or $2$, depending on how the oldest pirate chooses to propose. Thus this problem is ill-posed. (I also am confused why $97$ and $98$ are listed as answers with only $50$ gold total...)

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"you need to specify how a pirate will vote if he gets the same amount of money either way, and does not die either way." this is not correct. strategies are part of the answer not the question. in the worst case scenario, we can have multiple subgame perfect Nash equilibria. – Sergio Parreiras May 6 '14 at 10:34
@SergioParreiras This problem is not about Nash equilibria. The problem states: "Assuming that all 5 pirates are intelligent, rational, greedy and do not wish to die". The implicit assumption is that rational pirates will do one and only one thing, but the problem specifies no way to deal with cases of uncertainty. – 6005 May 6 '14 at 10:38
For your two pirate answer, since we know the pirates do not wish to die, the older pirate will split the coins as 0 for himself and 50 for the younger pirate. This way he doesn't have to die as the younger pirate gains nothing by rejecting his proposal. So I'm not convinced the logic you're applying holds. – postmortes May 6 '14 at 11:00
@postmortes Refer to my assumption at the beginning of this answer. Notice that the younger pirate gets $50$ coins either way, and certainly doesn't care whether the older pirate dies or not. So there is no logical reason to do one thing or the other. – 6005 May 6 '14 at 11:03

I will use Sub-Game Perfect Nash Equilibrium as a solution concept and solve the game by backwards induction (also known as Zermelo's algorithm).

In the last decision nodes of the game tree (round 4), there are only two pirates left. If the youngest pirate vote against the offer, he will get everything and and the oldest will be thrown overboard. Clearly the youngest is voting against the proposal if he does not get everything. If the proposal gives everything to the youngest pirate, he is indifferent between accepting or rejecting it (we already have least two equilibria!). Let's consider the case where the youngest votes for the proposal if he gets everything. In this case, the proposer offer everything to the youngest who then accepts it.

In the last but one nodes of the tree (round 3), there are only three pirates left. Clearly, the second youngest pirate strictly prefers to vote for any proposal that gives him any coins and is indifferent between voting for or against any proposal that gives him zero coins. So again we may have multiple equilibria. Let's consider the case he always votes for the proposal. In this case, the proposer offers zero to all the others and the proposal is accepted.

So in round 2, the proposer (the 2nd pirate) knows the 3rd pirate will require at least all the coins to vote for the proposal but both the 4th and 5th pirate will vote for any proposal that gives them any coins. Thus, the best he can do is to offer one coin to each of the 4th and 5th pirates and zero to the 3rd.

Finally, in round 1, the proposer knows the second pirate requires at least 48 coins to vote, the third requires at least zero, the forth and fifth require at least one each. So the first pirate will propose (47,0,1,1,1).

In sum, there is at least one subgame-perfect Nash equilibrium where the division proposed is (47,0,1,1,1) and all but the second pirate vote in favor. You can check the other cases (breaking the ties/indifferences in different ways) but for sure there are other equilibria (perhaps with similar payoffs).

This is a cute game, it reminds a bit "Hungry Lions" but with voting added.

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