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Deep Dive into the Pirate Game

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    Shumeng Liu


I came across the Pirate Game first in high school as an introductory problem to game theory. The solution to which I found incredibly couter-intuitive yet elegant.

To my surprise, after 5 years, I was asked this exact problem during an interview for a developer job. Thanks to my high school experience, I nailed the solution and impressed the interviewer.

Now, I want to share this exicting game-theory problem with you. Buckle up!

The Game

Amara is the captain of a pirate vessel. His mateys, Bart, Charlotte, Daniel, and Eliza, are the other members of the crew. The group has come upon a bounty of 100 coins coins, and now must divide it up among the group according to "the pirate's code".

The Pirate's Code

The code stipulates that Amara, as captain, gets to suggest the first plan for distributing the coins among the five pirates. For example, 40 for Amara, 30 for Bart, 20 for Charlotte, 10 for Daniel and 0 for Eliza (sorry, Eliza).

After that proposal, each pirate (including Amara) votes "yarr" or nay on whether to accept the proposal. If the proposal results in either a tied vote or a majority of "yarrs", it passes and the coins are immediately distributed. If it fails to meet this threshold, Amara must walk the plank (in other words, face death), making Bart the next captain. A dead pirate is not eligible for future votes and coin disburals.

This propose-and-vote process now repeats with Bart as captain, and the captain's hat will be passed on, in order, to Charlotte, Daniel, and finally Eliza. (If it gets all the way to Eliza without a passing proposal, she gets the booty.)

The Pirates' Rationale

Each of the pirates act in a SELFISH and EXTREMELY RATIONAL way, their action obeys the following rules:

  1. Their highest priority is to stay alive.
  2. Their next priority is to maximise their gold horde.
  3. They are bloodthirsty, and would love to see a fellow pirate walk the plank if they think it won't affect their own life and gold distribution.
  4. They distrust each other-there are no alliances and they cannot collaborate on a strategy. Again, each pirate has excellent logical deduction skills, and they're aware that everyone has the same skills.

The Problem

So we arrive at the key problem for Amara:

What distribution should he propose to ensure he lives and maximizes his own gold return?

Take some time to ponder your strategy before scolling down.


For simplicity, we will refer to the pirates as A, B, C, D and E.


According to Rule #1 in the Pirates' Rationale, A wants to gurantee that her proposal gets passed at all cost.

One might consider a proposal where A leaves no coins for herself and gives each other pirate 25 coins. I would give a round of applause for the generosity behind such thinking. However, this plan does not maximise Amara's gold gain and the problem remains unsolved.

In fact, to survive, A needs only 2 "yarrs" from other pirates (with a vote from A himself, 3 votes makes the majority for 5 pirates). So she needs to bribe 2 other pirates only.

Thus, the problem becomes: what is the minimum amount of coins required for each of B, C, D and E to vote "yarr"? If we can answer this question, Amara would bribe two of them with the least needs.

More Observations

According to the Pirate's Code, every pirate is aware that if the proposal from A gets rejected, B will take the captain's hat and make the next proposal. Because every pirate acts rationally and wants to maximise their gain, they will only agree the proposal of A if and only if A is giving them more gold than B would.

Figuring out B's optimal proposal has no difference from figuring out A's, except there are only 4 pirates left. Similarly, resolving B's proposal is not possible without knowing the proposal C would make.

In other words, we should determine the captain's proposal for nn pirates before deducing the reactions of each pirate towards the captain of n+1n+1 pirates.

Take a moment to let the logic above sink in. Next, we will explore what each pirate thinks exactly in each scenario.

Working Backwards

E as the Captain

As the only person on the vessel, E would give all 100 coins to herself. Unless E experiences a sudden moment of philosophical englightment realizing that life is all meaningless, the proposal would pass.

However, we will soon learn that E will never become the captain (sorry, Eliza).

D as the Captain

In this case, there are 2 pirates only, it is not hard to see D can force a tie no matter what he proposes. Therefore, he would give all 100 coins to himself, and E would reject, but the proposal gets passed anyway.

Notice how when D is the captain, E gets absolutely nothing. This is a basis for further analysis.

C as the Captain

Now the fun begins, as the captain, C needs another crew member to vote for her to win the majority. Because every pirate is very smart and rational, everyone is very well aware of the proposal that D would make. Most importantly, as we saw in the previous case, C and E acknowledges that E would get no coins if the proposal gets rejected and D becomes the captain. To win the vote from E, it is sufficient for C to give merely 1 coin to E.

Therefore, C will make the proposal of giving 99 coins to herself, nothing to D and 1 coin to E.

For E, getting only 1 coin is miserable, but it is better than the alternative—let D become captain and get 0. Hence E would say "yarrs". With another vote from the captain herself, the proposal gets passed.

Choosing the correct crew member to bribe is thematic to the solution of the Pirate Game. Bribing E only takes one coin for C. In contrast, trying to bribe the beneficiaries of the disapproval of the proposal is way more difficult. In fact, even if C were to give all 100 coins to D, D would still reject the proposal because his bloodthirstiness (see Rule #3 in the Rationale).

B as the Captain

This situtaion is similar to the previous one with C as the captain. As there are 4 pirates, B needs a vote from another pirate to force a tie.

Everyone would recognize that if the proposal gets rejected, C would get 99 coins and E would get 1. If B gave E 1 coin, because the pirates are bloodthirsty, E would not be satisfied, as she can get the same amount from C anyway. Again, B as the captain should choose the correct member to bribe. D would happily accept 1 coin from B, because the alternative is get 0 coins when C becomes the captain.

Therefore, B would propose giving 99 coins to himself, 0 to C, 1 to D, and 0 to E. B and D would agree while C and E would reject. The proposal would pass with a tie.

A as the Captain (Original Problem)

Now we have all the information to resolve the case for Amara. As the captain of 5 pirates, she would need 2 votes from other crew members. A is aware that if B becomes captain, C and E would get nothing. Therefore 1 coin is sufficient to bribe each of them.

Amara will propose giving 98 coins to herself, 0 to B, 1 to C, 0 to D and 1 to E. A, C, and E would agree while B and D would reject. The proposal will pass.

And that, my dear reader, is the solution to the Pirate Game with 5 players.


The following sketch illustrates the optimal coin distributions and voting results for each captain.



Above is the typical 5-pirate scenario of the Pirate Game. And that is also all my interviewer asked for. However, had we stopped here, I would not call this a "deep dive".

It is not hard to notice the pattern for more than 5 pirates. The captain would give a coin to every other crew member and keep the rest for themselves. Hoever, when there are more than 200 pirates on the vessel, the problem is no longer about maximising the gold gain, but the desire to survive.


We will use a slightly different annotation for this section. Suppose there are nn pirates on the vessel; we will call them #nn, #n1n-1, ..., #1 with decreasing captain priority. That is, #nn will be the first captain. We will refer to odd-numbered pirates with the pronoun "she" and even-numbered pirates with the pronoun "he".

Who Can Survive?

Pirate #201 can survive. He achieves this by giving 1 coin to every odd-numbered pirate. He gets no money but got to stay alive, which we will learn soon is a privilege in itself.

Similarly, Pirate #202 can survive by giving 1 coin to every pirate with an even number.

However, Pirate #203 will not be able to survive. She would need 102 votes to win. The 100 coins can only get 100 votes for her via bribing. Counting herself, she will be at 101 votes—1 vote short from survival.

So, is 202 the maximum number of living pirates on the vessel? We should not jump to the conclusion so quickly.

Interestingly, Pirate #204 actually gets to survive. Similar to #203, she would need 102 votes. In addition to the 101 votes #203 can get, critically, he can get a free-of-charge vote from #203! Because #203 knows, if #204 gets voted off, she would not be able to survive either as the captain.

Pirate #205 will not be able to survive. He can only get 101 votes because #203's survival issue is covered by #204, and nobody would voluntarily vote for her.

Similarly, #206 and #207 cannot survive.

Pirate #208 is able to survive. Because he gets 3 free votes from #205, #206 and #207, who deseprately want to survive. Therefore #208 can get 104 votes and force a tie.

Who Else Can Survive?

Who is the next captain that can get their proposal to pass? #216. He will get 100 votes from coins, and 8 votes from #209 through #216. This makes up the needed 108 votes to force a tie.

So far, we learnt #201, #202, #204, #208 and #216 are capable of making their proposals pass. Do you notice a pattern here? Indeed, the numbers of those captains (including #201 and #202) follows the pattern 200+2k200 + 2^k, where k is a non-negative integer.

Out of nowhere, the power of 2 appeared in the expression. This is another counter-intuitive and fun part of this problem. We can illustrate this in the following sketch.


Any pirate who does not have survival issue will reject the proposal out of their bloodthirstiness. The other pirates will deseprately vote for the incumbent captain to keep themselves alive. Since half of the crew's votes is needed for the proposal to pass.

Eventually, there will be enough new joiners to protect a new captain. Afterward, they all turn into auto-reject machines to vote against higher-numbered captains. Therefore, every time a new captain emerges, the voluntary voting crew has to double in size, hence the power of 2.

Determine the Bribable Pirates

In the 5-pirate scenario, we discussed the pattern: odd-numbered captains would bribe odd-numbered crews and even-numbered captain would bribe even-numbered crews. Let's investigate whether this still holds for 200+ pirates.

#201 has to give coins to odd-numbered pirates, because all the even-numbered pirates would receive a coin from #200.

#202 is in a pretty similar spot, she also needs 101 votes to survive. The odd-numbered pirates in the range #1-#200 are harder to bribe, because they will get gold from #201 anyway. So #202 will likely bribe the even-numbered pirates. But she has (just a little bit) more freedom than that. Notice how when #201 is the captain, he gets no money left for himself. Therefore, #201 is also willing to say "yarr" for the price of 1 coin. #202 can choose 100 pirates out of the 101 candidates and give 1 coin each.

Following this train of thought, one would argue #204 has to bribe odd-numbered crews. But now he has a lot more freedom. In fact, any pirate in the range #1-#202 is his candidate. Because #202 had the freedom to pick 100 pirates out of 101. The expected return for any even-numbered pirate in the range #1-#200 is 100/101 coins, which is just slightly under 1 coin. Therefore, when given the chance to get 1 full coin from #204, they no longer have to pray for chances and would happily vote for him.

General Form Solution

We now outline the survival chances and gold distribution for the pirates when the number of pirates and gold coins vary. Suppose NN pirates are dividing GG gold coins among them.

Scenario I: N2GN \le 2G

  • Every pirate can survive.
  • The captain will get G(N2)/2G - (N - 2) / 2 coins.
  • Other pirates whose numebr has the same parity as the captain will get 1 coin.

Scenario II: N>2GN > 2G

  • Pirates up to 2G+M2G + M can survive. Where M is the greatest power of 2 up to N2GN - 2G.
  • The captain will not get any coins.
  • Pirates who are bribable has an equal chance, Gbribable\frac{G}{|bribable|}, to get 1 coin. For a captain numbered CC, the bribable pirates are as follows.
    • C=2G+1C = 2G + 1: Odd-numbered pirates up to 2G2G are bribable.
    • C=2G+2C = 2G + 2: Even-numbered pirates up to 2G2G, and #2G+12G + 1 are bribable.
    • C2G+4C \ge 2G + 4: All pirates up to 2G+M/22G + M / 2 are bribable.


As a side note, this problem serves as an excellent choice for software engineer interviews, as the process of working out other captains' proposals closely resembles the concept of recursion.

The solution of the Pirate Game struck me as counter-intuitive because I did not expect Amara to get as much as 98 coins for herself. This is all possible because they know every pirate is so rational and emotionless that gaining just one coin outweights any frustration and anger caused by such a minuscule amount.

After all, the Pirate Game is just a game theory exercise that, like many other game-theory problems, may not perfectly represent the real-world behaviours. Often, it is the overlooked human emotions that become the deciding factor in similar situations.