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Pirate game
Pirate game
Pirate game
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The pirate game is a simple mathematical game. It is a multi-player version of the ultimatum game.

The game

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There are five rational pirates (in strict decreasing order of seniority A, B, C, D and E) who found 100 gold coins. They must decide how to distribute them.

The pirate world's rules of distribution say that the most senior pirate first proposes a plan of distribution. The pirates, including the proposer, then vote on whether to accept this distribution. If the majority accepts the plan, the coins are disbursed and the game ends. In case of a tie vote, the proposer has the casting vote. If the majority rejects the plan, the proposer is thrown overboard from the pirate ship and dies, and the next most senior pirate makes a new proposal to begin the system again. The process repeats until a plan is accepted or if there is one pirate left.[1]

Pirates base their decisions on four factors:

  1. Each pirate wants to survive.
  2. Given survival, each pirate wants to maximize the number of gold coins he receives.
  3. Each pirate would prefer to throw another overboard, if all other results would otherwise be equal.[2]
  4. The pirates do not trust each other, and will neither make nor honor any promises between pirates apart from a proposed distribution plan that gives a whole number of gold coins to each pirate.

The result

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To increase the chance of their plan being accepted, one might expect that Pirate A will have to offer the other pirates most of the gold. However, this is far from the theoretical result. When each of the pirates votes, they will not just be thinking about the current proposal, but also other outcomes down the line. In addition, the order of seniority is known in advance so each of them can accurately predict how the others might vote in any scenario. This becomes apparent if we work backwards.

The final possible scenario would have all the pirates except D and E thrown overboard. Since D is senior to E, they have the casting vote; so, D would propose to keep 100 for themself and 0 for E.

If there are three left (C, D and E), C knows that D will offer E 0 in the next round; therefore, C has to offer E one coin in this round to win E's vote. Therefore, when only three are left the allocation is C:99, D:0, E:1.

If B, C, D and E remain, B can offer 1 to D; because B has the casting vote, only D's vote is required. Thus, B proposes B:99, C:0, D:1, E:0.

(In the previous round, one might consider proposing B:99, C:0, D:0, E:1, as E knows it won't be possible to get more coins, if any, if E throws B overboard. But, as each pirate is eager to throw the others overboard, E would prefer to kill B, to get the same amount of gold from C.)

With this knowledge, A can count on C and E's support for the following allocation, which is the final solution:

  • A: 98 coins
  • B: 0 coins
  • C: 1 coin
  • D: 0 coins
  • E: 1 coin[2]

(Note: A:98, B:0, C:0, D:1, E:1 or other variants are not good enough, as D would rather throw A overboard to get the same amount of gold from B.)

Extension

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The solution follows the same general pattern for other numbers of pirates and/or coins. However, the game changes in character when it is extended beyond there being twice as many pirates as there are coins. Ian Stewart wrote about Steve Omohundro's extension to an arbitrary number of pirates in the May 1999 edition of Scientific American and described the rather intricate pattern that emerges in the solution.[2]

Supposing there are just 100 gold pieces, then:

  • Pirate #201 as captain can stay alive only by offering all the gold one each to the lowest odd-numbered pirates, keeping none.
  • Pirate #202 as captain can stay alive only by taking no gold and offering one gold each to 100 pirates who would not receive a gold coin from #201. Therefore, there are 101 possible recipients of these one gold coin bribes being the 100 even-numbered pirates up to 200 and number #201. Since there are no constraints as to which 100 of these 101 they will choose, any choice is equally good and they can be thought of as choosing at random. This is how chance begins to enter the considerations for higher-numbered pirates.
  • Pirate #203 as captain will not have enough gold available to bribe a majority, and so will die.
  • Pirate #204 as captain has #203's vote secured without bribes: #203 will only survive if #204 also survives. So #204 can remain safe by reaching 102 votes by bribing 100 pirates with one gold coin each. This seems most likely to work by bribing odd-numbered pirates optionally including #202, who will get nothing from #203. However, it may also be possible to bribe others instead as they only have a 100/101 chance of being offered a gold coin by pirate #202.
  • With 205 pirates, all pirates bar #205 prefer to kill #205 unless given gold, so #205 is doomed as captain.
  • Similarly with 206 or 207 pirates, only votes of #205 to #206/7 are secured without gold which is insufficient votes, so #206 and #207 are also doomed.
  • For 208 pirates, the votes of self-preservation from #205, #206, and #207 without any gold are enough to allow #208 to reach 104 votes and survive.

In general, if G is the number of gold pieces and N (> 2G) is the number of pirates, then

  • All pirates whose number is less than or equal to 2G + M will survive, where M is the highest power of 2 that does not exceed N – 2G.
  • Any pirates whose number exceeds 2G + M will die.
  • Any pirate whose number is greater than 2G + M/2 will receive no gold.
  • There is no unique solution as to who gets one gold coin and who does not if the number of pirates is 2G+2 or greater. A simple solution dishes out one gold to the odd or even pirates up to 2G depending whether M is an even or odd power of 2.

Another way to see this is to realize that every pirate M will have the vote of all the pirates from M/2 + 1 to M out of self preservation since their survival is secured only with the survival of the pirate M. Because the highest ranking pirate can break the tie, the captain only needs the votes of half of the pirates over 2G, which only happens each time (2G + a Power of 2) is reached. For instance, with 100 gold pieces and 500 pirates, pirates #500 through #457 die, and then #456 survives (as 456 = 200 + 28) as they have the 128 guaranteed self-preservation votes of pirates #329 through #456, plus 100 votes from the pirates they bribe, making up the 228 votes that they need. The numbers of pirates past #200 who can guarantee their survival as captain with 100 gold pieces are #201, #202, #204, #208, #216, #232, #264, #328, #456, #712, etc.: they are separated by longer and longer strings of pirates who are doomed no matter what division they propose.

See also

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Notes

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References

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Revisions and contributorsEdit on WikipediaRead on Wikipedia

Pirate game

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from Grokipedia
The Pirate game is a puzzle in that demonstrates and rational choice under uncertainty, typically involving five rational pirates dividing 100 indivisible gold coins through a sequential proposal and voting process. The pirates, ordered by seniority from the most senior (Pirate A) to the most junior (Pirate E), follow a strict protocol: the senior-most pirate proposes a distribution of the coins, and all pirates then vote on it, with the proposer included in the tally. A proposal passes if it receives at least a majority of votes (including the proposer's own); otherwise, the proposer is thrown overboard and presumed dead, and the next senior pirate repeats the process with the remaining crew and the same 100 coins. The game assumes all pirates are perfectly rational, prioritize survival above all else, seek to maximize their own gold holdings given survival, and—when indifferent—prefer to minimize the gold received by higher-ranking pirates or maximize deaths among them, while knowing that all others share these priorities. This setup serves as a multi-player extension of the , highlighting how perfect rationality can lead to counterintuitive outcomes where junior pirates wield significant influence despite their low status. Solving the game requires : beginning with the endgame where only two pirates remain (the senior taking all 100 coins, leaving the junior with zero), and iteratively determining optimal proposals for larger groups. For five pirates, the subgame-perfect equilibrium has Pirate A proposing 98 coins for themselves, 0 for B, 1 for C, 0 for D, and 1 for E; this secures votes from A, C, and E (a of three out of five), as C and E rationally accept one coin over the zero they would receive if A is rejected and the four-pirate proceeds. The puzzle, popularized in its modern form by mathematician Ian Stewart based on an earlier formulation by Steve Omohundro, has been extended to arbitrary numbers of pirates, revealing periodic patterns in winning strategies that depend on crew size modulo powers of 2. Beyond its instructional value in teaching , the Pirate game has inspired variants in academic literature, including analyses of requirements, quantum decision models, and to test human deviations from rationality. It underscores the importance of credible threats in sequential games, making it a staple in and curricula.

Game Setup

Rules of Division

The Pirate game involves five rational pirates, labeled A through E, with A being the most senior and E the most junior, who must divide 100 indivisible gold coins among themselves. The coins are to be allocated in nonnegative amounts summing exactly to 100, reflecting their indivisibility and the pirates' goal of maximizing their personal shares. The division process begins with the senior-most surviving pirate proposing a specific allocation of the coins to each pirate. All surviving pirates, including the proposer, then vote on the proposal, with each having one vote. Approval requires a vote, meaning at least half of the voters (rounded up) must vote yes; for the initial five pirates, this threshold is three yes votes. In the event of a voting tie, the proposer's vote counts as decisive, effectively breaking the tie in favor of approval. If the proposal receives sufficient yes votes, the allocation is implemented, and the process ends. However, if it is rejected, the proposing pirate is immediately thrown overboard and dies, leaving the remaining gold intact at 100 coins for the survivors. The next most senior pirate among the survivors then becomes the proposer and repeats the process with the reduced group. This sequential procedure continues iteratively until a proposal is approved or only one pirate remains, in which case that pirate claims all 100 coins without needing a vote. The pirates, seeking to maximize their individual gold holdings while ensuring survival, engage in this high-stakes voting mechanism that underscores the game's emphasis on strategic proposal-making.

Assumptions about Pirates

The pirate game in relies on a set of foundational assumptions about the pirates' behavior and motivations, which render the puzzle solvable through deterministic analysis. Central to these is the premise that all pirates are perfectly , meaning they make decisions solely to advance their while anticipating that others will do the same. This extends to among the pirates: each understands that every other pirate is equally logical and aware of this mutual understanding. Their priorities form a strict : is paramount, followed by maximizing personal gold acquisition, with a tertiary preference for minimizing the number of surviving pirates (or equivalently, maximizing deaths) only if it does not compromise the first two goals. This bloodthirsty inclination underscores their lack of ; pirates derive satisfaction from eliminating rivals but subordinate it to securing their own life and wealth. Consequently, no binding agreements or side deals are feasible, as pirates inherently one another and act only in immediate during proposals and votes. The model further assumes infinite intelligence on the part of all pirates, enabling them to perfectly foresee every possible outcome, counter-proposal, and voting reaction across the game's sequential stages. This foresight is crucial, as junior pirates—those lower in the —grow increasingly desperate with each reduction in crew size, making them more amenable to accepting minimal gold shares to avoid death, which in turn influences senior pirates' strategic bribes. These assumptions draw directly from game theory's rational , where agents optimize under and logic, ensuring the puzzle yields a unique, predictable equilibrium without probabilistic elements or irrational deviations.

Solution Process

Backward Induction Approach

Backward induction is a fundamental technique in for solving finite games of , where players move sequentially and observe all previous actions. It involves analyzing the game tree by starting from its terminal nodes—end positions where payoffs are determined—and working backwards to derive optimal strategies for each decision point, ensuring that no player has an incentive to deviate at any subgame. This method guarantees a , as formalized by , by eliminating non-credible threats and promises through iterative checks. In the pirate game, is applied by considering progressively larger s defined by the number of remaining pirates, beginning with the simplest cases. With one pirate left, that pirate claims the entire 100 gold coins, as there are no others to vote or object. For two pirates, the senior pirate proposes keeping all 100 coins for themselves and offering zero to the junior, which passes with the senior's own vote (sufficient for a under the game's rules requiring at least half the votes in favor). The junior accepts zero rather than rejecting, as rejection would result in the senior's overboard fate and the junior receiving all 100 anyway—but the proposal passes regardless due to the voting threshold. This process continues to three pirates, where the senior anticipates the two-pirate outcome and offers just enough (one coin) to secure the junior-most pirate's vote, as that pirate expects zero in the ; the analysis builds iteratively up to five pirates, determining each proposal based on expected rejections leading to solved s. The approach works because it assumes all pirates possess perfect foresight and of rationality, allowing each to predict outcomes in any potential with and propose divisions that minimize concessions while ensuring passage. This eliminates in sequential , yielding a unique equilibrium where earlier pirates exploit the structure to maximize their share without fear of implausible deviations later in the game tree. While powerful for theoretical analysis, relies on idealized assumptions of error-free prediction and unbounded computational ability, which may not hold in real-world scenarios where or experimental deviations lead players to forgo the predicted equilibrium.

Specific Outcome for Five Pirates

In the standard pirate game with five rational pirates labeled A (most senior) through E (most junior) and 100 indivisible gold coins, reveals the unique division. The analysis begins with the base case of one pirate (E). E proposes to take all 100 coins, and the proposal automatically passes since there are no other voters. With two pirates (D and E), D proposes 100 coins to D and 0 to E. The voting rule requires at least 50% yes votes (including the proposer's) for passage, so D needs 1 out of 2 votes. D votes yes, securing passage regardless of E's vote, as E expects 0 in this scenario and cannot improve their outcome by rejecting. For three pirates (C, D, E), rejection would lead to the two-pirate outcome where D gets 100 and E gets 0, with C dying. C needs at least 2 out of 3 votes (≥50%). C proposes 99 to C, 0 to D, and 1 to E, bribing E—who expects 0 if rejected—with the minimal amount to secure E's yes vote alongside C's own. D votes no, expecting 100 otherwise, but the proposal passes with 2 yes votes. In the four-pirate case (B, C, D, E), rejection yields the three-pirate outcome: C gets 99, D gets 0, E gets 1, and B dies. B needs at least 2 out of 4 votes. B proposes 99 to B, 0 to C, 1 to D, and 0 to E, bribing only D—who expects 0—with 1 coin to gain D's yes vote plus B's own. C and E vote no (C expects 99, E expects 1), but 2 yes votes meet the threshold, so the proposal passes. B offers 0 to E rather than bribing E because only one additional vote is needed, minimizing B's payout. For the full five-pirate scenario (A, B, C, D, E), rejection leads to the four-pirate outcome: B gets 99, C gets 0, D gets 1, E gets 0, and A dies. A needs at least 3 out of 5 votes (≥50%). A proposes 98 to A, 0 to B, 1 to C, 0 to D, and 1 to E, bribing C and E—who each expect 0 if rejected—with 1 each to secure their yes votes alongside A's own. B and D vote no (B expects 99, D expects 1), but the three yes votes ensure passage. This division is stable because each pirate, being rational, anticipates the worse outcomes from rejection: C and E prefer 1 over 0, while B and D cannot block it despite their incentives.

Generalization and Extensions

Solution for Arbitrary Number of Pirates

The solution to the pirate game generalizes to an arbitrary number nn of pirates through , building on the outcomes for fewer pirates. In this setup, the senior pirate proposes a division of the 100 indivisible gold coins, requiring at least half the pirates (including themselves) to vote in favor for acceptance; otherwise, they are thrown overboard, and the next senior proposes among the remaining n−1n-1 pirates. The pirates are perfectly rational, prioritizing more gold, then survival, then the death of others. By induction, assume the equilibrium proposal is known for n−1n-1 pirates. In the nn-pirate game, the senior pirate needs ⌈n/2⌉−1=⌊(n−1)/2⌋\lceil n/2 \rceil - 1 = \lfloor (n-1)/2 \rfloor additional votes beyond their own to secure the minimal . These votes are obtained by offering 1 each to exactly ⌊(n−1)/2⌋\lfloor (n-1)/2 \rfloor junior pirates who expect 0 coins in the (n−1)(n-1)-pirate , as they will accept the minimal positive offer over the alternative of rejection. The senior gives 0 coins to the remaining juniors (who expect at least 1 in the subgame and thus reject) and retains the rest: 100−⌊(n−1)/2⌋100 - \lfloor (n-1)/2 \rfloor coins. This pattern holds by induction. For the base case n=2n=2, the senior proposes 100 to themselves and 0 to the junior, securing the required 1 vote (their own, as ≥1\geq 1 suffices for half of 2). For n=3n=3, the subgame yields 1 pirate expecting 0, so the senior bribes that one with 1 , retaining 99. Assuming the result for n−1n-1, the subgame has exactly ⌊(n−2)/2⌋\lfloor (n-2)/2 \rfloor pirates receiving positive amounts (bribed) besides the sub-senior, leaving ⌊(n−1)/2⌋\lfloor (n-1)/2 \rfloor expecting 0—precisely enough to bribe minimally without excess cost. This ensures the proposal passes, as the bribed pirates prefer 1 over 0, while others prefer rejection but lack . For example, in the 5-pirate case, ⌊4/2⌋=2\lfloor 4/2 \rfloor = 2, so the senior retains 98 coins and bribes 2 juniors expecting 0 in the 4-pirate . As nn increases up to , the senior's share decreases to 100−99=1100 - 99 = 1 coin for n=200n=200, since ⌊199/2⌋=99\lfloor 199/2 \rfloor = 99. For n>200n > 200, the senior cannot afford ⌊(n−1)/2⌋>100\lfloor (n-1)/2 \rfloor > 100 bribes of 1 coin each with only 100 coins available, so they propose 0 for themselves and 1 coin each to 100 juniors expecting 0 in the (selected from the ample number available), securing the votes while the remaining juniors get 0.

Common Variants and Modifications

One common modification to the pirate game involves varying the number of pirates and the total number of coins, highlighting how limited resources alter the equilibrium when the number of participants grows large. In a variant analyzed by Steve Omohundro, 500 pirates divide 100 indivisible gold coins under the standard rules. reveals that the dynamics shift dramatically beyond approximately 200 pirates, as the proposer lacks sufficient coins to bribe the required number of supporters with at least one coin each. For instance, the 201st pirate offers zero to themselves and one coin each to the 100 lowest-ranked pirates who would receive nothing in the subgame, securing approval and surviving, while the proposer in the 203rd position fails due to insufficient bribes and is thrown overboard. This pattern continues, with "winning" proposers (such as the 201st, 202nd, 204th, and higher powers-of-two offsets) able to distribute the coins effectively, but ultimately, the 456th pirate proposes the successful division, keeping zero coins while bribing 100 juniors. Historical presentations of the puzzle often adjust the crew size and majority threshold to emphasize different strategic insights. Ian Stewart's 1999 article in popularized the game with five pirates and 100 coins, requiring the support of at least half the pirates (including the proposer), leading to the senior pirate retaining 98 coins by bribing two juniors with one each. Earlier formulations by Steve Omohundro sometimes featured 10 pirates or adjusted voting rules, such as requiring support from half the crew excluding the proposer, which reduces the number of necessary bribes and increases the senior pirate's share. These variations demonstrate the puzzle's flexibility in illustrating perfect equilibria under changing parameters. A key extension examines different quorum requirements for approval, effectively introducing unequal voting power or modified majority rules. In the standard setup, the proposer needs at least 50% of votes including their own (with a casting vote in ties), but analyses of variable quorums Q (the fraction of yes votes required) show how outcomes change. For Q = 0.5 excluding the proposer, the senior pirate needs fewer bribes—roughly (n-1)/2 supporters—altering the distribution to favor the proposer more heavily, as juniors anticipate lower subgame payoffs. For higher Q (e.g., 2/3), the proposer must bribe more pirates, often leading to more even splits or failure in coin-limited scenarios. Payouts cycle in patterns based on whether the coin total M is divisible by 3 or 4, with pirates grouped into types receiving 0, 1, 2, or 3 coins depending on rank and Q. This generalization underscores the game's sensitivity to institutional rules.
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