Pirate game
Source: Pirate game

Pirate game #2
Source: en.wikibooks.org

Pirate game puzzles

The pirate game is a simple mathematical game. It illustrates how, if assumptions conforming to a homo economicus model of human behaviour hold, outcomes may be surprising. It is a multi-player version of the ultimatum game. (Wikipedia)


Pirate game puzzle #1 - 5 pirates.

There are five rational pirates, A, B, C, D and E. They find 100 gold coins. They must decide how to distribute them.

The pirates have a strict order of seniority: A is superior to B, who is superior to C, who is superior to D, who is superior to E.

The pirate world's rules of distribution are thus: that the most senior pirate should propose a distribution of coins.
The pirates, including the proposer, then vote on whether to accept this distribution.
If the proposed allocation is approved by a majority or a tie vote, it happens.
If not, 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.

Pirates base their decisions on three factors.
First of all, each pirate wants to survive.
Secondly, each pirate wants to maximize the number of gold coins he receives.
Thirdly, each pirate would prefer to throw another overboard, if all other results would otherwise be equal.


Answer: It might be expected intuitively that Pirate A will have to allocate little if any to himself for fear of being voted off so that there are fewer pirates to share between. However, this is as far from the theoretical result as is possible.

This is apparent if we work backwards: if all except D and E have been thrown overboard, D proposes 100 for himself and 0 for E. He has the casting vote, and so this is the allocation.

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 1 coin in this round to make E vote with him, and get his allocation through. Therefore, when only three are left the allocation is C:99, D:0, E:1.

If B, C, D and E remain, B knows this when he makes his decision. To avoid being thrown overboard, he can simply offer 1 to D. Because he has the casting vote, the support only by D is sufficient. Thus he proposes B:99, C:0, D:1, E:0.
One might consider proposing B:99, C:0, D:0, E:1, as E knows he won't get more, if any, if he throws B overboard. But, as each pirate is eager to throw each other overboard, E would prefer to kill B, to get the same amount of gold from C.

Assuming A knows all these things, he 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
Also, 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.

This is the puzzle in precise words from the source: wikipedia.org Pirate game
Similar puzzle can also be seen at www.folj.com/ 100 Gold Coins & www.techinterview.org Pirates

Pirate game puzzle #2 - 5 pirates.

5 pirates of different ages have a treasure of 100 gold coins.
On their ship, they decide to split the coins using this scheme:

the oldest pirate alive will propose how the coins will be distributed among the pirates, and all pirates remaining will vote for or against it.

If more than 50% of the pirates vote for it, then the coins will be distributed as such.
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 extremely intelligent, rational, greedy, and do not wish to die, (and are rather good at math for pirates) what will happen?

Please note that in this #2 version of the pirate game, the proposed allocation has to be approved by a majority (more than 50%) vote.


Answer: The eldest pirate will propose a 97, 0, 1, 0, 2 split.

Working backwards, splits in terms of younger to older:

2 Pirates: Pirate two splits at 100, 0. Otherwise, and perhaps even then, pirate 1 (the youngest) votes against him and over he goes!

3 Pirates: Pirate three splits at 0, 1, 99. Pirate 1 (the youngest) is going to vote against him no matter what (see above), but this way, pirate 2 will vote for him, to get at least one gold out of it.

4 Pirates: Pirate four splits of 1, 2, 0, 97. This way, pirate 1 will vote for him, and so will pirate 2 - they're getting more than they would under 3 pirates.

5 Pirates: Pirate five splits of 2, 0, 1, 0, 97. This way, pirate 1 will vote for him, and so will pirate 3 - they're both getting better than they would under 4.


Pirate game #3
Source: Pirate game

Pirate game #3 - Ten pirates or more.

Ten pirates have got their hands on a hoard of 100 goldpieces, and wish to divide the loot between them.
They are democratic pirates, in their own way, and it is their custom to make such divisions in the following manner.
The fiercest pirate makes a proposal about the division, and everybody votes --- one vote each including the proposer.
If 50% or more are in favour, the proposal passes and is implemented forthwith.
Otherwise the proposer is thrown overboard and the procedure is repeated with the next fiercest pirate.

All the pirates enjoy throwing people overboard, but given the choice they prefer hard cash. They dislike being thrown overboard themselves.
All pirates are rational, know that the other pirates are rational, know that they know that... and so on.-----
Moreover, no two pirates are equally fierce, so there is a precise 'pecking order' --- and it is known to them all.
Finally: gold pieces are indivisible and arrangements to share pieces are not permitted (since no pirate trusts his fellows to stick to such an arrangement). It's every man for himself.


Pirate game #4
Source: Flickr.com/kevindooley

Pirate game #4 - Six Pirates Fight for 1 Gold Coin.

Six pirates discover a chest containing 1 gold coin. They decide to sit down and devise a distribution strategy.

The pirates are ranked based on their experience (Pirate 1 to Pirate 6, where Pirate 6 is the most experienced).

The most experienced pirate gets to propose a plan and then all the pirates vote on it.

If at least half of the pirates agree on the plan, the gold is split according to the proposal.

If not, the most experienced pirate is thrown off the ship and this process continues with the remaining pirates until a proposal is accepted.

The first priority of the pirates is to stay alive and second to maximize the gold they get.

Pirate 6 devises a plan which he knows will keep him alive. What is his plan?


Answer: If there is only one pirate, he takes the coin.

If there are 2 pirates, pirate 2 takes the coin leaving pirate 1 with nothing.

If there are 3 pirates, pirate 3 needs one more vote. There is no way we can convince pirate 2 since he benefits if pirate 3 dies.
Pirate 3 gives the gold coin to pirate 1 to get his vote so he survives.

If there are 4 pirates, pirate 4 needs one more vote. He can’t give it to pirate 1 since pirate 1 will get the gold coin even if pirate 4 dies.
He can give it either to pirate 2 or 3 which will get him another vote to ensure his survival.

If there are 5 pirates, pirate 5 needs two more votes. There is no way he can get two more votes since there is only one gold coin to bribe with.
Pirate 5 dies regardless of his proposal.

If there are 6 pirates, pirate 6 needs two more votes.
He will surely get pirate 5’s vote since pirate 5 will die if pirate 6 dies.
If pirate 6 and pirate 5 die, pirate 4 will survive but will not get the gold coin. Pirate 6 can bribe pirate 4 with the gold coin and get his vote. That makes a total of 3 votes allowing pirate 6 to survive.
He can also give the gold coin to pirate 1, 2 or 3 although giving it to pirate 2 or 3 is a little confusing since
either of them could get the gold coin in the case of 4 pirates.