Talk:Pirate game

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[edit] overly simple

This analysis doesn't mention the fact that A gets nothing if E decides that his vote is worth more than 1 coin, for example.

The game assumes that each pirate is entirely rational. No other pirate will offer him more, and some would offer him less, so E has no motivation to refuse A's offer. If he did, he would end up with nothing when B and D outvoted C and E to split it 99-0-1-0. Even if he got lucky and B went overboard and C made an offer, E would once again have to accept a mere 1 coin (ganing nothing over A's offer) or C would go overboard and D would keep everything for himself. --Icarus (Hi!) 21:38, 11 July 2006 (UTC)
If this was a real scenario with real people the game would have close relationship with Ultimatum Game. In which case if A takes vast majority of the gold the rest are willing to gain nothing (loss of small amount of gold < punishing distributor A) in order to punish A for greediness. In reality, assuming greedy yet real pirates solution 50-30-20-0-0, or similar, would be much more likely with D & E as a minority are left without a dime. :-)
Absolutely! In the three pirate game (C, D, E), the logic that C can offer E just a single coin, and can be confident that E will accept it, seems flawed. It is in E's interest to be vindictive, but not too vindictive, in order that C doesn't think that an offer of 1 to E is acceptable. If I were E, I would be thinking (because E can't propose offers, only vote) that C had better offer me something good (like C:50 D:0 E:50), or else I'd vote to kill him. If E is steadfast in this attitude, then they must call C's bluff of a single-coin offer and vote to kill. E risks losing 50 coins by adopting this strategy, but C risks losing their life! C's argument of "E, vote for me or you get nothing" doesn't seem as persuasive as E's (implicit) counter-argument of "C, give me X coins or you die!". The best strategy, i.e. quantifying X, requires knowledge of the coin value of each pirate's life to themselves.
The whole problem with the pirate game is that the logic is somewhat circular: "The best strategy for C is to offer E just 1 coin, because the E *knows* that the best strategy for E is to maximize their coinage, and they will take 1 over 0". However, if the best strategy for E is to reject offers below 50, then, because C *knows* this, the best strategy for C is to offer E 50 and stay alive! So if indeed the best strategy for E is to accept a paltry offer, E gets left with 1. If the best strategy for E is to reject offers below 50, then E gets left with 50! So E accepting 1 does not sound like the 'best' strategy, because there exists a strategy that will get them more. That is the problem with trying to answer semantic questions like 'best' using something like Nash equilibria. Nash equilibia do NOT always give the 'best' strategies for intelligent agents. In answering the pirate game, you need to quantify what is meant by 'best'; i.e., the strategy that maximizes some utility function. In particular, the pirate game does a poor job of this by assuming that a pirate thinks that getting 0 coins is the same as losing their life!

[edit] Assumptions Missing

It needs to be assumed that all pirates are perfectly logical and that each of them knows this about all the rest.

Also that the pirates believe they will not be cheated in the procedure anywhere.

It also needs to be clarified what a pirate would do when offered the same amount of gold from one proposal or another. (I.e. would they rather have pirates thrown overboard or not, all other things being equal).

And if a pirate would rather accept 0 gold pieces as opposed to being thrown overboard.

[edit] Conclusion is suspect

I'm with you all the way up until you get to B's decision. Rather than offering one gold piece to E, he should offer it to D, who would otherwise get 0 if B's proposal fails. That means B's proposal should be (B,C,D,E) as (99,0,1,0).


With that in mind, working back to A, he would need 2 more votes, which he can buy from C and E for 1 gold each, as they would get 0 gold if A's proposal fails. So the solution I see is (A,B,C,D,E) as (98,0,1,0,1).

Yup. Fixed. EdC 10:57, 25 June 2006 (UTC)
Whoops! Back to Core Micro for me...wrote the original article in something of a hurry, thanks for the heads up. Oxymoron 21:06, 26 June 2006 (UTC)
DN: No, you are wrong! D always says no, beacuse he wants to get 100 when only D and E are there. Therefore the previous solution (that I already edited) is correct.
No, because D knows if B goes overboard, C will offer one coin to E and nothing to D. Therefore he will accept one coin from B. Right? 192.75.48.150 16:06, 29 June 2006 (UTC)
That's how I see it. Fermium 23:55, 29 June 2006 (UTC)
I don't see the advantage of giving D the 1 coin instead of E... The game works either way and the outcome is the same... (DN)
Because if B offers 1 coin to E, E knows that that is also what C would offer him. E might then decide that he might as well accept B's offer because no one will offer more, or he might decide that he'll vote to throw B overboard just out of spite because C's offer won't be less. So B's survival is still uncertain. The only way he can guarantee his own survival is to offer the 1 coin to D, who knows that he must accept B's offer because if B dies and C offers, he'll end up with nothing. ----Icarus (Hi!) 10:56, 7 July 2006 (UTC)
This might be true if you don't apply the homo economicus model, which we do in this case (see top of article). And according to this model each individual only acts to maximise its own benefit and does NOT do anything to harm others "out of spite"... So it does make no difference if B offers the 1 coin to D or E! (DN)

I don't think that is quite right, DN. Each individual acts to maximize its own benefit, yes. But faced with equally beneficial options, there are NO requirements on its behaviour. It is not required to refrain from harming others. 192.75.48.150 13:49, 12 July 2006 (UTC)

[edit] Extension of game

If you begin with more than 200 pirates and cannot divide the individual gold pieces, an interesting pattern develops. Here is when it is important whether a pirate would take 0 gold rather than being thrown overboard. (For the sake of discussion, I assume they would).

The pirates start voting 0 for themselves, with the other 100 for either the even or odd "numbered" pirates at the end of the list. Still, many will get thrown overboard no mattter what. A pattern develops that has 100 of the first 200 getting 1 coin (alternating between odds and evens as you add pirates), then the next pirates up to the highest power of 2 available getting 0 coins, and anyone beyond that getting thrown overboard no matter what.

For example, the solution for 500 pirates is 44 getting thrown overboard, the next 256 (2^8) getting 0 coins, and the remaining "even" numbered pirates each getting 1 gold piece.

This is evidently mentioned in Scientific American in May, 1999 but I cannot confirm that. Fermium 01:35, 24 June 2006 (UTC)

I am not so sure about that. The rules of homo economicus fail to determine what happens after 201 pirates. Try it with one indivisible prize and five pirates:
  • With A, B, and C overboard, D takes the prize.
  • Therefore with A and B overboard, C offers E the prize. C votes yes to save his life, and E votes yes to get the prize.
  • Therefore with A overboard, B... what? He can offer the prize to C or D. B votes yes to save his life, and C or D votes yes to get the prize. Nothing determines who B will offer the prize to. Also, nothing determines which way the one who doesn't get the prize votes, but the vote doesn't matter, it passes.
  • Therefore A ... what? If he goes overboard, B, E, and one of C or D will get nothing (but survive). A can only offer the prize to one of them. The other two will still get nothing (but survive). Again, their behaviour is not determined, and this time, it affects the vote outcome.
If we assume that pirates will, if offered nothing, vote to throw someone overboard, then A goes overboard. In the 100 coin case, the 203rd pirate goes overboard. And it's still not determined who the 202nd pirate offers the coins to. If, on the other hand, we assume that pirates will, if offered nothing either way, vote not to throw someone overboard, then nobody ever goes overboard. I'm not sure what assumption generates the pattern you mention. 192.75.48.150 16:49, 29 June 2006 (UTC)
One prize 5 pirates is easy...You work it backwards:
  • With D & E remaining D gets the prize.
  • With C, D & E remaining either D or E gets the prize as then C's decision has a majority.
  • With B, C, D & E remaining C, E (equilibrium) or D (majority as C knows he can't get the prize even if he objects) gets the prize.
  • With all A, B, C, D & E present C, D or E gets the prize as B knows there's no way he can get the prize and survive creating 3 (A,B & prize taker) - 2 majority.
-G3, 12:00, 28 November 2006 (UTC)


Here's the article; couldn't find it before: [1] Fermium 23:55, 29 June 2006 (UTC)
Hmmm. The solution is flawed. Under the assumptions, a bribe is effective if the pirate being bribed was not going to receive a bribe under the next successful proposal. Yet, as the article admits, at a certain point, the proposals are not uniquely determined, so at some point there will be some bribed pirates who don't know whether they would be bribed anyway. Their votes cannot be counted on. 192.75.48.150 14:19, 30 June 2006 (UTC)
This comes a bit late, but I think that the missing piece is that the pirates, in the Sci Am article, like throwing people overboard. In the five pirate example this means that A can only bribe one of them, and the others will want to see him thrown overboard whether or not they can get another chance at the gold with B's proposal. Adam Faanes 03:36, 2 September 2006 (UTC)

[edit] Pirates like throwing pirates overboard

This problem is a little undecidable. If there are 3 pirates and the first one offers 100 coins to himself and 0 to the other two, the last pirates cannot decide how to vote (he would get 0 coins either way). If he likes throwing pirates overboard, he would say no and if he dislikes it he would say yes. --Petter 23:42, 8 March 2007 (UTC)

you have missed something. If the two vote to throw the proposer over-board this will mean that only 2 pirates are left, thus the higher ranked pirate will allocation 100 - 0 and always win (since he has the deciding vote). Thus it is in the best interest of pirate of least rank not to throw the proposer overboard.--Dacium 03:15, 26 July 2007 (UTC)
He'd get zero gold coins either way. That's why the third condition is needed. --Petter 04:10, 27 August 2007 (UTC)

[edit] Pirate loot problem

Messy Thinking 02:29, 7 April 2007 (UTC) Does this connect in any way to the pirate loot problem? Seems to me that, especially as the loot was used to pay certain fees, this was an issue in genuine pirate culture.

Yes, they are the same. They should be merged. Mlewan 21:11, 26 June 2007 (UTC)

[edit] Image

You do realize of course that right after the image in question the entire crew were marooned/killed...

[edit] Pulled references

I deleted two entries from the references list that seems doubtful. One (reed.edu) simply stated the problem for 10 pirates and asked for answers. No answer was given, no research or theory presented. As such it's useless as a reference. The other (rasmusen.org) looks like a book of game theory. I downloaded all of the TeX source files and found no references to pirates. Maybe there is valuable information, but it's not reasonably found, so it's gone. (If someone finds it and re-adds the citation, please give enough information to find the reference, ideally the chapter and maybe a a very, very brief snippit to search for.) Finally, I didn't delete but I'm suspicious of the hanovercollaboration.com link. It just looks like some guy's blog in which he poses the game and other people replied with analysis. It doesn't seem like much of a resource, especially compared to the article from Scientific American. — Alan De Smet | Talk 03:13, 10 August 2007 (UTC)

[edit] Possible sources for citations

Some possible sources for more citations. Please add more if you find some, but don't have the time to integrate them into the article. If someone takes these and adds more to the article, 1: thanks!, 2: please delete the ones you added from this list. — Alan De Smet | Talk 04:36, 10 August 2007 (UTC)

  • "Game Theory spring 2004 Problem Set 2 - solutions" by Eli Berger [2] page 2. (You use Google to read it if you can't handle PostScript format files [3]. Backup link to WebCite, since the author's home page has a warning that it may go away: [4]) - States the problem and offers some analysis. The author was an instructor (probably a professor?) at Princeton University.

[edit] B's possible strategies

When A offers 98 0 1 0 1, B will be incensed at the distribution and will try to think up strategies. A possible strategy for B is to announce that if A is voted off, he(B) will announce a distribution plan perhaps of 25 25 25 25 or perhaps 49 0 2 49. Now E has to decide whether to trust B, but being a pirate, it has to be unlikely and B is likely to go back on his word and offer 99 0 1 0 when he gets in. C is unlikely to be able to persuade D and E to punish B for going back on his word as C could offer 99 0 1 with impunity. E won't vote with D to get rid of C because E will then get none. If E thinks there is a 1 in 25/49 chance that B will keep his word or that there is a good chance B will actually offer something like 95 0 3 2 to stave off punishment action then E might be persuaded to vote off A. If E assesses 0% chance of B offering anything other than 99 0 1 0 then E would not be pursuaded to join B and D in voting off A.

The logic and solution (subject to some reasonable probability assessments) therefore survives this B strategy when there are 5 or fewer pirates. More than 5 pirates and such a B strategy might just work as the punishment action for breaking his word becomes more credible. Even with lots of pirates this B strategy would not work with strictly 'game theory rational' pirates. The article isn't clear whether the 5 rational pirates are 'game theory rational' or 'real world rational'.

Another strategy is for B to bribe E with a gold coin for his vote. E cannot then loose out as a result of voting off A. Therefore if A is afraid of this happening then A may be sensible to offer more to E. If this is not allowed or E could take the bribe then break his word and vote to accept A's offer the logic and solution will again survive.

Hope that makes some sort of sense even if the obvious answer is we cannot add original research. Nevertheless, is the article complete without explaining this? crandles 19:12, 10 August 2007 (UTC)

Given that it's a game theory problem, it's reasonable to assume that perfectly rational (that is "game theory rational") pirates are intended. Thus the above analysis isn't really relevant as the pirates won't trust each other in the slightest as upholding a promise has zero value within the realm of the puzzle. — Alan De Smet | Talk 22:52, 10 August 2007 (UTC)
Isn't the most interesting part of game theory the normative part of seeing whether the solution really applies? crandles 09:20, 11 August 2007 (UTC)

[edit] C Strategy

What happens if C or E rather oddly votes with B and D to get rid of A? Would this set a precident? Would this unsettle B and make him fear he will be voted off even if he makes a 'game theory rational' offer? Might this persuade B into making a better offer? C in this case is risking 1 in the hope of making different things happen. Could that be ultimately 'real world rational'? Should A be afraid of this and offer C and E more? crandles 19:12, 10 August 2007 (UTC)

[edit] Intuitive result

Article currently says

It might be expected intuitively that Pirate A will propose that the allocation shall be 20, 20, 20, 20, 20. However, this is not the theoretical result.

I don't think 20 20 20 20 20 is an intuitive result because this would imply that B,C,D and E should vote off A as there would then be fewer pirates to share the loot and judging by that allocation they may then get 25. Therefore the intuitive result may be that A can allocate little if any to himself. Should it be changed to:

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.

crandles 17:04, 27 August 2007 (UTC)

[edit] more for pirate E

logically pirate A would need to offer 97 , 0 , 1 ,0 , 2 in the example given .

Pirate E will need a greater money from Pirate A as he knows logically that Pirate C will offer him 1 coin also and he is also aware Pirate B will offer him nothing .

Now remember that these particualr pirates tertiary motivation is making pirates walk the plank .

Pirate A offers 1 gold and no plankwalkings Pirate C would also offer 1 gold but also 2 plankwalkings .

Knowing this Pirate A would have to offer more gold than Pirate C would do so that Pirate A offers pirate E 2 gold wheras pirate C would only logically need to offer him 1 gold . —Preceding unsigned comment added by 194.72.129.67 (talk) 15:34, August 29, 2007 (UTC)

Sorry but pirate B will make a rational offer sufficient that B (with rational votes) will survive so E needs to compare the A offer with the B offer. Comparing the A offer to the C offer (which will not be offered) is not rational. crandles 17:59, 29 August 2007 (UTC)

Just a reminder, since someone attempted to change the article in this way: the current numbers are supported by a citation. If you can find another source proposing different answers, feel free to add it and add a citation for it, but don't replace the existing numbers. — Alan De Smet | Talk 17:42, 17 April 2008 (UTC)

[edit] A is actually in a bad position

Actually I don't see why A should get the most coins as A is the pirate first to be at risk of dying. E doesn't risk dying at all. Remember the rule that everything being equal pirates would prefer that another pirate dies.

The official result to me appears to be just a preliminary result after an initial iteration. Taking it as a final result is incorrect.

Basically if B is willing to offer more to 2 pirates than A is, A will die.

If A attempts "99 0 1 0 1", B could propose "dead 33 33 33 0" or even "dead 0 50 0 50" or "dead 0 33 33 33", after all B would otherwise be getting zero anyway and in these proposals a pirate dies (pirate A). Assuming the pirates are Perfectly Rational Beings they will all know that too.

So in order to not die A might propose "0 50 50 0 0". B won't object since he's getting more than the zero in the 1st iteration. If A dies and it's his turn (to risk death as well) he might have to propose "dead 50 50 0 0" anyway to prevent C from proposing "dead dead 50 50 0".

Thus I suggest that "0 50 50 0 0" is a stable offer, but I am uncertain that this is indeed the best proposal for A.

While I'm not sure what the final result is (or if there's a single answer), I'm pretty sure it's not "99 0 1 0 1" :). 118.100.13.242 (talk) 17:13, 24 January 2008 (UTC) (Link).

This is a game-theory problem, not a realistic problem. As described, B doesn't get to make proposals at all until A is dead. If A dies, it's identical to the 4-pirate case and doesn't change anything. Ultimately, however, our theories aren't really relevant. Remember that Wikipedia's job is to report on what others have said, not to engage in original research. If you're quite confident that you have something to add, write up a paper on it and get it published. Indeed, if a reliable source were to publish your paper on the pirate game, we would be remiss in not including it in the article. — Alan De Smet | Talk 03:11, 25 January 2008 (UTC)

[edit] Extension

"Ian Stewart extended it to an arbitrary number of pirates in the May 1999 edition of Scientific American, with further interesting results.[1]" Anyone care to elaborate? As it currently stands, the reader is left wondering what the "further interesting results" are, and is forced to go to the references, download the article, and read through it themselves. Loggie (talk) 21:44, 3 February 2008 (UTC)