Talk:Pirate game

message: 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)
replies: text: 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)
replies: text: He'd get zero gold coins either way. That's why the third condition is needed. --Petter 04:10, 27 August 2007 (UTC)
thread_title:  Pirates like throwing pirates overboard 

message: Messy Thinking 02:29, 7 April 2007 (UTC)
replies: text: Yes, they are the same. They should be merged. Mlewan 21:11, 26 June 2007 (UTC)
thread_title:  Pirate loot problem 

message: 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)
thread_title:  Pulled references 

message: 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)
thread_title:  Possible sources for citations 

message: 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)
replies: text: 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)
replies: text: 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)
thread_title:  B's possible strategies 

message: 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)
thread_title: C Strategy

message: 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)
thread_title:  Intuitive result 

message: 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)
replies: text: Just because there is a 'citation' does not mean it is a 'logical' answer. I would expect E to refuse the offer of 1 coin. It is rational for E to assume that A would prefer to lose at least one more gold coin rather than die. So, if E refuses the offer, there is a very strong chance he will be offered one more coin. (If he refuses again, then D may accept 2 coins rather than the 1 he would get from B).  — Preceding unsigned comment added by 80.229.45.71 (talk) 11:34, 20 December 2012 (UTC)
replies: text:  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 05:33, 22 December 2012 (UTC)
thread_title:  more for pirate E  

message: 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)
replies: text: 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)
thread_title:  A is actually in a bad position 

message: "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)
thread_title: Extension

message: In the SciAm version (referenced), ties go to the proposer. The bit in the description about the "casting vote" is nonsense. 
In a less-used version, ties go against the proposer. In those cases the solution is similar, but not the same. Several people on this board have been bothered by the solution not being 97-0-1-2-0; this is because they have the solution for the "tie loses" variant.
The article should reflect both variants, and the reference to casting vote should be deleted. Jd2718 (talk) 14:27, 3 July 2009 (UTC)
thread_title:  Ties are screwed up 

message: A casting vote is made after all other votes, and only when those votes are tied.  For the case of 2 pirates, D and E, the initial vote has E voting by himself.  Only if the initial votes are tied (which is impossible with one voter!) is D's vote counted.  So, for two pirates, E should vote down any proposal by D and take all the loot.
I agree with Jd2718 that "the proposer wins ties" is the correct way out of this.  The part about "casting vote" should be removed and replaced with "proposer wins ties."
Mdnahas (talk) 21:47, 31 July 2014 (UTC)
thread_title:  I agree, "casting vote" is wrong and needs to be deleted 

message: A does 0-25-25-25-25 so no one wants to vote him off. If they vote off then either B gets the same or gets something less (if he gets more then others they will vote to throw him overboard, and if he gets less then he gets less), so A should do 0-25-25-25-25. 24.1.201.172 (talk) 00:51, 22 June 2010 (UTC)
replies: text: It's a game theory puzzle. In game theory, you generally assume that all the other players play according to theoretically perfect strategy; hence, their moves are predictable. The article has a mathematical proof as to why 98-0-1-0-1 is the best strategy. So if you're going to argue with it, you need to find a reason why the proof isn't adequate. - furrykef (Talk at me) 07:43, 27 November 2010 (UTC)
thread_title: Possible answer

message: I think a solution might be missing. If you have a very high number of pirates, for which at least half of them will die regardless of their strategy (as exposed in Extension of the game), then the first pirate may very well alocate all the gold to himself, as all the others who would have died will vote for him so they actually don't die.
I'm not too keen calculations, so if anyone wants to establish the formula... Alleria86 (talk) 09:22, 16 April 2014 (UTC)
replies: text: Establishing a formula would probably be original research  which couldn't be added. Once you get some dying, you have to start making assumptions about life expectancy, inheritance rules, time to carry out a vote, time to distribute booty and probably more. Too many variables for me. I don't think you have to have half dying before the solution changes. Once there are some unexpected deaths, pirates won't know if they are of odd or even rank so the normal solution fails. Low ranks may as well vote lots out to have chance of getting higher distribution. Suppose 10,000 is the most that can be efficiently voted on and distributed without expectation of deaths playing a role but you have 100,000 pirates and 100,000 gold. I can't think of anything much better than offering 6 gold to ranks 50003 through 66666 and 2 gold to ranks 1 to 4. 12 remaining gold given at random amongst those allocated 6. Why? Low ranks cannot be bribed as they can enjoy some killing til the numbers are lower for more chance of being offered a higher payout when there are fewer pirates. If the top third cannot think of a better distribution, they will suffer same fate so should agree just to survive without being bribed. The trouble with trying to bribe ranks 49999 through 50001 is that unexpected deaths may mean they do not get a payout so they prefer to vote out a few to make sure. While 50005 could enjoy a few deaths and is safe from losing gold for quite a few rounds of votes, if he votes against many rounds of voting out may occur. This leaves a few votes short. I chose 1-4 as targets to bribe as these seem least likely to end up as rank 10000 or gain a rank above 50% to get some gold. If previous lead pirate failed with trying to bribe 1-4 with 2 gold then these low ranks should be varied. That way they should take a bribe when it is offered. Not sure this is the right solution, there could be better solutions or many of similar level of sense. If ranks just above 66666 voted against, a subsequent lead pirate might spread the gold over a larger number of pirates above 50% ranking but then those just above 50% might vote against in the hope the gold is spread to fewer pirates. Such swings could mean there may not be a proposal that works til you get down to somewhere near 10,000 pirates. Tapering the gold downwards above 50% rankings may discourage having lots of rounds of voting against. Such tapers might work better with much more gold for example 100,000 pirates and 1,250,000,000 gold allows you to bribe rank 3rd highest with 1 gold, 4th highest with 2 gold, 5th highest with 3 gold, and so on. Anyway probably  crandles (talk) 22:01, 14 April 2015 (UTC)
thread_title:  Missing solution for high number of pirates 

message: Not intended for a section for much discussion, but please post here if you've decided the answer & solution are wrong and you'll edit them. I've just reverted back an anonymous edit by someone who had obviously read a slightly different version of the puzzle, so they missed that the different answer here was actually correct. Their solution contradicted the rules given here. - Zuytdorp Survivor (talk) 23:38, 17 March 2015 (UTC)
replies: text: Sorry yes.  I have tried to correct the extension section but there may be other clearer ways of explaining it. I guess a certain offer of 1 is better than an 100/101 chance of being offered a gold coin but giving much consideration to this might be . The source does say the solution in not unique so I hope what little I said about this is OK. crandles (talk) 15:17, 15 October 2015 (UTC)
thread_title:  Thinking of "correcting" the answer? 

message: Does the problem as described adequately rule out part gold coin distributions via some random procedure in the distribution proposal? E.g. A sets distribution so C and E each role a dice and if they roll a 6 they get a gold coin otherwise 0 gold coins. I.e. effectively bribed with 1/6 of a gold coin so expected gold for A is 99.666 ? Then a better solution is roll n die and get gold coin only if total is 6n with n being a high number. How best to edit in that restriction? crandles (talk) 10:05, 15 April 2015 (UTC)
replies: text: Changed from "apart from the main proposal" to "apart from a proposed distribution plan that gives a whole number of gold coins to each pirate" crandles (talk) 10:12, 15 April 2015 (UTC)
thread_title:  Part of a gold coin distributions