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. :-)

[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 hw would say yes. --Petter 23:42, 8 March 2007 (UTC)