To review the bidding:
Two days ago I posed a puzzle about 10 pirates dividing 100 coins.
Yesterday, I presented what appears to be an airtight argument that the coins must be divided 96-0-1-0-1-0-1-0-1-0.
But yesterday I also told you that the “airtight argument” is in fact not airtight, and that other outcomes are possible. I challenged you to find another possible outcome, and to pinpoint the gap in the “airtight argument”.
Our commenter Xan rose to the occasion. (Incidentally, his website looks pretty interesting.) Here’s his solution:
| The most ferocious pirate proposes to take all the coins, leaving everyone else with nothing. The other nine all want this proposal to fail. Nevertheless, they all vote yes and the proposal is accepted. |
Note that this is perfectly consistent with rational behavior. No matter how much you want the proposal to fail, as long as everyone else is voting “yes”, your vote can’t affect on the outcome, so voting “yes” can’t hurt you. (Neither, of course, can voting “no”.)
Another possible outcome is that one of the pirates (perhaps even the most ferocious pirate!) votes no, while the rest vote yes. Or three vote no while seven vote yes.
In each of these cases, no single voter can change the outcome so there’s no particular reason for anyone to change his vote. In other words, everyone is being perfectly rational, taking as given the choices made by everyone else. In the jargon of economics, Xan’s solution is a perfectly good Nash equilibrium.
Likewise, if the 2016 election comes down to Trump versus Sanders, and if every single voter prefers Sanders to Trump, it’s perfectly consistent with rational behavior for Trump to get 100% of the votes. If everyone else is voting for Trump anyway, you might as well vote for him too. (You also, of course, might as well vote for Sanders.)
That shows that yesterday’s conclusion was wrong (or at least incomplete). The next question is: What is the flaw in the logic that led to that conclusion?
Answer: The argument (as you’ll see if you read it carefully) makes the implicit (and quite unwarranted) assumption that people always vote for their preferred outcomes. Given that assumption, the argument really is airtight (despite some of the skepticism that was expressed in comments.) If pirates always vote for their preferred outcomes, then the only possible Nash equilibrium is indeed 96-0-1-0-1-0-1-0-1-0.
But if all we assume is that pirates behave rationally, then we can’t predict much about how they’ll behave in cases where their votes don’t matter. And that’s a lot of cases.
A historical note: I assigned this problem to classes for years, expecting (and usually getting) the 96-0-1-0-1-0-1-0-1-0 answer. One day in class, while I was reviewing that “correct” answer, my student Matt Wampler-Doty raised his hand to point out the plethora of alternative solutions and the unjustified assumption in the “official” analysis. I was sure he was wrong, and made some inane attempt to dismiss him, but to his credit, he persisted until I got the point. He made my day.
I actually understand this solution better than the 96-0-1-0-1-0-1-0-1-0 one. I had heard the latter before and could never really buy into it. In the example from yesterday, it is in both George’s and Igor’s interest to keep the game going. If they each vote no on Arlo’s and Bob’s proposal, they will already have seen two plank walks and are better off than at the start. George would vote no on Arlo if he thought Igor was thinking the same way. It is in Igor’s interest to keep the game going until Howard is up, and it is in George’s interest to keep the game going until he himself is up. It’s not clear why they should accept the proposals until one of those obtain.
I think my problem is that I see them as humans — pirates, no less — before seeing them as logical automatons. In another version of the puzzle, the pirates are described as smart, but if they are willing to take 1 gold-coin where an even distribution would have given them 10, they seem more like chumps to me.
I don’t see how assuming that the pirates vote for their preferred outcomes is a strong enough assumption.
Take Xan’s Nash equilibrium: The most ferocious pirate proposes to take all the coins and everybody else votes yes.
This is the most ferocious pirates preferred outcome and he votes for it. No one else has it as their most preferred outcome.
Did you mean something like “all pirates vote for their preferred outcomes and against everything else”?
Exactly how do you define preferred outcomes?
Naively, I would say that all the pirates prefer the outcome where they get all the gold and everyone else walk the plank.
I’m late to the party here, but is there another inconsistency in the 96-0-1-0-1-0-1-0-1-0 solution?
To wit, this solution assumes that if there are three pirates left (8,9,and 10) all three remaining pirates will act rationally.
But, how did the game get to the stage where there were only three pirates left? Answer: at least one pirate acted irrationally in the previous stages.
So, how is pirate 7 supposed to act? It would depend on what he concludes from the irrational actions. While pirate 7 may be able to solve this problem since he saw the irrationality, how is pirate 1 supposed to know what pirate 7 (and 6, 5, 4, 3, 2, and 1) will do?
So, pirate 1 is scared to death that he will propose 96-0-1-0-1-0-1-0-1-0, but some pirates will act irrationally. Since he has an infinite preference for staying alive, he proposes 0-11-11-11-11-11-11-11-11-12, implicitly saying “some of you may be crazy, please don’t kill me, take all the money, you won’t do better because you don’t know who is crazy and how they are crazy.” This is the best pirate 1 can do because he is giving as many coins as possible to the other pirates and he can’t differentiate between them as to rationality.
Seems like this equilibrium could be enforced by each pirate 2 through 10 voting no to any distribution that gives them less than 11.
This alternate solution has intuitive appeal. Pirate 1 is the pirate most in danger of being thrown overboard. Since going overboard is infinitely bad, he would give away all coins to avoid that. (Who he allots it to is not so clear.)
The infinitely bad outcome of being thrown overboard completely dwarfs the coins the pirate get, so each pirate cares only about not being thrown overboard. Pirates 2-10, therefore would all vote yes because they have an infinitely large motive to stop the game immediately.
This logic could lead to the conclusion that all pirates will always vote yes because all they care about is that the game stop as soon as possible.
I’m disappointed. This is a trivial point. Yes, of course the pirates could all go insane and vote against their interests en masse, and if everyone else went insane then your vote wouldn’t matter. But why in the world would everyone decide en masse to vote against their interests? Similarly, if everyone supports Sanders, why in the world would they decide to vote trump. Or if everyone thinks the Internet is a good thing, why would they all vote to destroy it?
But I’m at least happy that according to what I think actual humans would do, my preferred solution is the only solution.
Yes, I’m in agreement with you Zazooba. But so far, no one else has been.
I wouldn’t go so far as to say a pirate would act “irrationally” by rejecting 96-0-1-0-1-0-1-0-1-0, though. I would say perhaps “he rejects the logical line of thought that lead to the distribution (96-0-1-0-1-0-1-0-1-0) being offered”. If a pirate, as a result of rejecting 1 coin could get 11+ coins instead, that doesn’t seem irrational to me.
Zazooba: “This logic could lead to the conclusion that all pirates will always vote yes because all they care about is that the game stop as soon as possible”
Exactly why a rational Pirate 1 might also propose 100 – 0 – 0 …
I’m not convinced by any of the arguments that the official answer is wrong.
Lets get one thing straight – would the official answer be what actually happened in any believable scenario? No of course not. Try this with a real bunch of pirates and you’ll end up with chaos (likely very bloody chaos). The outcome will be entirely unpredictable and different every time. The only truthful answer is “nobody knows what would happen”.
It’s also true that you can construct any number of arbitrary strategies that would be consistent against any individual changing their mind.
But this is a logic puzzle and the ground rules of logic puzzles surely state that we assume all participants act as fully rational, logical, game-theory aware operators (ie not like real pirates at all) and they they don’t all follow some arbitrary sub-optimal strategy.
You’ve agreed that in the case of the first pirate suggesting 100 for herself it’s in all the other pirates’ interests for the proposal to fail. Therefore unless there’s a compelling reason for another pirate to believe that all the other pirates are voting yes – a pirate’s only rational option is to vote no. Any other option is based on circular causality and requires an external force to an impose an initial assumption that all the other pirates are voting yes – which is surely against the spirit of such a puzzle.
“The argument (as you’ll see if you read it carefully) makes the implicit (and quite unwarranted) assumption that people always vote for their preferred outcomes.”
In elections we usually get lots of people who don’t vote at all. These people are not voting for their prefered outcome. Many of these will have concluded that there is no point, since the chances of their vote making a difference are vanishingly small, and the cost of going to the voting station is not worth the rewards.
In a manner slightly reminiscent of the unexpected hanging, the victor is voted for only by irrational people – perhaps not the ones we should be relying on to make a good choice.
The alternative solutions seem unsatisfying as there is no logical path to arrive at them. If we assume each pirate votes for his prefered solution even when his vote cannot change the outcome, we can take it a step at a time and arrive at a solution. Without that assumption, we seem to have to start at a possible solution and then test if there is any reason for any votes to change change.
To explore a bit further, in the two pirate game, 100-0 is the “usual” answer, but 99-1, 98-2 etc all satisfy the requirements, since if pirate 9 votes yes (as he must do to his own proposal), then anything pirate 10 votes does not change anything. However, for pirate 9 to offer more than 0 is irrational, so we could enever arrive at this solution. For 2 pirates there is still only one outcome possible.
With three pirates, the usual solution is 99-0-1. Pirate 8 cannot offer less than 99 rationally, so the other possible offer is 100-0-0. If everyone is voting yes, then there is no reason to change one’s vote, so this is also a NE. We have to attach the votes to each pirate as well to see if it is a NE. So 100(Y)-0(Y)-0(Y) is a NE, but 100(Y)-0(Y)-0(N) for example is not. We have only two NE.
This is unsatisfying as we have no way to select between Y-Y-Y and Y-Y-N with the information in the puzzle. As far as we know each is possible, so we cannot reject the seemingly unlikely 100-0-0.
This does seem unsatisfying, and raises the question of whether the assumption is unwarranted. Whilst it is not stated in the question, does not Ken B’s “Good puzzle theorem” lead one to conclude that there must be a way to arrive at an answer? Either that or it is not a good puzzle.
People who do not vote are gaining by avoiding the effort. If people are forced to vote (as in Australia) I know of no evidence to suggest that people who believe their candidate will loose vote for the opposition anyway. The opportunity to register your intention is a small but positive value in voting for your prefered canditdate.
@Henri Hein “the pirates are described as smart, but if they are willing to take 1 gold-coin where an even distribution would have given them 10, they seem more like chumps to me.”
Well, a fairy might have given the pirate 10 gold coins *and* a unicorn ride. Sadly, the fairy wasn’t available, so this was not a relevant option. The even distribution is also not available to most pirates.
@Henri Hein If George and Igor vote no on Arlo’s proposal, then Bob’s proposal of 96-0-1-0-1-0-1-0-1 will be successful, because Bob, Dave, Freddy, Howard, and James will all vote for it. So they would be fools to vote no on Arlo’s proposal, since they would get no coins instead of 1 coin.
Prof. Landsburg offers an interesting hole in the [96,0,1..] proof. I certainly like this one better than any of the other proposed ones.
Yet, I am not entirely sure I am convinced.
Prof. L is right that we made the unexamined assumption that everybody who votes always votes their preference. A voter who, for some reason, votes even though he knows his vote will not affect the outcome cannot be predicted on the basis of his preference over outcomes.
But is that truly the case here?
Consider the problems in inverse order:
P10 proposes [100]. We really can predict that P10 is going to vote for 100 gold over his own death. P10 knows that his vote will be decisive.
P9 proposes [100,0]. Failure is P10’s preferred outcome; success P9’s. If P10 knows that P9 is going to vote aye, then P10’s vote is meaningless and might go either way. Similarly, if P9 knew that P10 was (for some reason) going to vote aye, then P9’s vote would be free to go either way.
But it is just because neither of their votes are predictable to each other, neither P9 nor P10 can conclude that their vote will not be decisive.
Is it too strong an assumption that in a situation where the voter cannot *prove* that his vote will be indecisive, he’d vote for his preferred outcome, just as he would if he knew his vote was decisive?
If we can make that assumption, P9 will vote Aye. The vote of P10 still cannot be predicted, but it does not matter for the outcome.
In this manner, I believe, we can build all the way back to [96,0,1…] as the only solution.
Don’t believe I missed these posts. There is a really nice variation on this puzzle (from Ian Stewart I believe) where we now have 500 pirates vying for the 100 coins.
To apply this to the Sanders-Trump election:
A Trump-preferring voter T might vote Sanders if he knew for a certainty that his vote was meaningless. For example, if Sanders was listed alphabetically on the ballot before Trump then the tiny additional effort of moving his hand down one spot further down the ballot might be enough to make T vote Sanders.
But the T cannot know that his vote is not decisive for a certainty. It is true that in a mass election, the probability that T’s vote will be decisive is truly tiny. But it is not zero. If T’s utility for Trump over Sanders is sufficiently large, then–even multiplied by the tiny probability of decisiveness–it may be rational for T to expend the tiny disutility of moving his hand one spot further down on the ballot to vote for Trump.
@Jonathan Kariv
I believe in this case the solution is either for the first 300 pirates to die. Alternatively, if P1 can throw some additional coin from his personal stash into the pot, he will offer [-200,0,1,0,1…] and live.
@Jonathan Kariv
Correction: P1 will offer [-150,0,1,0,1…]
Sub Specie (#11 and #13): In the problems with 1 or 2 pirates, there are, in fact, unique Nash equilibria. That breaks down when you add more pirates.
You suggest that if a pirate cannot prove his vote will be indecisive, he’ll vote for his preferred outcome. I agree with this. But if there nine other pirates all voting yes, and if (as has been assumed) you know their strategies, then it’s easy to prove that your vote will be indecisive, and therefore perfectly rational to vote yes.
I think this seems counterintuitive largely because, in real life, we rarely cast votes in situations where we *know* everyone else’s strategy. But of course, there’s a whole lot in this problem that differs from what we encounter in real life.
And since we were speaking of Robot Pirates earlier, I should have supplied this: http://media.animevice.com/uploads/0/3346/470849-bwabbp4.jpg
Nathan (#7)
But this is a logic puzzle and the ground rules of logic puzzles surely state that we assume all participants act as fully rational, logical, game-theory aware operators (ie not like real pirates at all) and they they don’t all follow some arbitrary sub-optimal strategy.
Agreed. But of course casting a “yes” vote cannot be sub-optimal in a situation where your vote has no effect on the outcome.
“But if all we assume is that pirates behave rationally, then we can’t predict much about how they’ll behave in cases where their votes don’t matter. And that’s a lot of cases.”
If you can’t predict how rational pirates would behave, doesn’t that imply that you can’t predict that your vote won’t matter?
It’s circular to stipulate knowledge of another’s strategy when that strategy depends on simultaneously acquiring knowledge of mine.
The claim that “your vote has no effect on the outcome” does not apply to this magical world. Of course it has an effect! Switching my vote will likely collapse the wave function right into 96…1.
I saw a movie just last night (“Time Lapse”) where every day, a painter magically acquired a photo of tomorrow’s painting. He then felt compelled to paint it using the photo as reference. Where did the painting come from?
@Steve “Agreed. But of course casting a “yes” vote cannot be sub-optimal in a situation where your vote has no effect on the outcome.” But why would you have a situation where everyone else was voting yes in the first place?
@ Prof L. (#16)
Let’s consider the 3 pirate problem:
P8 proposes [99,0,1]. Success is the preferred outcome of P8 and P10; failure, for P9. Two ayes are needed for success. How are they going to vote?
In the first instance, unpredictable. If P9 and P10 were known to vote Aye or P9 and P10 were known to vote Nay, P8’s vote is meaningless and hence unpredictable. Similarly for the permutations.
But if the make what is admittedly is an assumption, but approved by the puzzle maker, that a voter who cannot prove that his vote is meaningless will vote his preference over outcomes, resolution becomes possible.
P8 cannot prove that his vote is meaningless and will hence vote Aye. P9 can now prove that P8 will vote Aye but still cannot prove that his vote is meaningless, so P9 will vote Nay. P10 now knows that his vote is decisive and votes Aye.
If we change the order of decision-making with P9 going last, P8 and P10 will still for the same reasons vote Aye. At this point, P9’s vote is meaningless and it cannot be predicted how he will vote. But–because his vote is meaningless–the outcome will still be the same [99,0,1] passes.
It is unclear to me how this reasoning is any different from the one that gave us a definitive outcome in the two-pirate game. Nor why it does not carry over to the n-pirate game. It is true that–depending on the ordering of votes–many votes may become meaningless and hence unpredictable, but the outcome remains predictable.
One may object that this presumes an ordering of pirate decision-making or voting. But is that not necessarily implied?
The premise that all pirates know how all other pirates are going to vote on each proposal before deciding how they are going to vote is not logically possible. If it was PX’s strategy to always vote the same way as PY and it was PY’s strategy to always vote the opposite way from PX, universal pre-knowledge would lead to an impossible loop.
The strongest pre-knowledge condition we can impose is that there exists a voting order among pirates and each pirate knows only how any previous pirate in this order voted and can only predict the votes of others if they logically follow from the premises and the situation. This will often leave a pirate unable to predict the votes of subsequent pirates.
Sub Specie:
The premise that all pirates know how all other pirates are going to vote on each proposal before deciding how they are going to vote is not logically possible.
This is not a criticism of the puzzle solution. It is a criticism of the notion of Nash equilibrium.
@Steve Sub Specie Aeternitatis isn’t criticizing the notion of a Nash equilibrium. He’s just talking about the process by which a Nash equilibrium is arrived at. He’s saying that there’s no rational process by which, e.g. everyone will decide to vote yes on 100-0-0 or by which everyone will decide to vote no on 99-0-1.
@Steve Landsburg “This is not a criticism of the puzzle solution. It is a criticism of the notion of Nash equilibrium.”
Perhaps you are right. It has been many years since I seriously studied game theory and my memory may be fuzzy. But let me give it a shot at the problem in those terms.
10-pirate game. P1 proposes [96,0,1,0,1,0,1,0,1,0]. The pure strategy for each pirate is to vote Aye or Nay. The mixed strategies are to vote Aye with probability p. Let’s represent all of these strategies by the vector [p_10,p_9,..,p_1].
Is the vector [1.0,1.0,1.0…1.0] a stable Nash equilibrium? It is certainly true that for any pirate to go to 0.99 would in no sense improve that pirate’s expected outcome nor would it cause any other pirate to choose a different strategy.
But, if I recall correctly, for an equilibrium to be stable, it is necessary that any player who deviated from the equilibrium would not only be no better off, but would have to be strictly worse off. In that sense, this vector is not a Nash equilibrium.
What about the vector representing the predicted votes [1,0,1,0,1,0,1,0,1,0]?
It is stable to deviations by the even pirates, P10, P8, P6, P4, and P2. Deviating to 0.99 for any of them would make them strictly worse off and while the other pirates would not find a better strategy.
It is not stable to deviations by the odd pirates, P9, P7, P5, P3, and P1. Any of them can deviate without being worse off and without giving the even pirate any better option.
Given that this is the closest thing to a stable equilibrium and that all permissible deviations from it still lead to the same outcome, is it really wrong to call that outcome the only correct one?
The Good Puzzle Theorem apparently vindicated again:
http://www.thebigquestions.com/2015/09/08/are-you-smarter-than-google-redux/#comment-160741
I did not see that coming!
I understand that [100,0,…,0] is a Nash equilibrium, but I don’t see how a Nash equilibrium is a solution of the original problem. Initially none of the pirates know how the other pirates will vote. If the votes are cast simultaneously, then none of the pirates know yet that their vote doesn’t matter, so the original argument applies. If they vote sequentially, then only after 5 yeses are cast will votes cease to matter. Where will those 5 yes votes come from?
I assume I’m missing something here as otherwise Steve would not have bothered doing these 3 posts
BUT:
The assumptions stated yesterday were:
“A pirate’s first priority is to stay alive, his second is to amass as many gold coins as possible, and his third is to see others thrown overboard. “.
There are 5 pirates who stand a chance of gaining at least 1 gold coin and having at least 1 pirate thrown overboard if they vote no to Pirate A’s proposal.
Given this, why is the “implicit assumption that people always vote for their preferred outcome” quite unwarranted ?
Any individual pirate will not know for sure that his vote will not be the decisive one so it will be rational for him to vote for his self-interest even if he ends on the losing side of the vote.
In addition: If Pirate A’s top priority is to stay alive, and he assumes that the other pirates vote rationally – what motivation would he have for proposing something that will get him thrown overboard at least some occasions ?
“‘Note that this is perfectly consistent with rational behavior. No matter how much you want the proposal to fail, as long as everyone else is voting “yes”, your vote can’t affect on the outcome, so voting “yes” can’t hurt you. (Neither, of course, can voting “no”.)”
You can’t truly know with 100% certainty how others will vote even if they told you in advance, and you don’t know how your vote may influence others to vote (for example if its a show of hands).
Given this: It can never do any harm (and can sometimes do you some good) to vote with your self-interests.
Daniel R. Grayson:
I understand that [100,0,…,0] is a Nash equilibrium, but I don’t see how a Nash equilibrium is a solution of the original problem. Initially none of the pirates know how the other pirates will vote.
It was an explicit assumption that each pirate optimizes, taking all the other pirates’ strategies as given. If each pirate’s strategy includes “Always vote yes on the first round” as a component, then each pirate must take it as given that all other pirates will vote yes on the first round.
This does seem to raise the question of how they can take each others’ strategies as given for the purposes of formulating strategies that others are already taking as given. But as I said in an earlier comment, that’s an issue with the notion of Nash equilibrium, not with the puzzle. The puzzle as stated does, I think, clearly equate “solution” with “Nash equilibrium”.
@Jonatan
“Yes, I’m in agreement with you Zazooba. But so far, no one else has been.
I wouldn’t go so far as to say a pirate would act “irrationally” by rejecting 96-0-1-0-1-0-1-0-1-0, though. I would say perhaps “he rejects the logical line of thought that lead to the distribution (96-0-1-0-1-0-1-0-1-0) being offered”. If a pirate, as a result of rejecting 1 coin could get 11+ coins instead, that doesn’t seem irrational to me.”
I agree. Perhaps we are overlooking that there are many Nash equilibria. Then, we need to be clearer about what each pirate knows about the others’ strategies. If we assume no information (no announcements or negotiations), then each pirate has to guess about the others’ strategies.
So, why not use our knowledge of human nature to assign probabilities to the various strategies? Giving only one coin to a bunch of ferocious pirates sounds like a really good way to tick them off so much they want to throw you overboard. So, giving away all the coins in a “fair split” is a reasonable way to minimize the probability of being tossed overboard.
@6maznak @Zazooba: “This logic could lead to the conclusion that all pirates will always vote yes because all they care about is that the game stop as soon as possible”
Exactly why a rational Pirate 1 might also propose 100 – 0 – 0 …”
I Agree. There may be infinitely many equilibria. Your proposed equilibrium is a game of chicken. But it may not work if pirates adopt strategies of being easily provoked into dangerous fits of violence.
We also haven’t paid much attention to mixed strategies.
What if each pirate votes per the 96-0-1-0-1-0-1-0-1-0 strategy 90% of the time and votes opposite 10% of the time?
Then, there is a non-zero probability that pirate 1 suffers the infinitely bad result of being thrown overboard if he uses the 96-0-1-0-1-0-1-0-1-0 strategy, because if even one of the 1-coin pirates votes no, pirate 1 goes overboard. Hence, pirate 1 may be better off proposing 96-1-1-0-1-0-1-0-1-0 if this means that pirate 2 will vote yes 90% of the time rather than just 10% of the time, so that two “yes” pirates have to switch to “no” for pirate 1 to be thrown overboard. By the same logic, 96-1-1-1-1-0-1-0-1-0 is better still, etc.
Sounds like there are lots and lots of equilibria to choose from and no good criterion for choosing. Then, perhaps our prior knowledge of human behavior and pirate-behavior in particular is as good a criterion as any for choosing the best strategy.
Perhaps the first pirate, being the most ferocious is able to intimidate the others into letting him take all the coins. Or, perhaps the ferocity of 9 less-ferocious pirates outweighs the ferocity of the most ferocious pirate.
Obviously I am confused. Multiple commenters have referred to proposals (like 96-0-1-0-1-0-1-0-1-0) as strategies in a game-theoretic, nash-equilibrium sense.
But the proposals are purely sequential. Each previous proposal is known and fixed at the time the next one is made. There is no simultaneous choice. Hence, the proposals are not strategies in the game-theoretic sense and we do not need to analyse them as such, no more than we need calculate Nash Equilibria or chess.
The simultaneous game for which game theory provides answers is not the proposal, but the voting with each pirate have just two pure strategies (aye and nay) and an infinitude of mixed strategies in between.
So can somebody set me straight?
@SSA 33
Assume that is a solution. As has been pointed out, then so is the same distribution with the last two pirates’ amounts changed, i.e. 0-1.
That will garner 5 votes I think.
Second, the conditions of the puzzle do stipulate Igor knows Fred’s votes in all cases, etc. This corresponds to a known or filed and published set of trees. Hence talk about strategies.
Here’s a suggestion. Rephrase the puzzle with the additional stipulation that pirates prefer to vote no, unless a yes vote gets them some reward. Now what happens?
@Ken B
I don’t think 96-0-1-0-1-0-1-0-0-1 is a solution. Even though P10 get a coin, he will be a Nay, because the next alternative 96-0-1-0-1-0-1-0-1 gives him just as much coin plus the pleasure of drowning P1.
Also, I don’t think that it is possible to stipulate that everybody knows everybody else’s vote. Imagine trying to analyse a two-player game of rock-paper-scissors under that stipulation.
What it is possible to stipulate is that all pirates know all other pirates’ game-theoretic strategies (i.e., probability of voting Aye).
@Steve “This does seem to raise the question of how they can take each others’ strategies as given for the purposes of formulating strategies that others are already taking as given. But as I said in an earlier comment, that’s an issue with the notion of Nash equilibrium, not with the puzzle.” No, it’s not an issue with the notion of Nash equilibrium. The definition of Nash equilibrium is a perfectly coherent definition. The issue is that just because something is a Nash equilibrium doesn’t mean that it’s possible for rational actors to arrive at it.
I disagree with your statement “The puzzle as stated does, I think, clearly equate “solution” with “Nash equilibrium””. The puzzle statement is quite clear that any solution must be a Nash equilibrium. But that doesn’t mean that every Nash equilibrium is a valid solution.
1. I believe the spurious 100-0-0… Nash Equilibrium only exists if voting is simultaneous (or secret ballot). If voting is sequential and public, from most ferocious remaining pirate to least ferocious pirate, the standard solution is the only one. I’m sure it is true for the three pirate case and I’m pretty sure it is true for the four pirate case (Comments are too short to show my work). I see nothing in the problem statement that specifies the voting procedure.
2. Only the standard solution is a “trembling-hand perfect” Nash Equilibrium. That is, if there is an epsilon chance (arbitrarily small but greater than zero) that a pirate accidentally deviates from his optimal strategy, there is now a non-zero chance (again arbitrarily small) that a particular pirate’s vote becomes decisive and he should vote self-interest.
By the way, the possible three-pirate solutions in the sequential voting case are a proposal of 99-0-1 followed by votes of Y-Y-Y, Y-Y-N or Y-N-Y. Only the Y-N-Y vote is trembling-hand perfect.
@Steve Let’s consider a simpler game, just to illustrate my point. Look at this payoff table:
1 2
1 (1,1) (0,0)
2 (0,0) (0,0)
Both (1,1) and (0,0) are Nash equilibria, but rational actors would clearly choose (1,1). It’s true that if the first player were to choose option 2, the second player might be indifferent between choosing options 1 and 2. But there’s no reason why the first player would choose option 2. Both players would choose option 1, end of story.
In comment 40 the 1 and 2 should be spaced out more, to be column headings.
@Keshav Srinivasan Quite right, I think.
Also consider this game:
A\B Aye Nay
Aye 1\-1 1\-1
Nay 0 0
(Note that the two columns for B are identical.)
It is certainly true that there is no fixed, predictable strategy for B. Any pure or mixed strategy for B will result in the same pay-off. However, that does not mean that there is no predictable outcome. A will play the pure strategy of Aye and the outcome will always be 1\-1.
Similarly, in the pirate game, while the votes of opposed pirates are not predictable, the outcome of the game is.
The comment system did something funny with the text. The “0” in A’s Nay row should be “0” backslash “0”.
Keshav Srinivasan:
Both (1,1) and (0,0) are Nash equilibria, but rational actors would clearly choose (1,1).
This is the kind of statement that requires both a) a rigorous definition of “what rational actors would do” and b) a proof.
You’ve provided neither.
Eric (#38): I did not know the term “trembling hand perfect”, but it’s of course applicable here, and a good addition to my vocabulary.
The terminology varies between authorities, but perhaps the description will be more comprehensible if couched in the following terms:
The voting game has many Nash Equlibiria. But all of them are Weak Nash Equlibiria. For the ten-pirate voting game with offer [96-0-1-0-1-0-1-0-1-0], the Nash Equilibria of the form [1,A,1,B,1,C,1,D,1,E] (where A,B,C,D,E are any mixed Aye/Nay strategies and 1 represents the pure Aye strategy) are relatively the most strict and all of them have the same pay-off.
@Steve Landsburg re: 44
To be fair, neither have you. You say something like “everyone is being perfectly rational… in the jargon of game theory, it is a Nash equilibrium.” But Nash equilibrium does not correspond to ‘rational’, neither in common usage, nor in game theory. Nash equilibrium is a statement about beliefs of the players, namely that the strategies of the players is common knowledge, while rationality is a statement about players choosing what’s ‘best’ for them, given what they believe other players will be playing.
In your original problem, I agree that you were unambiguous about what players know about what other players are playing. I’d identify the ambiguity as being in your use of the term ‘as happy as possible’, which gets at the meaning of ‘rationality’. One perfectly reasonable interpretation of this is that players do not play weakly dominated strategies. That seems to be the favored interpretation of a lot of commentators. You might say ‘as possible’ rules out consideration of zero probability events, but you’ve already assumed some sort of lexicographic preferences for the pirates, in that 1 gold coin seems to outweigh seeing any number of pirates thrown overboard. So why not lexicographic probability assessments?
Anyway, I agree it’s a good and interesting point that the game has many Nash equilibria, but I also think it would be more interesting if you took some of the points commentators are raising, like Keshav Srinivasan, more seriously, and discussed how game theorists think about these issues. Their comments may not be rigorous or stated with the jargon of game theory but they are still getting at something interesting.
Just to add: It’s for this reason that much of the literature on voting games rules out weakly dominated strategies, precisely because the Nash equilibrium concept results in predicted behavior which in common experience we would call ‘irrational’.
@Steve To use the terminology of this webpage, my definition of what rational actors would do is that they would select a weakly dominant strategy (assuming such a thing exists).
http://www.econ.ucdavis.edu/faculty/bonanno/teaching/122/Dominance.html
And I think a set of rational actors would always select a weak dominant-strategy equilibrium (if such a thing exists), not merely any Nash equilibrium.
(1,1) is a weak dominant-strategy equilibrium, (0,0) is not.
@Steve Landsburg #16 “In the problems with 1 or 2 pirates, there are, in fact, unique Nash equilibria.”
On reflection, I do not believe that is true for the 2-pirate voting game. For it has infinitely many (Weak) Nash Equilibria. If 1 denotes the pure Aye strategy for the first pirate and X denotes a pure (Aye or Nay) or mixed (stochastic mix between the two pure) strategy for the second pirate, then any [1,X] is a Nash Equilibrium.
As there are infinitely many X, there is–contrary to the statement above–no unique Nash Equilibrium.
The argument today is that there are many Nash equilibria, but why is a pirate going to vote for one that gives him a lousy deal?
I partially agree with Zarooba that the less ferocius pirates will demand and get a better deal. Only I would not call it irrational, because he gives a rationale for them voting that way. It is rational for a pirate to reject any deal that does not give him at least 11 coins.
I do think that both 96-0-1-0-1-0-1-0-1-0 and 100-0-0-0-0-0-0-0-0-0 are incorrect, because they would not happen with real human pirates. I gave this argument yesterday.
Keshav:
(1,1) is a weak dominant-strategy equilibrium, (0,0) is not.
Agreed.
Here is an argument in support of my solution yesterday.
Suppose pirates E-J announce that they want at least 11 coins each. Arlo offers 96-0-1-0-1-0-1-0-1-0 and gives a proof that it is optimal. E-J vote him overboard. Bob then offers 96-0-1-0-1-0-1-0-1, shows his Google badge, and also gets voted overboard. What do you think Charlie is going to do? Is he going to stick with the 97-0-1-0-1-0-1-0 Google-Landsburg logic, or will he offer something like 23-11-11-11-11-11-11-11 or 0-12-12-12-12-13-13-13-13? I say the latter. The later pirates will all get at least 10 coins, and maybe more if they bargain harder.
@ Roger
“Suppose pirates E-J announce that they want at least 11 coins each. Arlo offers 96-0-1-0-1-0-1-0-1-0 and gives a proof that it is optimal. E-J vote him overboard. Bob then offers 96-0-1-0-1-0-1-0-1, shows his Google badge, and also gets voted overboard. What do you think Charlie is going to do? Is he going to stick with the 97-0-1-0-1-0-1-0 Google-Landsburg logic, or will he offer something like 23-11-11-11-11-11-11-11 or 0-12-12-12-12-13-13-13-13? I say the latter. The later pirates will all get at least 10 coins, and maybe more if they bargain harder.”
Good point.
What does P3 think when P1 and P2 got thrown overboard for going full-Google? The 96-0-1-0-1-0-1-0-1-0 solution assumes P3 will stick to the Google strategy. But will he? P3 has learned that the full-Google strategy leaves some positive probability of being thrown overboard. Since that is infinitely bad, he will do anything he can to reduce that likelihood.
Hence, to rigorously prove the 96-0-1-0-1-0-1-0-1-0 strategy, you have to explain why each successive pirate will stick to the strategy despite every previous pirate having gone overboard. To do this, the strategy needs to examine all the possible reasons previous pirates were overboarded, and the strategy must tell us why, in each of those cases the next pirate still goes full-Google. Until someone does this, the full-Google strategy has not been rigorously proven.
@Sub Specie, Keshav,
I don’t think we need to stipulate a fairy to show that Igor and George can do better than 1 gold-coin.
If Arlo and Bob are both voted down, the game will end somewhere between Charlie and George. If the pirates stick to their strategy, then I concede that the expected outcome is less than 1 gold-coin. However, as Roger points out, there is no good reason to assume the pirates will stick to the standard proposal (adjusted for number of pirates).
If they do, and Igor and George consider the game landing on their respective payout slots as 50% (which it roughly is), then their expected payout in an open game is .5 gold-coins, and a number of plank-walks. To keep this payout past the first 2 proposals, the chance of Charlie, et al, to not propose more generous offers, would have to be less than 50%. That seems implausible.
Consider Igor mulling over Bob’s proposal. George already voted against Arlo, so there is a non-zero chance that he will also vote against Charlie. Likewise, there is a non-zero chance that Charlie, et al, will increase their proposals. At worst, Igor stands to gain his expected payout (2+ gold-coins * 50%) plus extra plank-walks. Thus, a profit-maximizing Igor will vote against Bob.
George knows this. Therefore, he should vote against Arlo.
The last pirate, who is not in mortal danger, has a very simple happiness maximization strategy of always rejecting any split that gives him nothing, and always accepting any split that gives him something (which is why getting something is stipulated to be preferred to seeing a pirate thrown overboard). This is his only rational approach – any attempt to hold out or coordinate will fail because when there are only two pirates left, he gets nothing. That’s really the point of this exercise – without the ability to make binding promises, you can’t coordinate and you get outcomes that seem absurdly unfair even though everyone is acting rationally
@Sub #14/15 If I remember right without the head pirate adding his own coin, 456 pirates live. 456=200 (from the coin) +256 (who’re just trying to stay alive)
@Ben The trouble with that reasoning is that the last pirate is not going to worry about there being only 2 pirates left. For that to happen, 8 consecutive pirates would have to offer a foolish deal and get thrown overboad. Not likely. And even if it does, the last pirate is not getting thrown overboard anyway. So he would be a fool to settle for 1 coin. He can do much better.
@Henri Hein
Sorry about the fairy comment. But you are still wrong in your analysis.
For clarity, one should think of the pirate game as composed of an iteration of two sub-games: the proposal game (where the most ferocious pirate decides what offers to make) and the voting (where the pirates vote on the proposal). Each pair of these games may be followed by another pair (i.e., if the proposal is voted down).
The proposal game and the voting game are two very different theoretical sorts of games.
The proposal game is a game of the first type. The games, like Chess or Go or Nim, are sequential and have perfect information (i.e., the player knows exactly what the entire state and history of the world is). To a perfect player (as we assume the pirates to be) of a game of this type, there are no probabilities. The set of all possible consequences of every move can, in theory, be calculated.
The voting game is a game of the second type. These, like poker or rock-paper-scissors, are simultaneous or, equivalently, have hidden information. Game theory was invented to deal with this sort of game. Within such a game, rational players may pick a mixed strategy (i.e., one in which there is a random distribution of his possible moves). But, if the game is well-specified, the distribution of mixed strategies can be fully calculated and hence its outcome reduced to a single number for each player.
There is some disagreement here on whether the voting game is fully specified (with our host taking the position, no, and other in the affirmative). If it is not, the outcome of the whole pirate game cannot be calculated. If it is, then the entire pirate game with its sub-games becomes just a game of the first type.
And, as said above, in games of the first type, to perfect players there are probabilities and no psychology. Only winning moves and those that are not.
@Jonathan Kariv
Interesting. I’ll need to think about why my guess was wrong.
“So he would be a fool to settle for 1 coin. He can do much better.”
Suppose there are three pirates, and last pirate declares “I will only accept 2 coins for my vote!” The first pirate says, “I don’t care what you think, I still propose a 99-0-1 split”. A rational third pirate would *still* accept the split according to the rules, because 1 coin is preferable to the 0 coins he will get when the second pirate is in charge. In this game, there are no happiness point for standing on principle
@Ben 61
But cannot the middle pirate be voting Yes to 98 0 2 and No to 99 0 1, since he is indifferent?
This is Xan’s point. You cannot assume that when a pirate is indifferent he will always vote no. Why can’t he sometimes be voting Yes in the case where he’s indifferent? And that makes a pig’s breakfast of the official solution.
“It is rational for a pirate to reject any deal that does not give him at least 11 coins.” No, it’s not. If Pirate 10 were to do that when offered 99-0-1, then he would get no coins rather than 1 coin.
Comment 63 was addressed to Roger.
@Ben Actually psychology researchers have done very similar experiments on random people, and found that most people do indeed prefer to walk away with nothing rather than to accept a grossly unfair deal.
Yes, a rational pirate could accept a sure 1 coin instead of taking a chance on 10 coins or more. But a rational pirate could also take the chance on more coins.
@Keshav: That is not correct. P8 is the irrational one, if he offers the same deal that just caused 7 other pirates to get killed. He would also get killed, and P10 stands a chance of getting 50 coins from a grateful P9.
The more likely scenario is that those 8 pirates will not all die out of intellectual Google stubbornness. At least one will offer P10 a better deal.
@KeshavSrinivasan
“It is rational for a pirate to reject any deal that does not give him at least 11 coins.” No, it’s not. If Pirate 10 were to do that when offered 99-0-1, then he would get no coins rather than 1 coin.””
But, you have to explain why Pirate 8 would offer 99-0-1 when that strategy got 7 prior pirates overboarded. Why were the prior 7 pirates overboarded when they shouldn’t have been? Since Pirate 8 began the game assuming NONE of the prior 7 would be overboarded, how does this new information affect his thinking about what Pirate 9 and Pirate 10 will do?
I’m not saying there isn’t some way to get to the solution you want, I’m just saying you haven’t proven your point until you answer these questions.
Someone else mentioned Trembling Hand Perfect Equilibrium as the way to resolve this. But, we’ve been told to assume only Nash Equilibria, and in the Nash Equilibrium, Pirate 8 observes 7 prior outcomes that contradict the Nash Equilibrium. Plus, the Trembling Hand route raises complicated questions about the specifications of the trembles and requires an explanation of how there could have been 7 infinitesimally probably trembles in a row prior to Pirate 8.
“He would also get killed, and P10 stands a chance of getting 50 coins from a grateful P9.” We’re assuming rational pirates whose only goal is to maximize their number of gold coins. There’s no room for gratitude here. So P9 can’t credibly commit to offer P10 a nonzero number of coins. And for that reason, P10 can’t credibly commit to reject P8’s offer, because he knows for a fact that rejecting it would hurt him. And for that reason, P8 has no reason to offer anything other than 99-0-1, because he knows that P10 will accept 99-0-1.
Remember, Steve said that people can’t credibly precommit to do sub-optimal things in some branches of their decision tree in order to persuade people to avoid those branches.
Comment 68 was addressed to Roger. Argh, second time in a row.
“But cannot the middle pirate be voting Yes to 98 0 2 and No to 99 0 1, since he is indifferent?
This is Xan’s point. You cannot assume that when a pirate is indifferent he will always vote no. Why can’t he sometimes be voting Yes in the case where he’s indifferent? And that makes a pig’s breakfast of the official solution.”
The part of “what happens” is that matters is the final split that is accepted – people are getting lost contemplating the nature of hopeless voting. A happiness maximizing first pirate will not offer 98-0-2 (which has a 100% chance passing) because that would bring less happiness then 99-0-1 (which also has a 100% chance of passing). I agree that it doesn’t matter how the second pirate votes, but that’s really the entire point – sometimes people are forced to take a bad deal
@Ben: “You cannot assume that when a pirate is indifferent he will always vote no.”
Actually, that is an assumption to the problem. It is the pirate’s third priority.
@Keshav: “because he knows that P10 will accept …”
And yet to get to that point, he just watched 7 of his fellow pirates die because they were also so sure of same knowledge, only to see it proved wrong.
I still don’t think that the “official” solution is airtight given that the problem doesn’t make it clear how the order is really determined or that the pirates know their place in line. It would have been better to simply say that the pirates have drawn their positions out of a hat or something to that effect. Since there is no measure of ferocity that is predetermined, and all pirates have the same strategy, all the pirates will want to suggest the solution first. If the pirates always “win” this competition in their order of ferocity and the fifth most ferocious pirate (for example) would rather see two other pirates thrown overboard than receive one extra coin, then he cannot be bought for 1 coin by the most ferocious pirate. Since the problem doesn’t specify exactly how much pirates enjoy seeing other pirates thrown overboard in the form of extra coins, you cannot know the solution. I think that’s thrown in to throw people off track, but they don’t give enough detail to support the favored/official solution.
Bob Murphy pointed out an interesting outcome from PD. If we limit it to a fixed number of plays, say 1000, then the NE is for each player to defect every time. This is from a similar recursive analysis. On the 1000th play you might as well defect, as your payoff will be bigger that way whatever the other player does. With the last play now fixed, we have a 999 play game. The same argument now applies to the 999th play, and so on back to play 1. Real people often cooperate for extended periods and do much better than the NE would give them.
The NE only gives the best outcome if the players are all informed and rational. Against such a player, a “normal” play of offering cooperation at some point will fail, because the other player will always defect.
Only if both players are not apparently rational can you arrive at the much bigger payoff.
This was a great series of posts, Steve. I think because I’m used to voting literature that includes a (small) cost of voting, I’ve never even thought of this quirk.
Anyway, for those who like this stuff, in general there is a counterintuitive thing in relying “backwards induction” to reach an answer that sounds crazy. This is a separate issue from the quirk of mass voting leading to a paralysis among the voters.
For example check out my article here on this stuff.
Steve, FWIW, I would’ve dismissed such a student too, and then I hope would’ve recovered in the class period. I almost fell out of my chair last night when reading Xan in your previous post bringing this up.
Oh one last thing: It’s pretty common in multi-stage games for there to be “weird” outcomes supported in a Nash equilibrium, but not a subgame perfect Nash equilibrium. E.g. if someone walks into a bank with a bomb attached to him, and threatens to blow everybody up unless they give him money. That could be a Nash equilibrium but not subgame perfect, assuming the guy values his life more than the money (i.e. not Jack Benny).
But to repeat, that’s not what’s going on here. As Xan pointed out, having the pirates do “weird” things even in the third last stage is possible, in a Nash equilibrium at that point. So the “weird” game play isn’t due to non-optimal threats off-the-equilibrium path, as in my bank robber example.
@Roger, you may very well be right about the psychology of real people. But this is not about real people. It is about perfectly rational actors.
You keep attacking the standard solution because we think about what, e.g., P8 would do and try to predict it using rational analysis. But, you say, that is how you analyzed P1 through P7 too and, if P8 is making an offer, P1 through P7 must have died! Surely P8 must realize that all your rational analysis is just bollocks and try something different! Like what I think ought to do.
That reflects a misunderstanding of our method.
The reason we analyse what, e.g., P8 would rationally do is not that we assume that it will ever get to this point. Rather, we analyse that *because* that what P7 would have to analyse. Again, P7 wouldn’t perform this analysis because he expected the event, including his death, to come about. Rather P7 must perform this analysis in order to make an offer that makes absolutely sure that the event does *not* happen.
And so on up the chain until P1 makes the 96-0-1-.. offer that prevents any of the other pirates ever making a proposal.
This is perfectly ordinary reasoning. To take a simple example: I do not want to be run over by a car. In order to make sure that I do not, I have to think of ways that would get me run over by a car–running out into the middle of busy highway, for example. I need to think about these things, not so that I can do them, but so that I know what to avoid.
Your argument here is the equivalent of shouting at me that I must be stupid for thinking about ways to get run over with a car, when that is exactly what I do not want.
Sub Specie (#77): Great comment.
@Harold #73
That n-times repeated PD results in total defection is a nice result I knew about.
What I was wondering about is the outcome in this scenario where players are not given n, but told that after every single PD there is a 1/(n-1) chance that the game will terminate.
This will still in practice be a finitely repeated PD (with excepted number of games being n), but the backward induction no longer works to predict total defection. After all, at the start of every PD game, each player would always except another n games, regardless of how games have already been played.
@Zazooba I think you’re misunderstanding the logical structure of the argument and the role that thinking about smaller numbers of pirates plays in the argument. Maybe the best way to clear things up would be to try to present proof as rigorously as possible.
Theorem 1: If 1 offers 96-0-1-0-1-0-1-0-1-0, then it will pass.
Proof of Theorem 1:
[Suppose for sake of contradiction that it is rational for 3, 5, 7, or 9 to reject 96-0-1-0-1-0-1-0-1-0. Now let us prove the following.
Theorem 2: If 2 offers 96-0-1-0-1-0-1-0-1, then it will pass.
Proof of Theorem 2:
[[Suppose for sake of contradiction that it is rational for 4, 6, 8, or 10 to reject 96-0-1-0-1-0-1-0-1. Now let us prove the following.
Theorem 3: If 3 offers 97-0-1-0-1-0-1-0, then it will pass.
Proof of Theorem 3:
[[[Suppose for sake of contradiction that it is rational for 5, 7, or 9 to reject 97-0-1-0-1-0-1-0. Now let us prove the following.
Theorem 4: If 4 offers 97-0-1-0-1-0-1, then it will pass.
Proof of Theorem 4:
[[[[Suppose for sake of contradiction that it is rational for 6, 8, or 10 to reject 97-0-1-0-1-0-1. Now let us prove the following.
Theorem 5: If 5 offers 98-0-1-0-1-0, then it will pass.
Proof of Theorem 5:
[[[[[Suppose for sake of contradiction that it is rational for 7 or 9 to reject 98-0-1-0-1-0. Now let us prove the following.
Theorem 6: If 6 offers 98-0-1-0-1, then it will pass.
Proof of Theorem 6:
[[[[[[Suppose for sake of contradiction that it is rational for 8 or 10 to reject 98-0-1-0-1. Now let us prove the following.
Theorem 7: If 7 offers 99-0-1-0, then it will pass.
Proof of Theorem 7:
[[[[[[[Suppose for sake of contradiction that it is rational for 9 to reject 99-0-1-0. Now let us prove the following.
Theorem 8: If 8 offers 99-0-1, then it will pass.
Proof of Theorem 8:
[[[[[[[[Suppose for sake of contradiction that it is rational for 10 to reject 99-0-1. Now let us prove the following.
Theorem 9: If 9 offers 100-0, then it will pass.
Proof of Theorem 9:
[[[[[[[[[If 9 offers 100-0, then obviously he will vote for his own proposal, so it will pass.]]]]]]]]]
By Theorem 9, if 10 rejects 99-0-1, then 9 will offer 100-0 and it will pass. So it is irrational for 10 to reject 99-0-1. Thus if 8 offers 99-0-1, 8 will accept it and thus it will pass.]]]]]]]]
By Theorem 8, if 9 rejects 99-0-1-0, then 8 will offer 99-0-1 and it will pass. So it is irrational for 9 to reject 99-0-1-0. Thus if 7 offers 99-0-1-0, 9 will accept it and thus it will pass.]]]]]]]
By Theorem 7, if 8 or 10 reject 98-0-1-0-1, then 7 will offer 99-0-1-0 and it will pass. So it is irrational for 8 or 10 reject 98-0-1-0-1. Thus if 6 offers 98-0-1-0-1, 8 and 10 will accept it and thus it will pass.]]]]]]
By Theorem 6, if 7 or 9 reject 98-0-1-0-1-0, then 6 will offer 98-0-1-0-1 and it will pass. So it is irrational for 7 or 9 to reject 98-0-1-0-1-0. Thus if 5 offers 98-0-1-0-1-0, 7 and 9 will accept it and thus it will pass.]]]]]
By Theorem 5, if 6, 8, or 10 reject 97-0-1-0-1-0-1, then 5 will offer 98-0-1-0-1-0 and it will pass. So it is irrational for 6, 8, or 10 to reject 97-0-1-0-1-0-1. Thus if 4 offers 97-0-1-0-1-0-1, 6, 8, and 10 will accept it and thus it will pass.]]]]
By Theorem 4, if 5, 7, or 9 reject 97-0-1-0-1-0-1-0, then 4 will offer 97-0-1-0-1-0-1 and it will pass. Thus it is irrational for 5, 7, or 9 to reject 97-0-1-0-1-0-1-0. Thus if 3 offers 97-0-1-0-1-0-1-0, 5, 7, and 9 will accept it and thus it will pass.]]]
By Theorem 3, if 4, 6, 8, or 10 reject 96-0-1-0-1-0-1-0-1, then 3 will offer 97-0-1-0-1-0-1-0 and it will pass. Thus it is irrational for 4, 6, 8, or 10 to reject 96-0-1-0-1-0-1-0-1. Thus if 2 offers 96-0-1-0-1-0-1-0-1, 4, 6, 8, and 10 will accept it and thus it will pass.]]
By Theorem 2, if 3, 5, 7, or 9 to reject 96-0-1-0-1-0-1-0-1-0, then 2 offers 96-0-1-0-1-0-1-0-1. Thus it is irrational for 3, 5, 7, or 9 to reject 96-0-1-0-1-0-1-0-1-0. Thus if 1 offers 96-0-1-0-1-0-1-0-1-0, 3, 5, 7, and 9 will accept it and thus it will pass.]
By Theorem 1, 1 will offer 96-0-1-0-1-0-1-0-1-0 and it will pass. Whew!
Notice the structure of the proof. Never at any time in any of the subproofs is there any indication that someone had to do something irrational in order to get you to that point. That is deduced after the fact.
So for instance, theorem 8 is proven under the assumption that it would be rational to throw pirates 1-7 overboard, so the fact that we’ve gotten down to three pirates doesn’t give any indication that anyone’s acted irrationally. It’s only after we’ve proven theorem 8 and deduced what pirate 8 is going to do that we show, using theorem 8, that it would be irrational to throw the first seven pirates overboard.
Does that make sense? Please tell me if my notation was unclear.
@Sub Specie Aeternitatis Your comment 77 is a much simpler way to make the point I took so many lines to make in comment 80.
@Sub: You (and Steve) misunderstand rational actors. I am not the one ignoring the hypothetical, you are. When P8 makes his decision, the hypothetical scenario is that the previous 7 pirates were killed for following your logic. If he is rational, he uses the info to inform his decision. Your reasoning does not use that info at all.
You keep saying that P7 “makes absolutely sure” he will live, and yet the previous 6 pirates were similarly absolutely sure, and died anyway.
As Roystgnr pointed out, you are like the guy who says that the 2-box choice is the only rational answer to Newcomb’s paradox. While there is an argument for that, there are plenty of rational actors who prefer the 1-box choice. That is, it is rational to observe that those who make the 2-box choice are poorer, and make a different choice. Likewise, it is rational to observe that those who chose the Google-Landsburg logic are dead, and make a more reasonable choice.
Roger: Just to help me understand what you’re saying —-
1) Suppose there were never ten pirates in the first place. Suppose there are only two. Are you arguing against the 100-0 solution in that case?
2) Suppose there were never ten pirates in the first place. Suppose there are only three. Are you arguing against the 99-0-1 solution in that case?
@Roger “You keep saying that P7 “makes absolutely sure” he will live, and yet the previous 6 pirates were similarly absolutely sure, and died anyway.”
No, they didn’t. In the standard solution, P1 makes the standard offer and nobody dies. P2, P3, … P7 never actually make any choices. P1 merely thought about what choices they would make in a hypothetical situation and then makes an offer so as to avoid those hypothetical situations from happening.
@Roger “When P8 makes his decision, the hypothetical scenario is that the previous 7 pirates were killed for following your logic. If he is rational, he uses the info to inform his decision. Your reasoning does not use that info at all.”
In this game there is no new information. All the pirates know all they need to know–the rules of the games, the preferences of the other pirates, and that everybody is rational–from the beginning.
So P8 will act exactly as if the game had begun with just P8, P9, and P10 on board.
It is true that if P8 had actually seen P1 through P7 tossed overboard, P8 would realize that either the premises or the reasoning was faulty. And they would be. But that the reasoning would be wrong in a scenario which the reasoning says would never happen does not prove the reasoning wrong until the scenario actually happens.
If tomorrow morning I actually woke up and found that I could fly by flapping my arms, I would indeed conclude that my previous understanding of the laws of physics, anatomy, or most likely my perception of reality was wrong. But merely imagining that hypothetical does not prove them wrong.
@Steve
“1) Suppose there were never ten pirates in the first place. Suppose there are only two. Are you arguing against the 100-0 solution in that case?
2) Suppose there were never ten pirates in the first place. Suppose there are only three. Are you arguing against the 99-0-1 solution in that case?”
Not arguing with 1). P9 can guarantee passage simply by voting for 100-0.
Not sure if arguing with 2). P8 assumes all pirates will act rationally, and has no evidence to contradict that assumption.
Move on to the next step, though. You claim P7 will play the Google strategy because P7 assumes all pirates act rationally. But, if play passes to P8, we know that P7 got overboarded because at least one pirate did NOT act rationally. So, if play passes to P8, P8 knows that pirates do not always act rationally, so we don’t know what P8 will do in the 4-pirate scenario (at the very least, you haven’t proven what P8 will do in the 4-pirate scenario).
This unravels P7’s logic because P7 doesn’t know what P8 will play, and so P8 does not necessarily play the Google strategy (or, at the very least, you haven’t demonstrated what P8 will play).
For your logic to work, you have to demonstrate that P8 will act as if all pirates will act rationally in the future despite the fact that P8 knows some pirates have not acted rationally in the past.
For what it’s worth, I think Roger is making a valid point and that Sub Species’ #77 comment–notwithstanding Steve’s praise of it–is not doing it justice.
It is indeed weird in these backward induction chains of reasoning to say something like, “If we ever got to stage 5, then we know the outcome would be X,” when in fact if we “know” that, we will never have gotten to stage 5.
So I don’t think this is analogous to Sub Species’ traffic reasoning in #77.
The reasoning in these things is a bit like saying, “Suppose you see a car coming at you faster than light, what do you do?!” And then saying, “Well, I know from special relativity that such-and-such, so therefore my optimal response in this situation is to do so-and-so…”
@Steve
Maybe the flaw in my argument and Roger’s argument is that the strategies are assumed to be set in stone at the outset. Therefore, we know for certain that P8 will play 99-0-1 even if that strategy wouldn’t make sense if play ever did, in fact, pass to P8.
But, would P8’s strategy be subgame perfect given that it would be known at the time that not all pirates are rational?
Let me throw this out there, to try to illustrate what (I think) Roger is complaining about:
Suppose Alice and Bob are playing a prisoner’s dilemma that will be repeated exactly 1000 times, and this is common knowledge. Clearly, the unique Nash equilibrium is for them to defect every round.
But now suppose Alice comes to The Big Questions blog seeking advice, and asks, “I know I’m supposed to Defect in Round 1, but what if I cooperate? Won’t that send a signal to Bob that maybe he should cooperate too, so we can at least get in perhaps several hundred rounds of payments that are much higher than when we both defect?”
Subspecies (to applause from Steve) says, “No no Alice, if you Cooperate in Round 1, you will do worse than if you Defect, because you know that Bob is definitely going to Defect in every round.”
Alice says, “Well how do you know that?”
Subspecies explains, “Because Bob is rational, he knows you are rational, and he thinks this is common knowledge. So he knows that you will defect in Rounds 3 through 1000 regardless of his move in Round 2, therefore he shouldn’t cooperate in round 2 and earn less that round compared to what he gets if he Defects.”
Alice says, “But how can Bob be so sure I will defect in Rounds 3 through 1000, regardless of what he does in Round 2, since by that very same logic, he should never have seen me cooperate in Round 1?”
I don’t see that Subspecies has given a good answer to that question.
Someone above was also asking about a prisoner’s dilemma where the game continues with probability p each round (or ends with probability 1-p). I am virtually certain you can sustain a cooperation outcome with an appropriate strategy mix (like each player says, “Cooperate until I see a Defect, after which I always Defect”), at least for certain ranges of discount rates compared to the p. But, I can’t find anything with a quick google search to verify this.
But yes, the intuition is that when you evaluate the expected payoff from that strategy, there is a huge string of possible high payoffs from the bursts of cooperation, and so long as the discount on those isn’t too high compared to their probability of occurring, that can outweigh the one-shot gain from defecting when the other player is cooperating.
But you can’t do this with a finitely repeated PD, because then you can always go to the last round and know it’s better to Defect there, which means there is no gain from cooperation in the 2nd last round, etc. This logic can’t be deployed if there is no definitive last round, however.
@Sub Specie Aeternitatis I’m on your side of the argument, but I agree with Bob Murphy that you’re not giving quite enough credit to Roger and Zazooba. It’s not that you are engaging in hypothetical reasoning and your opponents are not. They’re just engaging in different hypothetical reasoning than you.
Let me summarize what their argument is (even though I think it’s fallacious). They’re saying that 7 will reason to himself, “If I offer 99-0-1-0 and it gets rejected, then 8 will conclude that an offer of 99-0-1 might similarly get rejected, so he’ll make an offer that’s different from 99-0-1. So I should take that into account and make an offer that’s different from 99-0-1-0.”
@SubSpecie
“This is perfectly ordinary reasoning. To take a simple example: I do not want to be run over by a car. In order to make sure that I do not, I have to think of ways that would get me run over by a car–running out into the middle of busy highway, for example. I need to think about these things, not so that I can do them, but so that I know what to avoid.
Your argument here is the equivalent of shouting at me that I must be stupid for thinking about ways to get run over with a car, when that is exactly what I do not want.”
I don’t agree that this argument addresses the issue being raised by Roger, Jonatan, and me. In your example, you have no contradictory assumptions about physics or any of the rules of the game. You are merely going through a series of scenarios thinking “well if this happens, I will do that, and I will or won’t get hit by a car.” We know that these are hypotheticals and may never actually happen. In your example, though, there are no incorrect assumptions.
We are pointing out that the Google answer is based on an incorrect assumption — the Google answer implausibly assumes that all pirates in all hypothetical situations will assume that all pirates always act rationally. This is not true for P8 because if play should hypothetically pass to P8, P8 will then know for a fact that not all pirates act rationally all the time. Therefore, the Google answer’s assumption is based on an incorrect assumption. You have to prove that the Google answer holds even when P8 makes his choice knowing that some pirates do not act rationally.
We understand that it is only hypothetical that play will actually pass to P8. Nevertheless, P7’s thinking depends on correctly anticipating P8’s behavior in that hypothetical, and P7 cannot do that with an incorrect assumption.
@Keshav
“Let me summarize what their argument is (even though I think it’s fallacious). They’re saying that 7 will reason to himself, “If I offer 99-0-1-0 and it gets rejected, then 8 will conclude that an offer of 99-0-1 might similarly get rejected, so he’ll make an offer that’s different from 99-0-1. So I should take that into account and make an offer that’s different from 99-0-1-0.”
Good summary.
To save the Google answer, you have to show that P8 will play 99-0-1 even though P8 knows pirates do not always act as the Google answer assumes they will.
The burden is on the people arguing in favor of the Google answer to do this. We are just pointing out that no one has done this (or even tried to), so the Google answer has not been proven.
Steve, thanks for pointing out the (nonstandard) proviso in the statement of the puzzle, which reads: “Each pirate devises a strategy that is designed to make himself as happy as possible, taking the other pirates’ strategies as given.” The puzzle is better without it, for with it, it describes a situation that can never arise in real life, where causality rules.
@Zazooba Have you taken a look at my comment 80? It shows where I think you’re going wrong.
—–> Aside to the Proprietor:
Any chance we could get MathJax https://www.mathjax.org/ here in the comments section?
MathJax allows you to enter formulas, tables, and other nifty stuff into your text and it will display in publication quality formatting to the reader. MathJax uses TeX format, which judging by recent comments, is probably already in the muscle memory of half of us.
With MathJax, not only would many things be more readable and pretty, it would also avoid some recent stylistic mishaps when I and others tried and failed to format things in ASCII text.
I recently added it to my blog and it’s great. All it required was adding a few lines to the blog template. And of course it is free.
@BobMurphy
“Let me throw this out there, to try to illustrate what (I think) Roger is complaining about:”
Interesting. Your argument suggests that the argument of Roger, Jonatan, and me may prove too much, because it would also “disprove” well-known results about the Prisoner’s Dilemma.
This also suggests that we can’t be the first ones to think of this.
But, is the repeated PD different? Assume Alice cooperates in the first round. Bob then formulates an expectation about Alices’s future behavior (say that Alice will follow a tit-for-tat strategy). I suspect that Bob would take advantage of Alice’s strategy to the point where Alice is better off always defecting. (I don’t know this for a fact, but I strongly suspect it is true. Otherwise, tit-for-tat would dominate defection and so would be a NE.)
The pirate problem has a different problem, though. It makes the unproven assumption that all pirates in all scenarios will assume that all pirates will act rationally in the future even when their is concrete evidence to the contrary. No one has even tried to prove this.
Daniel R. Grayson:
The puzzle is better without it, for with it, it describes a situation that can never arise in real life, where causality rules.
Well, one can imagine (perhaps just barely) a real-life game in which each pirate simultaneously submits a complete contingency plan, specifying what division he’d propose, and how he’d vote on any proposed division, should the occasions arise. They then get to examine each others’ contingency plans and revise their own, and the process repeats until it reaches a fixed point. The Nash equilibria (including the weird ones) are the possible fixed points.
Of course if your criterion is realism, then the notion of pirates agreeing on (and sticking to) this mechanism in the first place seems as unallowable as anything else you’re objecting to.
@ Keshav 90, Now that I’ve been emboldened by Bob Murphy 88 I would like to ask if you might be kind enough to respond again to a proposed non-google solution that you invalidated for me previously. I’ll make an argument that I believe brings up Roger’s critique as it was described by Bob Murphy in 88.
I propose 96-0-(one coin each to four of any of the weakest 8 pirates) is a solution. So Pirate Arlo proposes 96-0-1-0-1-0-1-0-0-1. You responded that James (the weakest pirate who would not get a coin in the standard 10 pirate google solution) will reject Arlo’s proposal because then he can be sure to accept Pirate Bob’s standard google of 96-0-1-0-1-0-1-0-1 and then he still gets the coin and in addition he gets to see Arlo go overboard.
Two Questions: 1) Why is it Rational for James to conclude that Bob is sure to offer the standard google solution given that Arlo did not offer it? 2)If the answer to my question 1 in some way involves James discounting Arlo’s offer as that of an irrational fool that would never be repeated, does not that answer require that the google solution is the only solution – a claim not yet proved?
@Keshav
Thanks for all the effort you put into your Comment 80. I am just getting around to parsing it.
“It’s only after we’ve proven theorem 8 and deduced what pirate 8 is going to do that we show, using theorem 8, that it would be irrational to throw the first seven pirates overboard.”
I’m not convinced. All our analysis occurs before the game starts, so when P1 plays, everyone assumes P1’s proposal will win, so if P1 is overboarded, the analysis has been proven wrong. Given that the analysis is known to be wrong, why would P8 behave as if it were correct (if play ever gets to P8)? So how can P1 know what P8 will do?
You have to prove P8 would play the Google answer GIVEN THE SITUATION P8 WOULD FIND HIMSELF IN if play ever got to him. But, if play ever got to P8, P8 would know with certainty that pirates sometimes act contrary to the Google answer. So it is a leap to assume P8 would play the Google strategy.
To put it another way, there seems to be no basis for your statement that:
“… theorem 8 is proven under the assumption that it would be rational to throw pirates 1-7 overboard, so the fact that we’ve gotten down to three pirates doesn’t give any indication that anyone’s acted irrationally”
This seems to be contradicted by the pre-game analysis that concludes rationality guarantees P1 does not get overboarded. In crude symbols: if [rational] then [P1 survives]; hence, if ~[P1 lives] then ~[rational]. So P8 should assume ~[rational].
@Bob Murphy #86 and @Keshav Srinivasan #90 say that I’m not doing justice to @Zazooba’s and @Roger’s arguments. While obviously I thought I was, I could be wrong and if @Bob and @Keshav say so, that is good evidence that I am. I shall endeavor to do better below.
@Bob Murphy #88
I think I understand your Alice/Bob PD hypothetical and would say that the difference is that in your hypo, Alice and Bob are not given as an undeniable premise that their opponent will always play with perfect rationality. So evidence that the other player has a particular type of irrationality (such an offering a cooperate in a 1000-long series of PDs) may be a perfectly valid basis for forming judgments about their future conduct and perhaps even cooperating in return.
The difference is that in our puzzle we *are* given that all the pirates are perfectly rational. Unless that leads to an internal contradiction–and it has not, as far as I can tell–we can safely rely on it.
@Bob Murphy #89 That was me asking about that stochastically finite PD. Thanks for your answer.
@Zazooba #92 “To save the Google answer, you have to show that P8 will play 99-0-1 even though P8 knows pirates do not always act as the Google answer assumes they will.”
No, P8 does not know that. P8 *would* know that if the game reached him. But we have good reason to think it won’t, so he would not know that.
@Keshav Srinivasan #90 “It’s not that you are engaging in hypothetical reasoning and your opponents are not. They’re just engaging in different hypothetical reasoning than you.”
Yes, all of us engage in hypothetical reasoning. A reasoner entering the modus ponens is allowed to will into temporary truth any proposition p of his choice. Then, using p and any other propositons at hand, the reasoner may be able to prove another proposition q. Upon leaving the modus ponens it is correct to claim that one has proven p=>q. It is fallacious to say that one has proven q.
For example, both sides use as proposition p “The game will reach Pirate 7.”
On our side, we use p to prove our q1 “Pirate 7 will offer 99-0-1-0.” And then we conclude that p=>q1 or “If the game reached Pirate 7, then Pirate 7 would offer 99-0-1-0.”
@Roger et al., use the p to prove their q2 “The pirates are not acting in accordance with the presumed rationality.” And then they conclude q2 “The pirates are not acting in accordance with the presumed rationality.”
The former is a valid use of the modus ponens. The latter is not. All they really proved was that “If the game reached Pirate 7, then the pirates are not acting in accordance with the presumed rationality.” The latter is certainly true, but it does not support their conclusion.
@SubSpecie:
“@Zazooba #92 “To save the Google answer, you have to show that P8 will play 99-0-1 even though P8 knows pirates do not always act as the Google answer assumes they will.”
“No, P8 does not know that. P8 *would* know that if the game reached him. But we have good reason to think it won’t, so he would not know that.”
Sounds like you are conceding the argument here — you are conceding that P8 would know the Google answer is wrong if play were to reach him. Therefore, we do not know how P8 would act if play reaches him. This is the crux of the matter because P1’s analysis depends on his assumption about how P8 will act should play reach him.
Here is an illustration of how P1’s strategy depends very much on P8’s assumed strategy even if play does not necessarily reach P8. Let’s assume that (for some unexplained reason) P8 commits to playing 80-20-20 if play reaches him. Then P9 and P10 will vote down the first seven proposals if they stick to the Google answer. Therefore, the first seven pirates will change their strategies, even if their changed strategies mean play will never reach P8.
@SubSpecie
“All they really proved was that “If the game reached Pirate 7, then the pirates are not acting in accordance with the presumed rationality.” The latter is certainly true, but it does not support their conclusion.”
We agree on a lot. We agree that if the game reaches P7 or P8, then the pirates are not acting according to the assumptions of the Google analysis. Now it is up to the defenders of the Google analysis to prove that P8 will follow the Google analysis even though he knows that the Google analysis is a bad predictor of pirate behavior.
Also, you say that “both sides use as proposition p “The game will reach Pirate 7”” I don’t assume this. I only ask the question, “how will P7 (or P8) act if the game reaches him.”
@Sub Specie Aeternitatis “@Roger et al., use the p to prove their q2 “The pirates are not acting in accordance with the presumed rationality.” And then they conclude q2 “The pirates are not acting in accordance with the presumed rationality.”” No, I don’t think that’s what they’re doing.
They’re saying that 7 reasons as follows: “If my offer of 99-0-1-0 fails and the game reaches player 8, then player 8 will conclude that pirates are not acting according to presumed rationality, and thus pirate 8 will make a different decision then 99-0-1. So I should take that into account when formulating my offer.”
They’re not at all trying to reach the statement “the pirates are not acting according to presumed rationality” outside of an if-then statement.
I believe Steve and Keshav’s arguments about rational actors — which is the standard position — has a gaping hole. Rational actors can and do form and honor deals. Rational actors win repeated prisoners’ dilemma problems. So stipulating that they be rational EXCEPT that they don’t do the things rational actors do means that you are using the word as jargon, with a different meaning. So let’s call it kational. I am willing to believe kational actors behave as Keshav and Steve would have them do; they defined kational to work that way. But arguments about kationality do not answer Roger’s or Bob Murphy’s claims about rationality.
Twice in one day we agree Bob Murphy!
@Steve (#97) Can you please explain to me why, when the pirates submit their contingency plans for the first time, anyone would possibly plan to vote for something other than their preferred outcome? I think the first contingency plan of each pirate would definitely assign all votes to preferred outcomes. And that would rule out all the weird Nash equilibria as fixed points of the process.
@Ken B “Rational actors can and do form and honor deals.” Why in the world would a rational actor honor deals? Rational actors only care about pursuing their goals as best they can, and if their goals don’t include morality then moral considerations are irrelevant. If a pirate’s goals are only staying alive, maximizing gold, and maximizing number of people thrown overboard, what is the benefit of honoring a deal if it doesn’t benefit you ex-post?
@Keshav “how could anyone would possibly plan to vote for something other than their preferred outcome?”
Each pirate’s preferred outcome is that they get all 100 coins, and all more ferocious pirates are thrown overboard. Are you suggesting that’s the only outcome they’d vote for?
@106
Who said anything about morality?
People win repeated games, generous tit-for-tat is ESS, etc.
if your kational agents cannot form agreements then they lose out in the long term to Roger’s and Bob’s rational ones.
What was your criterion again? Self-interest?
@Mike H I just mean the preferred outcome of the vote, i.e. voting yes if you want the offer to succeed, and voting no if you want the offer to fail. I’m asking why anyone would plan to do something other than that when they submit their first contingency plans.
Keshav:
voting yes if you want the offer to succeed, and voting no if you want the offer to fail. I’m asking why anyone would plan to do something other than that when they submit their first contingency plans.
Of course, one could just as easily ask why they wouldn’t do exactly the opposite.
@Steve If you were in that situation and you were writing up your first contingency plan, what would you do? Would you plan to vote the same way as your preferred outcomes on every vote, or would you plan your votes in a different manner? I find it absolutely inconceivable that a rational agent would plan their votes in a different manner for their first contingency plan.
Keshav, “how could anyone would possibly plan to vote for something other than their preferred outcome?”
I thought this question was asked and answered earlier and the the answer was: Each pirate concludes that their one vote cannot change the outcome. If they are correct in this assumption then it is rational for them to be indifferent to how they themselves vote. Like you, I don’t think agree with this conclusion. I think my reasoning is that while it is true that one pirates vote cannot change the outcome of one given instance of voting by all pirates it is totally unclear to me that if each pirate considers all possible voting strategies for themselves and the other pirates and considers all possible voting outcomes using those strategies that the inevitable result is that each pirate will still conclude they can do no better than by simply adopting the random vote strategy.
Keshav: “I find it absolutely inconceivable” is not an argument. This is a truth that transcends this particular discussion.
@Capt. J Parker To be clear, I’m talking about the procedure Steve outlined in comment #97 for arriving at a fixed point, where everyone simultaneously submits an initial contingency plan and then the contingency plans are all revised until arriving at a fixed point. In that context, I’m claiming that a rational agent, when submitting their first contingency plan, would align their votes with their preferred outcomes.
@Harold #8 “I know of no evidence to suggest that people who believe their candidate will loose vote for the opposition anyway”
My dad did this once, but that’s more because Australia uses a preferential voting system, rather than because he felt his vote didn’t matter. So it’s off-topic, but interesting anyway:
His preferred candidate was A, from party X, who was sure to lose. Party Y’s candidate had always been the well-loved B. One year, Y decided to dump B him in favor of the young and energetic C.
B decided to run as an independent.
My dad figured that the vote for Y would be split – a lot of people voting for B, with C as their second preference, and a lot for C, with B as their second preference. A would get the most votes, but not a majority, and very few second preferences.
My dad figured “If B gets fewer votes than C, he will be out. Then, the second preferences of his supporters will go to C, and C will win”. So, he voted for B instead of his preferred A.
In the end, B got more votes than C, so party Y lost the seat, C’s supporters’ second preference votes ran to B, and B won.
@Steve Let me put it this way: by aligning your votes with yorur preferred outcomes in your first contingency plan, you increase the probability that the fixed point that the group ultimately arrives at will be a Nash equilibrium that is more favorable to you than some of the other Nssh equilibria. Isn’t that a good enough reason note to vote against your preferred outcomes in your first contingency plan?
@Keshav If you’re thirsty to the point of desperation, and you’re in front of a broken vending machine full of delicious cool drinks, do you kick it?
It’s not clear to me what a rational agent will do. Whether they kick it or not makes no difference to their utility function. Either way, there’s no drink available.
If the machine wasn’t broken, it would be a different matter. Kicking the machine would be silly, because the machine might break, and then there’d be nothing to drink.
The pirates in the alternative scenario are in this situation. They individually hate the fact that all the other pirates are kicking the machine. However, some of them will kick it anyway.
@Mike H “The pirates in the alternative scenario are in this situation. They individually hate the fact that all the other pirates are kicking the machine. However, some of them will kick it anyway.” Actually, I think those weird scenarios may be impossible if we use the process Steve discussed in comment #97. I think which contingency plan you choose the first time has an effect on the path of revisions the group undertakes, and thus has an effect on which fixed point the group settles on.
” Actually, I think those weird scenarios may be impossible if we use the process Steve discussed in comment #97″
Sure, but it’s not guaranteed. And the process wasn’t specified in the puzzle.
So, suppose they already found themselves all planning to vote for 100-0-0-…-0. Would you agree it makes no difference to Pirate Ethel how she votes? And therefore being a perfectly rational, self-interested pirate, unable to collude, fully aware of how the others are voting, she has no particular incentive to change?
Keshav 114, would it be irrational for James to say he would vote yes to the standard 10 pirate google solution?
Your description seems to indicate that only possibilities where all pirates survive are possible, but that doesn’t seem right.
As I understand it, Nash equilibrium only entails that given a set of player-strategy pairs, no one player can improve their strategy if all other strategies are held equal. There is no requirement about the rationality of the given strategies in question.
Thus, if all pirates have the strategy to vote “No” on the first vote and “Yes” on the second regardless of coin division. No single pirate who knows this can change the final outcome of the game, and therefore cannot improve their strategy.
Keshav “by aligning your votes with yorur preferred outcomes in your first contingency plan, you increase the probability that the fixed point that the group ultimately arrives at will be a Nash equilibrium that is more favorable to you than some of the other Nssh equilibria.”
I like concept of increasing the probability of your preferred outcome but how are you sure aligning your votes with your preferred outcomes actually does this?
For example let’s assume that there are two and only two solutions – the Google standard and the Arlo takes all Nash. What is James preferred outcome and how does his aligning his vote increase the probability of reaching it?
Kyle Hughart: yes, you’ve described yet another nash equilibrium.
Just a few loose ends:
==> I am sometimes a “funny guy” in blog comments, but I hope it is clear that I think this series of posts was great.
==> If you want to see something explain Nash equilibrium (first in pure, then in mixed, strategies) from scratch, and put it in the historical context, try my article from earlier this year when Nash and his wife died in a car accident.
==> Someone above was asking whether my siding with Roger et al. meant that their reasoning overturned standard results regarding the Prisoner’s Dilemma. Well, yes and no. The definition of Nash equilibrium is what it is. It is indisputably true that you can’t sustain cooperation in any Nash equilibrium of the finitely repeated PD. However, if you then go on to say, “And therefore rational people will always Defect in a finitely repeated PD, at least if their rationality is common knowledge,” *then* things are not nearly so clear.
==> In the pirate game as Steve laid it out, I (and I think Roger et al.) have no problem saying that in the last round, the 9th pirate gets 100 coins and the 10th pirate gets 0.
I also have no problem saying that in the second-to-last round, pirates can all look ahead and be sure of what would happen in the last round, if their actions led to an overthrow and continuation of the game.
However, I *do* have a problem saying that in the third-last round, the pirates can all look ahead and be sure of what would happen in the second-last round. And this puts aside the (quite amazing) point that Steve’s student (and Xan) brought up.
Specifically, I think there is a problem in treating the game from the third-last round as if it started from scratch at that point, versus a longer game that arrived at that point.
The reason this distinction strikes here, as opposed to the second-last or last rounds, is that at this point, we have to rely on assumptions based on other people’s reasoning about other people’s reasoning. And since the logic we would be using at that point is clearly defective, then I think it would be IRrational to proceed in the way standard game theory suggests.
==> Last one: In real life, if you had to play a 1,000-round Prisoner’s Dilemma where the payoff from (C,C) was $10,000 each round but from (D, D) was $1 each round, do you honestly want to tell me you would start off with D? How irrational are you?
Sorry, let me amend my last point:
In real life, if you had to play a 1,000 round Prisoner’s Dilemma where (C,C) pays $10,000 and (D,D) pays $1, and you initially play “D” and see that the other guy plays “C” (and you earn $10,001 let’s say), are you going to think, “Idiot!” and keep playing “D,” since you’re sure that his strategy is always going to be a string of Ds from the next round until 1000? Maybe after seeing the other guy play a third or fourth “C” against your initial “Ds,” you wouldn’t even entertain a 10% probability that the other guy was trying to coax you into playing a bunch of (C,C) rounds, even though (admittedly) you could both be pretty sure this would break down near the end?
@Bob Murphy
“==> In the pirate game as Steve laid it out, I (and I think Roger et al.) have no problem saying that in the last round, the 9th pirate gets 100 coins and the 10th pirate gets 0.”
“I also have no problem saying that in the second-to-last round, pirates can all look ahead and be sure of what would happen in the last round, if their actions led to an overthrow and continuation of the game.”
I disagree. In the 10-pirate game, it hasn’t been proven what will happen in any of the rounds, including the last ones — we can’t even be sure that P10 will vote in favor of giving himself 100 coins, because in the second round it has been revealed that pirates are sometimes irrational. So, it is possible that P10 is irrational and will irrationally vote against his own plan.
Roger, Jonatan, and Zazooba are, at most, willing to agree that in a 2-pirate game (not a 10-pirate game) that the second-to-last pirate will propose 100-0 and vote for it.
@Sub Specie,
Absolutely no offense taken at the fairy comment. I know I deserved it, if not this time, then for some other asinine comment I have made. The discussion has certainly been educational.
You mentioned chess as a game of the first type, the deterministic, open type. I thought that was interesting, because there is definitely psychology in chess. For instance, if we are playing, and I offer you a chance to swap queens, normally you should take it, because it would gain you the initiative. You also have to consider, though, why I was offering the swap in the first place. If I’m willing to cede the initiative, it must be worth it to me, so maybe you shouldn’t take it.
Another example is the opening move. E4 is the optimal opening move. If all Chess players acted like Google pirates, no other opening move would be used by white. Yet, experienced chess players do sometimes use other opening moves. To make this even more interesting, the opening move may even depend on the opponent.
For the record, I agree that the student’s point (and Xan’s point) is valid and that there are a lot of unintuitive Nash Equilibria in the game. This is a common problem with NE.
This particular set of NE bothers me a bit because many of the pirates are indifferent about which strategy they choose, so the tiniest of incentives to behave more intuitively destroys the equilibrium. For instance, if there is the slightest possibility that a pirate’s vote will be decisive (perhaps due to trembles by the other pirates) the pirate will vote his self-interest.
@Bob Murphy #125
I recall reading about an anecdote about Alchian, this must have been written somewhere at the time of his dead, where he and a mathematician were playing a finitely repeated PD. Alchian (initially) kept defecting whilst the mathematician tried to coax him into cooperating and was wondering why “this idiot” wouldn’t cooperate as both would do better by cooperating.
Sidebar on psychology in chess:
In the 1997 match between Deep Blue and Kasparov, Deep Blue’s move #44 was completely random, due to a glitch. Although Kasparov won that game, the move unsettled him, because it looked to him as if Deep Blue could look farther into the game than he himself could. Kasparov subsequently underperformed in the second game and lost it.
Mike H 115. I was thinking of a two candidate vote, but your example demonstrates the value of an AV type system. As I understand it, he voted B (with A as second preference) to deny C his second preference vote in the case that B got more votes than C. In a first past the post election he would have had no motivation voting for his least prefered candidate. Tactical voting is another matter, which was why I was thinking two candidates.
Bob Murphy makes some interesting points on game theory and PD. The classic solution to the pirate game is subgame perfectr equilibrium. This measn that it represent a NE in all sub games – that is you can do the backward induction thingy.
From wiki on Subgame Perfect Equilibrium: “strategies exist which are superior to subgame perfect strategies, but which are not credible in the sense that a threat to carry them out will harm the player making the threat and prevent that combination of strategies.” The example they use on wikipeia is the game of chicken, where a rational player will rip off his steering wheel if he can do it first. Threats by the second rational player to rip off his own wheel are then not credible. In real life, I suggest that this fails because you do not know the second player is rational, thus the premise of game theory is innapplicable.
In the PD example, the strategy of offering cooperation is superior, but not credible in the terms of game theory. Is that right?
In real life, particularly if you stack up the rewards for cooperation as in your example, it is both superior and credible in a way that the pirates voting “Yes” in the 100-0-0-… case is not crdible.
Is there a flaw with game theory that fails to give credibility to credible results? Or is it that the assumptions of rationality and perfect information are wrong, as I suugest in the game of chicken?
First, let me confess to two errors (or at least two misdemeanor counts of inexactitude) in the last section of #100.
1. The Google-Answerers do not actually use the hypothetical “If the game reached Pirate 7.”
Rather, they use the similar, but significantly different, hypothetical “If there were only Pirates 7, 8, 9, and 10”
2. The Anti-Google-Answerers cannot actually use the hypothetical “If the game reached Pirate 7” to prove that “The pirates are not acting in accordance with the presumed rationality.”
Rather, they can use the this hypothetical to prove that one of the stated premises is wrong. Because, by the Google proof, it follows from the stated premises that the the game will not reach Pirate 7. So, by modus tollens, if the game reaches Pirate 7, the one of the stated premises is wrong.
@Zazooba #102 “We agree on a lot. We agree that if the game reaches P7 or P8, then the pirates are not acting according to the assumptions of the Google analysis.”
I don’t think that is right (as corrected above). It is correct to say that if the game reached P7, then one of the stated premises is wrong.
But that does not mean that the false stated premise is rationality. It could be any one of the stated premises. Perhaps pirates don’t like gold. Perhaps they are suicidal or just yearn for a good swim. Or, perhaps, they are not rational. Any one of these changes to the premises could result in the game reaching P7.
That is good reasoning in real life where we have not been given a list of undeniable premises by a cosmic puzzle maker. Hence we get all our premises from limited observation and so by necessity all of them have a chance of error. If we observe an event which, according to our premises, is impossible, we know that one of our premises must be false and try to figure out which one. (A good Bayesian will be given pause even by observing an event which, according to his premises, is possible but exceedingly unlikely.)
That is not good reasoning in a puzzle where everybody has been given a list of undeniable premises. If you can prove that these premises lead to a contradiction, then you have proven that the puzzle has no solution and that’s it. If you go down the path of trying to alter the premises so that the contradiction disappears, you are no longer trying to solve the original puzzle; you are trying to make a better one.
@Henri Hein #127 & #129
Thank you and the things you say about the psychology of chess are absolutely true. But they are only true because no chess player (not even Kasparov or Deep Blue) has yet fully solved chess (in the way, e.g., nim or checkers have been) which is what we mean by perfect rationality.
A perfectly rational chess player would have no psychology. In every situation, he’d just make one of the moves that guarantees victory. If there was no such move (and he knew that the opponent was also perfectly rational), he’d just resign.
I get the point about why all the pirates could end up voting yes when the first pirate proposes 100-0-0-0-0-0-0-0-0-0, but the problem states:
“Each pirate devises a strategy that is designed to make himself as happy as possible, taking the other pirates’ strategies as given.”
So doesn’t this preclude the first pirate from even making such a proposal, since he knows the other Pirates’ preferences and would assume each pirate will vote no, taking them as given? Surely the first pirate wouldn’t propose a resolution that gave himself a small chance of gaining 100 coins instead of a 100% chance of getting 96 coins because he is trying to make himself “as happy as possible.”
I’d be interested to hear where I’m wrong here.
Sub Specie 131,
If the game reaches P7 or P8 one of the assumptions of the Google analysis is wrong. One possible assumption that could be wrong is that 96-0-1-0-1-0-1-0-1-0 is the only solution. I think attempts have been made to use the assumed uniqueness of that solution to prove its uniqueness.
@Keshav Srinivasan #103
You are quite right. That is the reason I corrected my #100 in #131.
P7 does not consider the strong hypothetical “If I am tossed.” because P7 can prove that he will not be tossed. (Unless the number of coins in the pot was non-positive and P7 couldn’t supplement the pot from his own stash. If he could supplement and there were (-m) coins in the pot, he would offer [-m-1,0,1,0] and not be tossed.)
As the strong hypothetical is provably false, all propositions (and their negations) follow from it. That is the logical minefield on which commenters left and right are exploding.
So instead P7 considers the weaker hypothetical “If there were only P8, P9, and P10.” This is a weaker hypo because it follows from the strong hypo (“If P7 was tossed, there would be only P8, P9, and P10”).
The weaker hypo has the advantage of not being provably false, but still being sufficient to prove all that P7 needs to prove (i.e., P8 would propose [99,0,1]). Hence that is the hypo that P7 would use.
@Capt. J Parker #134
The Google-Proof does not make any assumptions beyond the given premises. If the game reached P8, the Google-Proof would indeed imply that one of the given premises (not necessarily rationality) is wrong.
Sub Specie (#136): Here, I think, you have absolutely pinpointed the locus of many people’s confusion. I hope all confused readers will take the time to digest this comment.
Re: @Henri Hein #127 & #129 & Sub Specie Æternitatis #133
This of course implies that every chess game between two perfect players (who each know that the other is perfect) would have at most one move.
If White’s opening position was winning (as most people believe), he’d just make a winning move and Black would resign.
If White’s opening position was not winning, White would always just resign on the first move.
A more interesting question is how a perfect player would act in a losing position if he did *not* know that the opponent was perfect. (In a winning position, he of course would still just make one of the winning moves.)
The answer must be that a perfect player in a losing position would make the move most likely to induce the other player to make a losing move. What the perfect player’s move would be depends on a complex model estimating the probability of the type of imperfect player he faces. That is, again, psychology.
Dave (#134):
So doesn’t this preclude the first pirate from even making such a proposal, since he knows the other Pirates’ preferences and would assume each pirate will vote no, taking them as given?
Pirates take each others’ strategies as given. If pirates B through J plan to vote yes on any plan whatsoever, then pirate A’s best option is to give himself all the coins.
#133. I presume if chess had been solved, every game would commence with a draw offer from white, which would be accepted. Either that or white would make a move and black would immediately resign.
#141. Posted the above before I saw Su Specie’s follow up.
@140
But why would a pirate take *that* strategy as given? It’s not consistent with the idea that the pirates devise strategies that make themselves “as happy as possible.” Pre-commiting to any answer on the first vote doesn’t maximize happiness given any proposal, so no pirate would do it.
I get that, in the event that a pirate’s vote isn’t decisive, he may vote against his best interest. But certainly the problem makes clear that each pirate’s *strategy* is designed to maximize happiness, and that rules out *strategies* that are suboptimal. The first pirate knows this, and so would not propose the 100-0-0-…
I’m sure I’m missing it, but it still sounds to me like you’re leaping from “Pirates can vote against their own self-interest when they don’t believe the vote will swing the outcome” to “pirates can commit to strategies they know may not maximize their happiness” which seems to contradict the problem statement.
@Harold 141
I like to believe that in the opening position White is in zugwang.
@Ken B #144 White would have to be in ZugZwang if the opening position is not a winning one.
Amendment to #139: I neglected the possibility of drawn situations, but the logic follows the same way.
Dave (143): Yes, you are still not getting it.\
Suppose each pirate commits to the following strategy: “On the first round, I will vote yes for any plan that is proposed, and if on the first round I am called upon to make a proposition, I will claim all the coins. I am the only pirate, I will take all the coins. On the second round … (fill in anything you like).”
Question: If you are the most ferocious pirate (A), can you make yourself any better off by deviating from this strategy (given that B,C,D,E,F,G,H,I,J all follow it)?
Question: If you are pirate B, can you make yourself any better off by deviating from this strategy (given that A,C,D,E,F,G,H,I,J) still all follow it)?
Likewise for pirates C,D,E,F,G,H,I,J.
If your answers are all “no”, then this is a Nash equilibrium. Each pirate is maximizing his own happiness, taking the others’ strategies as given.
If one of your answers is yes, please tell me which one so that I can talk you out of it.
@SubSpecie
I think you need to add one more assumption to your analysis, that “pirates always assume that, in the future, pirates will act as assumed by the Google game even when pirates have not acted as assumed by the Google game in the past.” Let’s call this hypothesis H*. This assumption is implicit in your reasoning, but by making it explicit, you make clear the issue at hand.
You come close to acknowledging H* when you say:
“So instead P7 considers the weaker hypothetical “If there were only P8, P9, and P10.” This is a weaker hypo because it follows from the strong hypo (“If P7 was tossed, there would be only P8, P9, and P10″).”
If you explicitly add H*, your argument would read:
“So instead P7 considers the weaker hypothetical “If there were only P8, P9, and P10.” [AND P8, P9, AND P10 ACT AS ASSUMED BY THE GOOGLE GAME.] This is a weaker hypo because it follows from the strong hypo (“If P7 was tossed, there would be only P8, P9, and P10″).”
If you explicitly add H*, then I agree your analysis is correct.
Also, I don’t agree with your argument that the “strong hypothetical” can be ignored. You say:
“P7 does not consider the strong hypothetical “If I am tossed.” because P7 can prove that he will not be tossed.”
P7 does, indeed, consider this hypothetical when he considers how P8 will act because P8 will act if and only if P7 is tossed. How P8 plays assuming P7 is tossed is absolutely critical to P7’s analysis.
You need something like H* to tell us how P8 will act in the zero-probability state where play reaches P8.
I agree with your statement that:
“As the strong hypothetical is provably false, all propositions (and their negations) follow from it. That is the logical minefield on which commenters left and right are exploding.”
Indeed, this is my argument in a nutshell. It is the Google analysis that explodes because the Google analysis depends on predicting behavior under a provably false hypothetical where “all propositions (and their negations) follow”.
Zazooba: No, Sub Specie does not need your extra hypothesis H* because he has already *proved* H*. He has proved first that if there is only P10, then P10 takes all the coins. Then he has proved that if there are only P9 and P10, P9 takes all the coins. He has then proved (given the hypothesis that people always vote for their most favored outcome) that if there are only P8,P9,P10, the coins are split 99-0-1. Having proved this, he can make use of it to analyze what happens when there are P7,P8,P9,P10. THere is no need for an extra hypothesis.
@Sub Specie Æternitatis 136
“P7 does not consider the strong hypothetical “If I am tossed.” because P7 can prove that he will not be tossed.
[…]
So instead P7 considers the weaker hypothetical “If there were only P8, P9, and P10.” This is a weaker hypo because it follows from the strong hypo (“If P7 was tossed, there would be only P8, P9, and P10″).”
This is a non-sequitur to me. If he doesn’t consider the real scenario that would occur (“P1-P7 have been thrown overboard, and there are now P8-P10 left”), why would he consider exactly this weaker scenario instead (“There are now P8-P10 left”)?
@Jonatan #149 “This is a non-sequitur to me. If he doesn’t consider the real scenario that would occur (“P1-P7 have been thrown overboard, and there are now P8-P10 left”), why would he consider exactly this weaker scenario instead (“There are now P8-P10 left”)?”
P1-P7 (or any pirate) being thrown overboard would be inconsistent with the premises from which you can prove that no pirate is thrown overboard.
So, if you assume both the premises and that a pirate had been thrown overboard, you are assuming a contradiction. And from a contradiction you can readily prove anything: that 2+2=5; that unicorns are real; that unicorns are real and that unicorns are not real; etc. Hence, assuming a contradiction leads you literally nowhere and everywhere.
The assumption that there are only three pirate (P8, P9, and P10) does not contradict the rules. Hence we can fruitfully reason from it.
@Steve(83): 1) No, 2 pirates, 100-0 solution is good.
2) 3 pirates, I disagreed with 99-0-1 yesterday.
With 10 pirates, I think that the most obvious and rational strategy for pirates 5-10 is to vote against any deal that gives them less than 10 coins each. Pirate 1 will realize this, and offer them 10 each. If not, then he dies, and pirate 2 gets a similar choice. Watching a couple of pirates get thrown overboard can be very convincing. Pirates 5-10 will all get their 10+ coins.
Following the Google-Landsburg solution only gets them 0 or 1 coin each. Explain to me how that is more rational than the above strategy, which gets them 10+ coins each?
This whole puzzle is based on pirates anticipating what others will do, but you refuse to consider what seems like the most obvious scenario to me.
Roger: Because you disagree already when there are three pirates, I think we should concentrate on that case. Arguing about ten pirates is unnecessarily confusing if we already disagree about the case of three.
So: We have Howard, Igor and James.
1) They all know that if Harold is thrown overboard, Igor will get 100 coins and James will get 0.
2) James prefers 1 coin to 0.
3) Therefore James prefers 99-0-1 to having Harold thrown overboard.
4) Therefore, if James always votes for what he prefers, he will vote “yes” on 99-0-1.
Do you disagree with any of these? If so, which is the first one you disagree with?
Steve @146
Thanks for taking the time to help walk me through it. I won’t blame you if you give up on me…
My problem in understanding is not whether you’ve found a Nash equilibrium — you have. It’s that the specific Nash equilibrium you’ve found violates the problem statement, as far as I can tell.
You write:
Suppose each pirate commits to the following strategy: “On the first round, I will vote yes for any plan that is proposed, and if on the first round I am called upon to make a proposition, I will claim all the coins. I am the only pirate, I will take all the coins. On the second round … (fill in anything you like).”
My response is:
No, you can’t suppose that each pirate commits to that strategy. The problem states:
“Each pirate devises a strategy that is designed to make himself as happy as possible” hence no pirate would commit to saying “yes” an unknown initial proposed division of gold. For any pirate (other than the first one who controls the initial proposal), pre-committing to saying “yes” to the first proposal is not a “strategy that is designed to make himself as happy as possible.”
I think what you’re saying is more or less: the pirates take the other strategies as given, and we can make those given strategies whatever we want, as long as we arrive at a Nash equilibrium at the end.
But the pirates are well aware of how many pirates there are, as well as the order of ferociousness, therefore they would never commit to saying yes to the first proposal, knowing that this may not make them “as happy as possible”. The problem is when you state “and if on the first round I am called upon to make a proposition.” There’s no “if” involved. The pirates know the order, and they know who is making the first proposal, so I don’t think the rest of what follows in your argument is valid (although I’m still certainly open to the possibility that I’m missing it!).
@SubSpecie:
“P1-P7 (or any pirate) being thrown overboard would be inconsistent with the premises from which you can prove that no pirate is thrown overboard.”
Agree strongly.
“So, if you assume both the premises and that a pirate had been thrown overboard, you are assuming a contradiction. And from a contradiction you can readily prove anything: that 2+2=5; that unicorns are real; that unicorns are real and that unicorns are not real; etc. Hence, assuming a contradiction leads you literally nowhere and everywhere.”
Agree strongly. You just proved that the Google solution is wrong, because this is exactly what the Google solution does. You are on our side.
“The assumption that there are only three pirate (P8, P9, and P10) does not contradict the rules. Hence we can fruitfully reason from it.”
Non-sequitur. Because the Google solution is self-contradictory, you are allowed to switch to analyzing another problem you like better? The fact remains that if play passes to P8, we know there were at least 7 deviations from the Google solution, and we can’t just wish that away and solve a different problem where that didn’t happen. Or … you can explicitly assume away the 7 deviations by invoking H*.
@SSA 145
No. It is zugzwang ONLY if it is a forced loss, not if it is a forced draw. By symmetry if it is a forced draw with W to move first then it is a forced draw with B to move first, hence there is no zugzwang as moving does no harm.
@Steve (#152)
“Do you disagree with any of these?”
“1) They all know that if Harold is thrown overboard, Igor will get 100 coins and James will get 0.”
Disagree. If Harold is thrown overboard, at least 1 pirate has deviated from the Google strategy. In a world where pirates sometimes deviate from the Google strategy, how do you know Igor will get 100 coins and James will get 0.?
“2) James prefers 1 coin to 0.”
Disagree. If Harold is thrown overboard, at least 1 pirate has deviated from the Google strategy. In a world where pirates sometimes deviate from the Google strategy, how do you know James prefers 1 coin to 0?
“3) Therefore James prefers 99-0-1 to having Harold thrown overboard.?
Disagree. If Harold is thrown overboard, at least 1 pirate has deviated from the Google strategy. In a world where pirates sometimes deviate from the Google strategy, how do you know James prefers 99-0-1 to having Harold thrown overboard?
“4) Therefore, if James always votes for what he prefers, he will vote “yes” on 99-0-1.”
Disagree. If Harold is thrown overboard, at least 1 pirate has deviated from the Google strategy. In a world where pirates sometimes deviate from the Google strategy, how do you know James always votes for what he prefers?
As SubSpecie pointed out, if Harold gets thrown overboard, all statements about the Google world become simultaneously true and false.
@Zazooba “Agree strongly [with if you assume both the premises and that a pirate had been thrown overboard, you are assuming a contradiction]. You just proved that the Google solution is wrong, because this is exactly what the Google solution does. You are on our side.”
How is that so? The Google solution states that P1 will offer [96,0,1,…], that this offer will be accepted, and nobody drowns. How is that solution inconsistent with no pirate drowning?
@Zazooba “Non-sequitur. Because the Google solution is self-contradictory, you are allowed to switch to analyzing another problem you like better? The fact remains that if play passes to P8, we know there were at least 7 deviations from the Google solution, and we can’t just wish that away and solve a different problem where that didn’t happen.”
Apart from the Google not being contradictory, you also err in denying that a reasoner is allowed to assume something. To the contrary, a reasoner is allowed to assume anything/examine any hypothetical/enter modus ponens on any proposition (these all mean the same thing).
It is true that some assumptions are useless (for example because they are irrelevant or contradictory) and others often prove fruitless.
But as long as, in the end, the proposition “if , then ” helps you solve the original problem, you are fine.
@Roger
“With 10 pirates, I think that the most obvious and rational strategy for pirates 5-10 is to vote against any deal that gives them less than 10 coins each. Pirate 1 will realize this, and offer them 10 each. If not, then he dies, and pirate 2 gets a similar choice. Watching a couple of pirates get thrown overboard can be very convincing. Pirates 5-10 will all get their 10+ coins.”
I haven’t analyzed this rigorously, but I think it highly likely that this proposed equilibrium could be broken by offering one of the hold-put pirates 11, the other hold-outs zero, and the remaining pirates 1.
Your solution requires coordination amongst the hold-out pirates which is not allowed.
@Ken B Thank you. I should stop talking about things I do not understand fully. But then I might as well be a Trappist.
Drat, part of my comment #156 was eaten by the system. The last sentence should read:
But as long as, in the end, the proposition “if ASSUMPTION, then ANYTHING_PROVEN_FROM_ASSUMPTION_AND_PREMISES” helps you solve the original problem, you are fine.
Zazooba:
1) I am attempting to prove by contradiction that Harold will not be thrown overboard. To that end, it is perfectly legitimate to start by asking what happens if Harold is thrown overboard, and to make use of things we’ve already proved.
2) “How do you know James prefers 1 coin to 0?” Because it is given in the problem that pirates prefer more coins to less.
3) “How do you know James prefers 99-0-1 to having Harold thrown overboard?” Because I took the trouble to read what was given in the problem. I suggest you do the same.
Zazooba: Try this:
I claim that 4 is an even number. Proof: If 4 were not even, then 4/2 would not be an integer. But 4/2 = 2, which is an integer, so 4 is even.
That is, of course, a perfectly good proof in the world we live in. In Zazooba-world, though, someone can come along and say “I don’t buy it. In a world where 4 is not even, how do we know 4/2 would not be an integer?”.
In other words, the standards you are attempting to apply would rule out all proofs by contradiction, and hence much of mathematics. Is that where you want to go?
I sense miscommunication. Let me try to rephrase the point I *think* Zazooba is making.
The idea is like the the claim that if
G: 96 0 1 0 1 0 1 0 1 0 is a solution then so should be
Z: 96 0 1 0 1 0 1 0 0 1
and many more
Here’s an argument. All the pirates see G works and know they will live. If Z is offered instead some of the guys now getting 0 will see if they let P9’s rejection cause this to be rejected and them to drown. So they will vote for Z.
This isn’t quite right as they might vote for something other than Z that keeps them alive. But the claim is this: the existence of Q as a NE implies the existence of other NE.
Does anyone deny that Z could be a NE, if some of the indifferents are voting yes for it?
One reason that there is such strong, continued, and vocal opposition to the Google solution [96,0,1…] is that it is seems so terribly unfair to most people’s senses.
All ten of those pirates raped, pillaged, and looted hard for that gold! Sure, their leader can get a little extra, but to keep almost all for himself is so unfair! Worse, what the others get depends entirely on an arbitrary criterion of whether their ordinal number is odd or even. That can’t be right!
A lot of people reject economics for the same reason.
I see it slightly differently from Zazooba I think.
I think that the 1-4 in Landsburg #152 are correct. But somewhere with more pirates it breaks down, and they don’t follow the ‘google logic’.
Here is what I think happens:
If we have 10 Æternitatis pirates:
Pirate 1 offers (96 0 1 0 1 0 1 0 1 0). It is accepted, with votes from himself and all the ones.
If we have 10 Zazooba pirates:
Pirate 1 know that (96 0 1 0 1 0 1 0 1 0) will be rejected. Therefore he offers something else, which is accepted.
Maybe it is useful to ask how another well-known game survives this problem with predicting behavior in zero-probability states of the world.
Perhaps I can talk myself into agreeing with Steve.
In the repeated Prisoner’s Dilemma, what if Alice deviates by playing Cooperate in the first round? This is a zero-probability event in equilibrium. If Alice deviates by cooperating in round 1, Bob simply continues his strategy of Defecting each round and will be made better off by Alice’s deviation, while Alice will be made worse off by her deviation. This reinforces the proposed equilibrium since the deviating party is punished and the non-deviating party’s strategy is unchanged.
Note that Bob doesn’t have to assume Alice will act rationally all the time because even if she acts irrationally, his strategy is the same, so the optimal strategy is robust to unexplained non-equilibrium moves. Implicitly perhaps, we are assuming the possibility that, by some fluke, someone will play off-equilibrium, so in a sense, Alice’s deviation isn’t really a zero-probability event.
That seems reasonable to me.
What about the pirate game? Does the same logic apply? There, we have ten pirates and deviations have complex effects. Take the case where P10 inexplicably votes no on P1’s proposal. P1 is then infinitely worse off, while P10 is punished by losing 1 coin, P2 gains a lot, and the other pirates’ outcomes change a little too.
So, is P1’s strategy robust to the deviation by P10, i.e., will P1’s strategy be unchanged if he thinks P10 might deviate? No! P1 will completely change his strategy in he thinks there is the slightest possibility P10 will deviate. So, the proposed equilibrium is not robust to off-equilibrium flukes in the same way that the repeated Prisoner’s Dilemma is. P1 cares very, very much about any possible deviations.
Foo. I wasn’t able to talk myself out of my objections. Sorry.
I think this is the motivation behind the Trembling Hand Perfect Equilibrium – it makes all possible scenarios non-zero probability so we can analyze them without the zero-probability problem conundrum.
@Steve(152): I reject (1). If James rejects the 99-0-1, then he is effectively offering a deal to Igor to split the coins. Igor may or may not accept the deal.
You could say that Igor would be irrational to accept the deal, but you are already saying that James is irrational anyway to offer the deal. So we are in this supposedly-irrational territory anyway.
I also reject (3). James prefers a chance at 50 coins over surely getting 1 lousy coin.
It is also irrational for Harold to even offer the 99-0-1, after having watched 7 other pirates die because James rejected similar offers. I know he is a ferocious pirate, but this whole puzzle is about anticipating what other pirates will do. If James votes to kill 7 straight times, then Harold should learn from that.
@Ken B(163): There is a flaw in your reasoning. 96 0 1 0 1 0 1 0 0 1 will get the votes of A,C,E,G, but not J, because J can vote no and do just as well with a 96 0 1 0 1 0 1 0 1 on the next round.
(I am assuming the logic of the proposed solution here. I don’t think that it will really get the votes of A,C,F,G for reasons explained elsewhere.)
@Steve
I am very sympathetic to your argument that my objections may prove too much. I am very concerned about this myself. I also find it very appealing to think of the Google logic as an exercise in proof-by-contradiction, so it is not valid to object that the argument leads to a contradiction (since that is what it is intended to do). So let me try to convince myself using your logic.
I get how mathematical proof-by-contradiction works. You assert the opposite of the statement we want to prove, hold all other truths constant, and prove that the assumed statement leads to a contradiction of some other statements known to be true. I’m on board with that.
Is that what happens with the pirates? Let’s see. Pirates are people who presumably make decisions based on all the information available, i.e., if you give them additional information, they will incorporate that information into their behavior. (Unless you invoke H*). When you assume that 4 is odd, the other mathematical principals do not try to incorporate that assumption because they are not pirates, they are mathematical principals. So maybe pirates are different.
So, how does Igor reason if Howard is thrown overboard? For Igor, it is a 100% incontrovertible true fact that Howard was thrown overboard. To use your proof-by-contradiction logic, Igor should reject anything that contradicts this fact. For Igor, Howard being thrown overboard is the accepted law of mathematics which disproves any hypothesis that contradicts it, so to use your approach, in Igor’s eyes, Howard’s overboarding disproves the Google solution, not the other way around. Hmmm.
(Invoking H* would make this go away.)
Let’s take a concrete example. Let’s assume Howard went overboard because James voted no to Howard’s 99-0-1 proposal. What should Igor conclude from this 100% incontrovertible true fact? Does he say “none of this is happening, so it doesn’t matter”? Does he say “I will ignore the implications of Igor’s vote and assume Igor will play according to the Google solution in the future because that is one of the assumptions of my world”? It seems you want to do the latter. This sounds to me like invoking H*.
If I am wrong, please tell me how Igor rationally gets to the Google solution despite Howard’s going overboard. (It is not adequate to say that Igor will never have to make that decision because Howard’s strategy depends on what Igor would do in that situation.)
I really don’t mean to be stubborn. I just can’t reason my way around this. Perhaps I am just irredeemable dense. That has happened.
I vote for keeping Harold on board. Howard, not sure yet.
@Sub Specie #164:
I have no doubt that is a part of the issue, at least with myself. Not so much in the “poor pirate” sense, as in “why would a greedy, bloodthirsty pirate accept this?” sense.
Zazooba (and others): I am itching to respond to this, but swamped with other commitments for the next few days. I’ll try hard to get to it, but if I don’t, it’s not because I’ve abandoned you.
@Steve
Oops.
In my last post, I misunderstood the number of pirates. I thought there were four and my point is clearest when there are four.
To be clear, I am thinking of the case where there are four pirates, Howard, Igor, James, and Zed.
Howard goes overboard, and we have to analyze Igor’s strategy of 99-0-1 in light of James’s vote against Howard’s previous proposal of 98-0-1-0. How does Igor reason to a strategy of 99-0-1?
(I think the points still apply to three pirates, but it is easier to see with four.)
Sorry about this. I wouldn’t blame you for calling it a day.
Thinking about this, I think the last pirate’s best strategy could be “give me 100% of the coins or I vote no”. True, he knows that if it comes down to two pirates, he gets nothing. However, he has the first pirate over a barrel, who would rather get no coins then get thrown overboard.
With just three pirates, this strategy would seem to ensure that the third pirate gets all the coins, as the first pirate would have to offer “0-0-100” not to get thrown over. Even if the first pirate tried to escape with a single coin by buying the second pirates vote with “1-99-0”, the second pirate would simple vote against it to pick up the last coin. So the first has to give it all to the third.
I’m not sure if the structure of the problem allows pirates to communicate their intentions like that, but it seems fair that the first pirate could at least infer that is the third pirate’s strategy. It is even somewhat intuitive, as the first pirate’s life is in the hands of the third pirate, it makes sense that the third pirate could be coercive in that manner
@Ben Kennedy: As you seem to concede in your comment, the last pirate cannot get away with demanding all 100, because the next-to-last pirate can be bought more cheaply.
With 10 pirates, no one pirate can make an unreasonable demand, because the first pirate only needs 4 other votes, and can ignore the last pirate (if he pleases 4 others).
@Roger – the next-to-last pirate can’t be bought, his optimal strategy is to vote “no” for everything. Even if offered “0-100-0”, he should vote no because then if the deal fails, he gets all the coins anyway and gets to see the first pirate tossed, which he likes. That’s why the third pirate can force a “0-0-100” offer from the first.
The google answer is that the first pirate tells the third pirate “Take the crappy deal of 1 coin because otherwise you get nothing”. I’m proposing what really would happen is the third pirate would tell the first “Take the crappy deal of 0 coins or you DIE”. I think the third pirate deal would take precedence because his loss (getting 0 coins) is a lot worse that the first pirate’s loss (getting tossed)
STEVE READ THIS FIRST
I think I understand now what you are getting at.
An equilibrium is defined to be a set of strategies by all players in all states of the world (zero-probability as well as non-zero probability) such that no one player can improve his outcome taking the other strategies as given.
There is no reasoning, no evolution, no learning, and no requirement that the strategies make “sense”, or be plausible. It is simply a set of strategies where a single player cannot benefit from deviating from the equilibrium.
The Google solution is supported by off-equilibrium strategies that appear to make no sense if the pirates learn from events, but that is not a necessary condition of this particular equilibrium concept. James will not vote “no” because doing so would make James worse off if no other player changes their strategy. Igor does not try to base his decision on James’s deviation because Igor only submits a strategy assuming James does not deviate.
My criticisms then, go to the plausibility of the result rather than the validity of the results under these assumptions. (IIRC, a problem with the Nash Equilibrium concept is that you can get goofy results by assuming goofy off-equilibrium strategies such as having everyone severely punish any deviators.)
But, the bottom line is that I was wrong, and Steve was right.
152: I disagree with 1 (again). I object to Harold being thrown overboard when Howard is the pirate making the offer. Thanks Henri Hein 170.
With 3 pirates only we have the google solution (with votes)
99(Y)-0(N)-1(Y)
Pirate 2 is indifferent to Y or N since it does not change his outcome, so we could also have 99(Y)-0(Y)-1(Y). Lets say that all pirates vote Y if their vote does not improve their outcome.
Pirate 2 changing his vote makes no difference, so he will vote Y. This does not change the google solution, but does have a different pattern of voting than expected.
The google solution to a different offer is
100(Y)-0(N)-0(N)
The offer fails, and pirate 1 is thrown overboard.
This assumes pirates vote according to their preference. Given different strategies, we could end up with 100(Y)-0(Y)-0(Y).
This would arise if pirate 2 voted Y due to his strategy. Pirate 3 would then vote Y as his vote would not affect the outcome.
Given that strategies must be consistent with the objective of making all pirates as happy as possible -i.e. take 1 pirate overboard over nothing, and 1 coin over a pirate overboard, what does our new strategy of voting Y if your vote makes no difference have?
First thought, since pirate 2 gets nothing in either case, he must vote Y on the first round if these are the only two offers that could occur (given the vote Y if you do not improve by voting N strategy). Pirate 2 gets 0 in both cases, so he must vote Y to both.
There is the possibility hypothetically of pirate 1 offering something different, say 0-100-0. Pirate 2 would then feel foolish voting N, but he believes this offer is inconceivable given the knowledge and strategies of the others. Pirate 1 will offer 99-0-1 if pirate 2 has a strategy of vote with preference, or 100-0-0 if pirate 2 has a strategy of vote Y if your vote makes no difference to your own outcome.
Therefore, the three pirate problem has two solutions consistent with all the stated preferences.
Pirate 2 realises that he cannot get anything, so voting Y to the first round only affects the rewards of pirate 1 and 3.
I a going to submit this without checking it another time. There could be a glaring error.
Ken B. White cannot be in zugzwang, I think. First move must offer an advantage – or at least will not be a disadvantage, so white must either win every time or draw every time. After his move he will not be in a worse position.
@Harold, I think the google solution is incorrect because the last pirate’s happiness maximizing strategy could be “give me all the coins or I vote no”. The first pirate, who is aware of the third pirates strategy, then gives him all the coins.
The rejoinder would be the first pirate thinks the third is bluffing, and that if offered one coin he will be forced to take it due to the preference of having one coin over no coins. Of course, this means that the first pirate is effectively committing suicide, which is against his own preferences. That’s why I think he has to make the 0-0-100 offer
The second pirate is just a spectator – his strategy is to vote “no” for any deal offered by the first, even 0-100-0
@Ben Kennedy “Of course, this means that the first pirate is effectively committing suicide”
No, he isn’t. Once the first pirate makes the [99,0,1] offer he knows for a certainty that the third pirate will accept it, regardless of what he said before about what he will accept. This offer gives the third pirate one more coin than he’d get otherwise. So under the rules of the game he is forced to accept the offer.
The confusion is caused by your thinking of pirates as real people. Real people all the time live up to their promises, either because they feel a moral obligation or in order to signal to others that in the future they will also live up to their words.
The pirates are not real people. They are perfectly rational and care only about three things, in decreasing order of importance: not getting tossed, getting as much coin as possible, and seeing other pirates tossed. They care nothing for their honesty or their reputation.
Under these premises, the third pirate will *always* accept [99,0,1] and the first pirate is perfectly save in making that offer.
To deny this and struggle to find some way to make the pirates act more like real people you know is really just fighting the premise of the puzzle.
@Roger 168,
“@Ken B(163): There is a flaw in your reasoning. 96 0 1 0 1 0 1 0 0 1 will get the votes of A,C,E,G, but not J, because J can vote no and do just as well with a 96 0 1 0 1 0 1 0 1 on the next round.”
Why is it RATIONAL for J to assume that the 9 pirate Google solution will be offered in the next round given that the 10 pirate Google solution was not offered in the first round.
@Harold
I don’t claim white is, only that I woukd like him to be.
It is certainly a logical possibility that he is. It is not hard to set up endgame positions where to move is to lose.
@Capt: That’s a good point. Or maybe you are just humoring me. I am just reciting the logic of the Google-Landsburg solution.
I think that the next pirate will offer something like 0-10-20-30-40-0-0-0-0, if he is rational. This will be unsatisfactory to J again, but he can safely vote no and hope for a better deal on a subsequent round.
Roger,
If the 8 pirate google solution can arise if the alternate I suggested is rejected then when offered the one coin pirate 10 should vote yes. The order in which plans are voted matters.
@ Roger 183,
I am not just humoring you. I’d never dare do that given the brainpower exhibited by the crew that posts here. I believe there is a flaw in the google logic that claims anything other than 96-0-1-0-1-0-1-0-1-0 will be rejected. I know you do as well, although I think we differ as to what the actual solution(s) are. See below:
Following the method of Steve Landsberg 146
Suppose each pirate commits to the following strategy: “On the first round, I will vote yes if I’m offered a coin and no if I am not offered a coin. If on the first round I am called upon to make a proposition, I will offer 96-0-1-(one coin each to three of the 7 weakest pirates, selected randomly). On the second round I will vote yes if I’m offered a coin and no if I am not offered a coin. If on the second round I am called upon to make a proposition, I will offer 96-0-(one coin each to four of the 7 weakest pirates who were not offered a coin in the first round). On the third round … (fill in what you like).”
Question: If you are the most ferocious pirate (A), can you make yourself any better off by deviating from this strategy (given that B,C,D,E,F,G,H,I,J all follow it)?
Question: If you are pirate B, can you make yourself any better off by deviating from this strategy (given that A,C,D,E,F,G,H,I,J) still all follow it)?
Likewise for pirates C,D,E,F,G,H,I,J.
If your answers are all “no”, then this is a Nash equilibrium. Each pirate is maximizing his own happiness, taking the others’ strategies as given.
So 96-0-1-(one coin to each to any 3 of the 7 weakest pirates) are all NE solutions.
Having read the three “google” threads and all the comments therein, I have three things to say which might bring a different perspective to the issue:
1) Ouch
2) It is friday night
3) I need a glass of wine
There is wisdom in Robert’s approach, but if Steve is going to write on this again, I will take another stab at explaining my confusion.
First, I want to make it clear that I don’t claim any irrationality on the part of any of the pirates. On the contrary, if a pirate can see a strategy with an expected payout of more than 1gc, then it’s rational to reject the 1gc proposal.
I did learn in the course of the discussion the more precise meaning of ‘rational’ in the context of the puzzle (thanks, Sub Specie Æternitatis). The rub is whether we should consider the pirate game solved. I don’t think it is.
In order to explain why, I want to go back to the chess comparison. In chess, we have puzzles that drive towards an end, usually a win, in a few moves. The moves are predetermined by the rules of the puzzle: the winner (let’s say) must make the move that drives towards the win, and the loser’s moves are determined by the winner. This is just like the pirate-game with 1, 2 or 3 pirates.
Once we start moving back from the chess puzzle, in the sense of considering what moves led to it, the determination breaks. The winner cannot force the situation in the puzzle. The further we move away from the puzzle situation, in terms of moves, the more possibilities there are and the less likely it is that the situation in the puzzle occurs. The game opens up.
The question for me is: why does the pirate game not open up as we add pirates? With one pirate, there is only one possible proposal. A quick back-of-the-envelope calculation gives me 13 million different proposals when there are 10 pirates. That is more than 3 chess moves, about the limit of what the better players see out.
Most of these proposals will award more than 1gc to five of the pirates. It would be too complicated for the pirates to calculate all the possibilities, so they would have to go by probability and averages. Once Arlo is voted out (and if I am right, I believe he always will be), each time an MFP makes a proposal, they will award 0gc to themselves, since not being voted out is infinitely more valuable than 1 extra gc. Therefore, whatever proposal is eventually accepted will pay considerably better than 1gc.
@Capt. J Parker #181 “Why is it RATIONAL for J to assume that the 9 pirate Google solution will be offered in the next round given that the 10 pirate Google solution was not offered in the first round.”
Quite so. If J had observed anything other than the 10-pirate solution offered in the first round, he’d know that the premises are false and no longer rely on them. Of course, the premises are by definiton true, so J will see the 10-pirate solution offered.
@Capt. J Parker #185 “Suppose each pirate commits to the following strategy”
As our host has explained, the pirates cannot precommit to anything. Sure, they can say anything they want. But when it comes down to proposing and voting, they will always pick the extra coin, regardless of what they said.
@Henri Hein #187 “A quick back-of-the-envelope calculation gives me 13 million different proposals when there are 10 pirates.”
I get exactly 4,263,421,511,271 different proposals in the 10-pirate, 100-coin game when no coins can be added to the pot.
@Sub: “No, he isn’t. Once the first pirate makes the [99,0,1] offer he knows for a certainty that the third pirate will accept it, regardless of what he said before about what he will accept. This offer gives the third pirate one more coin than he’d get otherwise. So under the rules of the game he is forced to accept the offer.
The confusion is caused by your thinking of pirates as real people.”
The google solution is making that mistake by assuming that in the face of a 99-0-1 offer the third pirate would deviate from his chosen strategy. According to the rules, the first pirate strategy has to take the third pirate’s strategy as a given. The first pirate cannot simply pick the third pirates strategy, the third pirate can do whatever he wants to regardless of how rational the first pirate thinks it is.
In my scenario (give me everything or I vote no), the third pirate is essentially picking a strategy of rational irrationality to achieve his number 1 preference of getting all the coins. Sure he is pre-committing to doing something irrational at a micro-level, but the goal is to maximize happiness, not to maximize individual rational decisions.
To take one element out of the problem, suppose the third pirate’s strategy was to create a voting robot proxy that always votes “no” unless the third pirate gets all the coins. The third pirate even shares the source code, so the first pirate knows *exactly* what the robot will do. The third pirate sets up the robot and leaves the room. Can we agree that if the first pirate were in this scenario, knowing there was a robot who would cheerfully vote down anything but “0-0-100”, that the first pirate would indeed make that offer?
After seeing the error of my ways last night, I have been reviewing why I fell so grievously into error.
I think it is because I jumped to conclusions after reading the beginning of the problem and didn’t realize that the most important information is at the end. We are told at the end that the pirates are very special pirates — they are endowed with perfect foresight about how the other pirates will behave in the future. (We are told, a bit obliquely, that “each pirate … tak[es] the other pirates’ [past and future] strategies as given”.)
Pirates with perfect foresight don’t think about the past, what the past implies for the future, or how other pirates will reason and react to their choices. Each pirate knows exactly what will happen given her actions, and makes her choices accordingly. The “pirates” are basically robots and each player knows the programs of all the other pirate-robots and uses only that knowledge to program his own pirate-robot
I erred because I prematurely assumed that the pirates knew only the past and the structure of the game with certainty and had to use that knowledge to predict future pirate actions.
So this game is really a 1-period game, not a 10-period game. Each player simultaneously submits his robot-pirate program, and if the programs are such that no player can improve his outcome by changing his program, the set of programs is declared to be an equilibrium. Thinking of this as a 10-period game can lead one to mistakenly think that learning and updating is relevant to the game.
HT to commentators who saw this as a robot game before I did, e.g., Bob Kennedy.
@SubSpecie (180)
“The confusion is caused by your thinking of pirates as real people. … To deny this and struggle to find some way to make the pirates act more like real people you know is really just fighting the premise of the puzzle.”
Good point.
Indeed, the whole pirate story is a red-herring that lures the careless into error.
This isn’t a game about pirates dividing coins, it is a game about programmers programming robots (anthropomorphized as pirates) that will interact with other robots.
The pirate red-herring tempts us to think we will learn something about pirates from this game, but if we learn anything at all, it will be about how interacting robot programmers behave.
If you look at all the possible combinations of strategies that pirates could adopt then some will be NE and some will not.
But isn’t the point of the challenge not just to find any NE but to find the one that pirates will adopt to maximize their happiness based on their preferences relating to staying on-board, obtaining coins and getting others thrown overboard?
For the 2 least ferocious pirates the strategy of “vote in your self interest on the assumption that everyone else will as well” will sometimes do better than “always vote ‘yes’ and allocate the coins to your self when you get to allocate” and never do worse no matter what strategies others adopt so will always (according to the rules) be adopted by them in preference to it. (NOTE:The 2 least ferocious pirates always have it in their power to avoid being thrown overboard so have less to worry about than other pirates in choosing their strategies).
Given that the 2 weakest pirates will never adopt the “always vote ‘yes'” rule – then that particular NE can be ruled out. And I suspect that similar logic will eliminate all the others apart from the original one.
Rob Rawlings:
But isn’t the point of the challenge not just to find any NE but to find the one that pirates will adopt to maximize their happiness based on their preferences relating to staying on-board, obtaining coins and getting others thrown overboard?
By definition, any NE has this property, so it makes no sense to talk about “finding the one” that does.
Zazooba: The distinction you’re making between pirates and real people speaks to the question “Is NE a good predictor of behavior?”. The answer is: Yes in some situations, no in others.
I always stress to my students that it’s important to separate the two questions: “What is NE behavior in this problem?” and “Do we expect to observe NE behavior in the real-world implementation of this problem?”
Roger:
I reject (1). If James rejects the 99-0-1, then he is effectively offering a deal to Igor to split the coins. Igor may or may not accept the deal.
Okay, then, my understanding is that you somehow simultaneously believe all these things:
1) If Igor and James both vote against 99-0-1, then Harold will be thrown overboard, Igor will offer 100-0, and James will get 0.
2) Igor will vote against 99-0-1.
3) If James votes against 99-0-1, then James will get 0. (This is a logical consequence of 1) and 2)).
4) If Harold and James both vote for 99-0-1, then James will get 1.
5) Harold will vote for 99-0-1.
6) If James votes for 99-0-1, then James will get 1. (This is a logical consequence of 4) and 5)).
7) James prefers 1 to 0. (This is given in the problem.)
8) James always acts so as to maximize his happiness. (Given in the problem.)
It follows from 4), 6), 7), and 8) that James will vote for 99-0-1.
Nevertheless you beliefs:
9) James will vote against 99-0-1.
In other words, your beliefs contradict each other.
@Steve #193
Wikipedia’s article on NE says “If each player has chosen a strategy and no player can benefit by changing strategies while the other players keep theirs unchanged, then the current set of strategy choices and the corresponding payoffs constitutes a Nash equilibrium”
This is definitely true of a situation where all players adopt the strategy of “always vote ‘yes’ and allocate all the coins to your self when you get to choose”.
My point was that at least 2 players (the 2 weakest) will never adopt this strategy since there is always at least one (“vote in your self-interest”) that will never do worse and sometimes do better. And therefor this cannot be the solution required by the original challenge.
@Ben Kennedy #189
Now I see the source of our disagreement. You are assuming that our pirates can credibly pre-commit. Such pre-commitments may be a rational strategy in some games (e.g., throwing the steering wheel out the window in Chicken).
If such pre-commitments were possible in our games, I readily agree that this could change the outcome. On first reflection, I think it makes the game unsolvable.
But that is not the puzzle we have here. As our host has stated explicitly several times, our pirates cannot make credible pre-commitments. That means the Google solution holds.
Now that the comment traffic has died down, let me (re-)raise the one question about the problem which still troubles me.
I *do* now understand that the voting game has more than one Nash Equilibrium (NE). In fact, it has many. In fact, it has infinitely many. In fact, it has not merely countably infinite NEs, but the set of NEs has the cardinality of the continuum.
Hence, there is a sense in which all of these are possible. But does our rational inquiry have to end at this point?
For example, it occurs that all of the NEs are, in the technical sense, weak or unstable. That is, in each of them, there exists at least one player (to be precise, at least five players) who could change his strategy (in the game-theoretic sense) without harming his expected outcome.
Of all of these weak, unstable NEs, one set of NEs is distinguished as the least weak and unstable NEs. That set is the NEs where in the first vote, the odd pirates pick the pure strategy Aye and the even pirates pick any pure or mixed strategy. So can it not be said that these NEs (which correspond to the Google solution) are the equlibria towards which rational pirates would gravitate?
As I said, I am no expert in game theory and the proprietor (who has published on K Theory–a subject which even trying to read has left me dumbfounded on every occasion–and hence is much smarter than me) seems to say that the existence of multiple NEs is the end of discussion.
Did I doze of that day in Game Theory class when it was shown that trying to choose between NEs is a fool’s errand akin to trying to define a rich, mathematical system in which all propositions are decidable? I honestly do not know.
@Sub Specie AE 188
Quite so. If J had observed anything other than the 10-pirate solution offered in the first round, he’d know that the premises are false and no longer rely on them. Of course, the premises are by definiton true, so J will see the 10-pirate solution offered.
This is true if and only if there is one and only one solution to the 10 pirate game. The statement “there is one and only one solution to the 10 pirate game” is not one of the premises.
?@Capt. J Parker #185 “Suppose each pirate commits to the following strategy”
As our host has explained, the pirates cannot precommit to anything.
I attempted to use the exact same language and methodology as Dr. Landsberg did in in 147. Substitute ‘devises’ ‘for commits to’ if you like.
Sure, they can say anything they want. But when it comes down to proposing and voting, they will always pick the extra coin, regardless of what they said.
J is not picking an extra coin by rejecting say 96-0-1-0-1-0-1-0-0-1 because he risks 96-0-1-0-1-0-1-1-0 in the 9 pirate game.
J cannot make him self better off by deviating from my stated strategy in 185 taken as given that the other pirates continue to use it. 96-0-1-0-1-0-1-0-0-1 is an NE solution.
Oops, XML quotes don’t work here. I’m re-posting 199 with quotes added. Hope that’s OK
@Sub Specie AE 188
“Quite so. If J had observed anything other than the 10-pirate solution offered in the first round, he’d know that the premises are false and no longer rely on them. Of course, the premises are by definiton true, so J will see the 10-pirate solution offered.”
This is true if and only if there is one and only one solution to the 10 pirate game. The statement “there is one and only one solution to the 10 pirate game” is not one of the premises.
“@Capt. J Parker #185 “Suppose each pirate commits to the following strategy”
As our host has explained, the pirates cannot precommit to anything.”
I attempted to use the exact same language and methodology as Dr. Landsberg did in in 147. Substitute ‘devises’ ‘for commits to’ if you like.
“Sure, they can say anything they want. But when it comes down to proposing and voting, they will always pick the extra coin, regardless of what they said.”
J is not picking an extra coin by rejecting say 96-0-1-0-1-0-1-0-0-1 because he risks 96-0-1-0-1-0-1-1-0 in the 9 pirate game.
J cannot make him self better off by deviating from my stated strategy in 185 taken as given that the other pirates continue to use it. 96-0-1-0-1-0-1-0-0-1 is an NE solution.