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Pirates Dividing the Spoil
This is a classic logic problem. Briefly, the setup runs like this:
A band of cutthroats have skedaddled with some loot. Now comes the time to divide the take. The captain proposes a split of the treasure and the assembled pirates vote on it. If a majority of them approve of the split (or the vote is tied) the lucre is so divided. If a majority disapprove they make the captain walk the plank. Then the next highest ranked pirate takes over as captain and proposes his own split of the booty. How do they divide up the spoil?
Now, this discussion also comes with some basic assumptions:
- The pirates are greedy – every pirate will do his best to maximize his own share of the swag and deny it to others.
- The pirates are ruthless. There’s nary a one of them but would slit his captain prow to stern if it’d profit him a single guinea. Matter of fact these pirates are pretty well fed up with each other and would gladly heave each other over the side so long as it’s nice and legal according to the rules of this game.
- The pirates are shifty. Any single one of ’em would go back on his word quicker ‘n the wind shifts in a hurricane.
- The pirates are flawlessly logical. This might not be as sound an assumption as the previous three.
The standard version of the puzzle has five pirates in it (whom I’ll refer to as captain, 1st, 2nd, and 3rd mate, and the swabbie), with a hundred gold pieces. That’s just to make the logic easier to follow. The standard solution runs like this:
Suppose the first three ballots fail. That is, the captain, and the first and second mate are shark bait. That leaves a vote between the third mate and the swabbie. Because any vote that ends in a tie passes then the 3rd mate can force any split through he’d like. Since he’s greedy he’ll vote for 100 pieces for himself, and zero for the swabbie. Therefore any other spit benefits the swabbie more. All he needs is to get more than zero in one of the previous three ballots for him to vote in favor of one of them rather than end up with nothing at the hands of the 3rd mate.
Now suppose that only first two ballots fail instead. The 2nd mate is proposing a split for him, the 3rd mate, and the swabbie to vote on. The swabbie knows that if this ballot fails he gets the shaft, (and the 2nd mate understands this as well), so the second mate proposes the following split: 99 coins for him, none for the third mate, and one measly penny for the swabbie. The swabbie figures one pence is better than none, so he reluctantly allows this measure to pass. (He could offer one to the third mate and none to the swabbie, but then neither the swabbie nor the third mate will have a monetary incentive to agree, and both already harbor the opinion that the second mate would look better with a second smile.)
But that leaves the third mate in the lurch. Consequently, he can’t let the vote get to the third ballot. So we back up to where only the Captain is taking tea with Davy Jones, and the 1st mate is proposing the split. Knowing that an even vote favors his split he knows that he only has to keep one of his minions happy. Specifically the 3rd mate. The 2nd mate would just as soon have his guts for garters and take the majority of the loot itself. He could bribe the swabbie instead, but the swabbie will just as soon spill the first mate’s blood and take his miserable one coin from the 2nd mate instead. So the 1st mate proposes 99 for himself, one for the third mate, and the other two get boned.
Naturally, the other two don’t like this much. But the time to act isn’t in the second ballot, it’s in the first, when they can manage a slim majority to lock the other pirates out. Again, because exactly one coin is better than none, the captain can thus propose the following split: 98 coins for himself, zero for the 1st mate, 1 for the 2nd mate, 0 for the 3rd mate, and 1 for the swabbie. Because all five pirates are acting flawlessly logically the motion passes.
What’s Wrong With This Picture?
We all have an intuitive sense of how this would fail miserably should any real pirate captain try it. Shocked by the blatant unfairness the crew votes him off the ship and try again. Supposing it gets down to the swabbie he grudgingly allows the 3rd mate to keep all the gold, then stabs him while he’s sleeping. If the third mate doesn’t stab him first. Actually, a basic animal sense of fairness indicates that, if a perfectly logical split like this were proposed the whole endeavor would end in blood.
Okay, but assume the pirates are perfectly logical, and abiding by the rules of the game. Next time the ship makes port all but the captain are going to jump ship; presumably for other pirate ships which provide a more even split of the loot. Supposing every pirate captain and crew is this logical that won’t help them (getting either one or zero coins on every split), but maybe the transaction costs of hiring a new crew each voyage alone will cause the Captains to pay out a little bit better.
What if the Pirates Can Rely Even a Little Bit upon the Bloody Oaths Sworn by Their Crewmates?
Suppose the 2nd mate, tired of getting such a miserable rate of miles to the galleon, pulls aside0 the third mate and the swabbie. They agree each to vote against the first two proposals, keelhaul their officers, and split the remaining loot by thirds. Now, maybe the 2nd mate will betray them in the end and propose his traditional 99-0-1 split, but at the very least the 3rd mate and the swabbie are no worse off than before, and they still got to do some murderin’.
That last bit is interesting; simply by proposing an exchange of murders the final three pirates better their lot. The 2nd mate by most of the loot, the 3rd mate and the swabbie by getting to cut their officers to ribbons. If you relax the default problem’s prohibition from “No deals made” to “no deals made that involve promising even a single extra coin” and suddenly the whole calculation shifts.
First, the bloc executes everyone ranking just higher than the median pirate. Then, if there are enough pirates still available, a new bloc forms, and another round of executions begin. (Instead of our familiar crew of five, imagine a crew of nine. First, the lower five conspire against the upper four, then, when those four are pools of blood in the sea foam, the lower three conspire against the upper two.) In the end, no bloc smaller than three would be possible, since it’s never in the swabbie’s interest to kill the 3rd lowest ranking pirate and get zero coins instead of one.
Now Assume Pirates Aren’t Averse to a Little Bribery
Supposing the captain gets wind of this lower decks voting bloc. His life then hangs upon the question of whether he can bribe one of the members away. He only needs one — the 1st mate’s life is worth less than a bent penny in a Barbados barroom if the bloc has its way — and as such he’ll go along with whatever plan saves his yellow hide. The captain proposes a 98-0-0-0-2 split. Since he’s got the 1st mate’s vote in the bag, he’s relying on the swabbie’s greed to trump his bloodlust and betray the 2nd and 3rd mate’s voting bloc.
“Avast ye, ye scurvy bilge rat!” Exclaims the 2nd mate, by which I take him to mean “I can make a bid too.” He offers three doubloons to the swabbie, and a bidding war ensues. The final bids are, from the captain, a split of 0-0-0-0-100, and from the 2nd mate, first they murder the captain and the first officer, then they split the gold X-X-0-0-100, the “X”s denoting dead men. The swabbie ends up rich either way, but his cutlass is thirsty so as it turns out the captain can’t buy out the voting bloc that way.
What if he tries to bribe the 3rd mate as well? He proposes a split of 97-0-0-1-2, thus bettering both of the bottom pirate’s bottom line, at the expense of the first mutineer. On this bid he’d win, except a similar bidding war ensues. I’m not quite sure how this turns out. If the captain bribes one person and the mate bribes another (Captain: 0-0-0-100-0; Mate: X-X-0-0-100) the captain survives. It’s in the interests of the pirate the captain is bribing to vote with the captain and come away rich. On the other hand, if they bribe the same pirate (Captain: 0-0-0-100-0; Mate: X-X-0-100-0) then that pirate will take the option that first lets him shiver some timbers, and the captain dies. If you try some kind of even split in the bribes the captain dies as well.
He does have one option though; and that’s to bribe the 2nd mate. Suppose the captain offers a split of 0-0-100-0-0; the first mate (yellow as a jaundiced canary) votes along with, and the 2nd mate, greedy as ever, double-crosses his fellows for the gold. Three votes is enough to win the day and no pirates dance the jig o’ death.
Wait a Tick
“Avast and belay! Hoist ‘m by the yarrrdarrrm and ta Davy Jones with his pet goldfish!” By which I take the swabbie to mean “I say; when I agreed to this particular bit of mayhem I was promised that my lot would be bettered. But now, instead of the single gold piece I was promised if there were no collusion involved, I get none. What’s more, I lack even the joy of stretching the captain’s neck. If that’s the case then I, being a perfectly logical sort, will refuse to join this particular syndicate in the first place. What say you to that?”
By Jove, that intemperate blackguard may be on to something. If the ultimate logic of the voting bloc leaves two pirates high and dry and no pirates slaughtered then these ultimately logical pirates won’t enter into it. What if, though, the 1st mate organizes a voting bloc instead? (Bout time that cowardly sea dog bestirred himself.)
Here the odds are four to one against the captain, and whatever bribe he tries to make can be better than matched by the 1st mate, who can always add the sweetener “and you get to kill the captain.” What’s more, he can no longer bribe the top-ranking pirate in the voting bloc, since a prospective split of 0-100-0-0-0 still leaves three unhappy pirates and hence the captain’s blood running from the scuppers.
But then what happens? You’ve got the lower ranked pirates with every incentive to make a voting bloc again, only with the 1st mate in the position of the captain from before. He’s got to surrender all the ill-gotten wealth to keep his skin intact, and you end up with a final tally of X-0-100-0-0. Once again, the 2nd mate profits the entire amount, by the logic of bribing the only member of the voting bloc who’ll then have an incentive to keep you alive. But once again the swabbie is left with no incentive to join the original bloc of four against the captain.
Perhaps you could get by with a bloc of the three mates; 1st 2nd and third, leaving the captain and the swabbie on the outs. Then when the captain is gone the 1st mate is confronted by the traditional lower deck bloc of the 2nd mate, the 3rd mate, and the swabbie, and once again we’re left with a X-0-100-0-0 split so the 1st mate can delay his trip to the infernal regions.
One Final Wrinkle, I Promise
Suppose that the captain has one more line of motivation; if he’s going to go down then he’s going to take as many of the traitorous bilge rats with him. When a voting bloc arises he can structure his bribes so as to put paid to as many of the members as he can. It won’t save his dirty hide but he’ll at least have company with which to greet Davy Jones. When confronted by the bloc of three mates he offers to divide the loot fifty-fifty between the third mate and the swabbie. 0-0-0-50-50.
This is good news for the third mate; he’s been getting zero all along. He’d willingly betray his erstwhile companions for that much loot. “Belay that!” cries the 1st mate, and he proposes a X-0-0-50-50 split. This is even better for the latter two, so they assent to that. The captain goes over the side and they prepare to have a vote. But now that the 1st mate has promised to bribe the 3rd and the swabbie he’s got nothing with which to bribe the 2nd mate. The 2nd mate proposes a more bloodthirsty solution; x-x-0-50-50, which is still better for the last two. But when the 1st mate is overboard there’s nothing to stop the 3rd mate from proposing x-x-x-50-50, and so the captain has his revenge; two of the scurvy dogs who did him in will receive justice at the point of a cutlass.
However the perfectly logical mates can see where that train ends, and they’d prefer to get off at an earlier stop. The 2nd mate can choose to end it by supporting the 1st mate’s vote, preserving both their worthless lives or neither. This leads to the curious loot split of X-0-0-50-50, with the pirates voting X-yea-yea-nay-nay, those who are getting the entirety of the loot voting against, and those getting none of it voting for. The pirate captain is denied his ghostly revenge, however, this is as fair a split as is practicable in any of these scenarios.
To Sum Up
The final scorecard looks like this:
With no collusion: 98-0-1-0-1
With bloc votes but no bribery: X-X-99-0-1
With bloc votes and bribery: X-0-100-0-0
With bloc votes, bribery, and revenge: X-0-0-50-50
Conclusion:
If you’re the captains, sure as you value your hide keep your pirates from colluding!
If you’re the 1st mate, whatever you do make sure that bribery remains on the table.
If you’re the 2nd mate, steer your captain’s thoughts away from revenge
If you’re the third mate, whisper bloody thoughts of vengeance to your captain
If you’re the swabbie, sod the rules and knife ’em all whilst they sleep.
Question for all y’all: What ways of piratical murder have I failed to mention in here?
Published in Law
But the rules on the original problem are clear, right? HRFP is trying to extend the problem in a quality way (meaning, clear rules that we all can understand) that will create more enjoyment.
The purpose of an exercise is exercise. It’s just supposed to be enjoyable thinking. It isn’t thinking to give you knowledge on the best stock trades, or how to handle an unruly teenager.
Pirates, teenagers, same t’ing.
No, pirates are organized and goal-directed.
Teenagers are more like locusts.
I see the base problem. The four assumptions listed here are not the same as those stated in the Wikipedia version of the problem. *This matters*.
The author introduced a previously published problem, but one that I had never seen. It had clear assumptions. It was fun, and the result was surprising at first–a big part of the fun! It was also an excellent exercise of rational problem-solving skills. Which the world desperately needs more of.
Then he introduced four new spin-off problems. But on reflection I guess he is thinking the assumptions (the arbitrary rules of each new game) were ambiguous.
So, of course the new problems are not the same as the old one, given in Wikipedia.
Do I have this all right?
I’m not talking about the expansions. I mean the “basic assumptions” allow closure in the Wikipedia version, but not here, because they open the door to other possibilities.
I’m not crapping on the problem, or the post, or the author. I’m just pointing out that even the initial form in this post is intractable and can hardly be debated, nevermind the expansive assumptions, which I agree are fun. Just not tractable.
I don’t believe you have compared the four basic assumptions here to the ones given in Wikipedia. I hadn’t until a few comments ago.
Sorry I misunderstood. I didn’t read carefully.
I will have to go back and do that.
Do me a favor and actually explain the difference and how that leads to a different result.
No sweat. I’m not throwing shade — I just think you have a resistor of the wrong value here.
And I’ll admit that at first, I was thinking of an arbitrary number of pirates, rather than a handy five — it would be impossible for the swabbie or the last mate to know how many were involved until the vote came to them (or else they always knew how many [Carl Sagan] billions there were, and tuned their votes to this twee number, which then seems like a special case).
EDIT: Word count limits, so here’s the short version:
Too much? But consider the florid phrasing of this assumption fromt ghe OP:
I think this is an attempted (and charming) re-statement of the third condition in the Wikipedia article, namely:
Third, each pirate would prefer to throw another overboard, if all other results would otherwise be equal.
There are two differences: first, a bias based on all else being equal, vs “if it profit a single [coin]”. Slight, but noticeable. Second, Captain vs. “another”, so that perhaps the second mate simply throws the swabbie overboard. Not in the Wikipedia problem.
And once again, I’m out of space. Not like I have lock on this anyway. I’m tracking the things that threw me off, the non-deterministic aspect, and which popped clear once I read the original Wikipedia article. Except it’s not so clear. Looking forward to the truth table.
I don’t mean to go all Fermat on this. I can smell it! It’s a relaxation in deal-making and breaking. I think.
As to the first point, sure. It’s obscured by the colorful language in the original post, but I was assuming deals wouldn’t be made for the original problem, then changing that assumption and iterating the problem for later, slightly more realistic scenarios. Another round of proofreading or two and I hope I would have made that clear.
The last pirate never gets killed in any formulation. The only legal time to murder a pirate (in this puzzle) is when he’s made a proposal and had it rejected, and it’s always the highest ranking pirate (alive) who makes the proposal. The swabbie would have to propose a deal to a set of one pirates, being himself, and then reject it, knowing that the price of rejection out of hand is suicide.
If any proposal passes the loot is split and the game ends. That is also something I never stated explicitly, and needed to do so. Switching votes is perfectly acceptable before that point, but the pirate has got to have a good reason to do it.
First Hayward and Mirengoff. Now BDB and Rhody. The lockdowns and the political chaos really are driving us crazy.
The second sentence in my condition maps to the quoted condition from Wikipedia.
When I originally defined the problem in that top paragraph I specified that each senior pirate would become captain in turn after the last one was thrown overboard. Pretty much immediately upon starting to explain the logic I abandoned that descriptive convention because it’d make things entirely too confusing to track. The vestigial promotion bit is also something I should have stripped out entirely in proofreading as confusing the issue.
Hey now. Don’t go picking fights where there isn’t one.
Hank, I think I owe you an apology. Looks like I was more confused than anything else. I thought I saw a big flaw, but whenever I squint, it goes away. I may just be looking at something I cannot grasp all at once.
Maybe the original problem is flawed — maybe it’s just me. Hmmm. Place your bets.