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I did not get this correct the first time it was presented to me. any takers? I can refer you to the correct answer.

There are 10 pirates in a rowing boat. Their ship has just sunk but they managed to save 1000
gold doubloons. Being greedy bastards they each want all the loot for themselves but they are
also more interested in saving themselves than they are in the gold.

They each pick a number, from one to 10, out of a hat. Each person in turn starting with
number one, decides how to divvy up the loot among the pirates in the boat. They then vote. If
the majority of pirates approve of the allocation then the loot is divided accordingly, otherwise
that particular pirate is thrown overboard into the shark-invested sea. In the latter case, the
next pirate in line gets his chance at divvying up the loot. The same rules apply, and either the
division of the filthy lucre gets the majority vote or the unfortunate soul ends up in Davy
Jones's locker.

Question, how should the first pirate share out the spoils so as to both guarantee his survival
and get a decent piece of the action?

Remember that each pirate is completely selfish with respect to his life and the gold.

TR's helpful hint. invert, always invert
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The first pirate offers to divvy up the loot equally (100 each), plus pay an additional bonus (1 doubloon?) out of his own stake to the first five pirates that votes yes in favor of his plan (assuming his own 6th yes vote costs nothing). He'll stress that unless this plan is adopted immediately, a vicious cycle will be initiated until all but 1 pirate is dead (whether this is true or not, it makes a good story). I think he's got to offer an incentive based on behavior, and the minimum cost I can envision is a single doubloon to the first 5 critical votes. Tit-for-tat retaliation doesn't work in this case because you can't tit anybody back for tatting on you.

If the pirates vote against this plan, they are assured of getting 111 doubloons on average, less a 1/9 chance that they will be selected to go next, which they'll want to avoid, because the marginal return for drowning pirate number 2 is 111/8, and the marginal value will keep climbing with successive drownings. The infinite value of your life doesn't change. Is 111 over infinity more than 100 over infinity?

Todd

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How are ties broken? I'll assume that if there are 2n+1 pirates and there are n in favor and n against the plan, then the plan is passed. But if there are 2n pirates and there are n-1 in favor and n against, the planner gets thrown overboard.

The first pirate doesn't have to stress or threaten a french revolution if his vote isn't accepted, since the pirates are rational. All he has to do is to make an offer they can't refuse.

I believe it's not allowed for the offer to be contingent on any other pirate's behavior: you have to specify which pirate and how much.

One key fact that wasn't stated in the problem was that every pirate knows the numbers of all the other pirates. So pirate #1 knows who is going to be pirate #2.

The key to the first pirate's offer being accepted is that for exactly five of the 9 other pirates, the offer is better than they could otherwise hope to receive if they voted "no".

If there is one pirate (#10) left, he says "I'll take all the gold." #10 will vote against all plans because he knows his life is not in danger. So we won't bother trying to bribe him.
If there are two pirates left, #10 votes against #9's plan, and #9 is thrown overboard. So #9 knows his doom is sealed if it gets to #9, so he'll be happy to get out of this alive. It's in his best interest to vote FOR some plan.
If there are three pirates left, #9 will vote for any plan, #10 will vote against it, the plan passes. #8 can say "Everything for me" and win. So we won't bother trying to bribe him either; he'll be voting against every plan.
If there are 4 pirates left (#7-10), #7 loses, as only #9 is on his side.
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todd, you and i think alike. my original answer was similar, but there is a problem, why shouldn't the pirates just toss pirate number 1 into the ocean? with your allocation strategy the fewer the pirates the better. your allocation strategy does not prevent a vicious ccyle from developing.

tr
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jrr7, you are on the path to the correct answer, that is, working through it backwards from the perspective of the last pirates.

tr
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sorry - a tie isn't good enough....splash!
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So you are saying:

- the pirate who proposes the plan doesn't get to vote
- if there are 2n+1 pirates, and n vote for the plan and n vote against it, then the planner sleeps with the fishes
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A long attempt:

This first part assumes that every pirate, including the proposer, gets a vote.

As has been said, #10 will vote against everything, because he might be the last alive and get everything.

#9 knows he'll lose if he and #10 are the last, unless he gives #10 everything and #10 has mercy. So he'd accept any offer from #8, even if it gives him nothing, because it gives him certainty of his life and leaves him free from the whims of #10. He'd also accept an earlier offer, but only if it gives him something, since he knows he can always wait for #8, accept his offer, and take nothing but his life. In fact, since #9 has effective swing-vote power, he might take a chance and try to extract a lot in offers from 1-7.

#8 knows he's in control if it's 8-10 left, so he'll vote against everything.

#7 knows he's doomed if 7-10 are left. So he'll accept any offer from #6, and any earlier offer that gives him something, with the same swing-vote extraction possibilities as mentioned for #9.

Similarly, #6, #4, and #2 are in control and vote against everything, while #5 and #3 try to use their swing-vote power to extract things from earlier bidders, or suck it up and vote for a plan that gives them nothing if they'd be up next).

So #1 looks doomed. The only chance is to figure out the deal that #2 would give, and give it himself. That means give #2 most of it, give tokens to the odd numbers, and keep nothing for himself.

Why does #2 vote for this instead of killing off #1 and proposing it himself? No economic reason.

This next part assumes that the proposing pirate does not get a vote.

#10 votes against everything, since he gets it all if he's left last.

#9 knows he'll lose, so he votes for anything 7 proposes. But for people before 7, he'll want something in return for his support (swing-vote deal).

#8 knows he'll lose (1-1 tie will be bad), so he votes for anything, with the same swing-vote deal before 7.

#7 knows he's got it locked up, so he gives everything to himself and wins 2-1. He thus votes against anything before him.

#6 knows he's doomed on a 2-2 tie, so he votes for anything 5 says, and uses swing-vote power before 5.

#5 knows he needs 6, 8, and 9. So he needs to give 8 and 9 something. If he thinks he'll succeed, he'll vote against everyone before him to get to that point.

#4 knows he's doomed, so he votes for anything 3 says, and tries to use swing-vote power before that.

#3 needs 4, 6, 8, and 9, so he'll give 6, 8, and 9 something, and vote against 1 and 2 if he thinks he'll get away with it.

#2 knows he's doomed, so he votes for anything 1 says. Except that if #2 comes up, he can try to offer the same deal that #3 would propose, and hope that #3 will have mercy on him and accept it rather than killing #2 off and proposing the same deal himself. Still, #2 would accept a #1 offer since it guarantees his life rather than leaving it to the whim of #3.

#1 needs 2, 4, 6, 8, and 9, so he'll give 4, 6, 8, and 9 something. He doesn't need to give 2 anything. How 1, 4, 6, 8, and 9 split the money depends on how effective each is in using their swing-vote power. Since 9 has the most chances to escape eventual death, a deal would probably give 9 the most, 8 the next most, 6 the next most, 4 the next most, and 1 the least.

dan
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Let's pick up where we left off...

#9 can save his life by giving #10 all the gold, but he doesn't want that to happen. He'd prefer to get 1 gp than make it to step 9.

If there are three pirates left (Step 8), #9 will vote for any plan where he gets gold, #10 will only vote for a plan where #10 gets all the gold. No plan that #8 gives can induce both of them to vote for him. (If he gives all the gold to #10, #9 is indifferent about letting #8 live; #8 doesn't want to be in that situation.) #8 will do *ANYTHING* to avoid getting here, even if he doesn't get gold as a result.

So #7 can offer the plan: "999 for me, 0 for #8, 1 for #9, none for #10." #8 votes in favor of the plan because it lets him live. #9 will vote in favor of the plan because he at least gets some gold out of it. #10 will vote against it, but it won't amount to anything. So #7 hopes things will get to step 7! He can win if he gets there!

When #6 offers his plan, #7 can only be induced to vote for it if #7 gets all the gold (he knows he'd get 999 if the plan fails), but if any of #8, #9, #10 get more gold than they would in step 7, they'll vote for it! #8 gets 1, #9 gets 2, #10 gets 1. That's more than they'd wind up with under #7's plan, so they all vote for #6's plan. 3-1. So #6 hopes it will get to that step; he'll get 996 gold.

If it ever gets to step 5, #6 will vote against any plan (so stiff him). #7 will vote for a plan if he gets any gold out of it (since he knows #6 intends to stiff him). The rest will only vote for a plan if they get more gold than they would under the plan offered in the next round. So #5's plan has to give two of (#8, #9, and #10) more gold than they'd get under #7's plan. It's cheapest to make friends with #8 and #10 and stiff #9. So #7, #8, and #10 in favor, #6 and #9 against. #5 walks off with $995.

Step 4... we need a 4-2 vote. #4 already know #5 is voting against it, he smells the big payday coming right up. Of the remaining five, four have to be bribed. Here there is a difficulty in that to be most rational you have to arbitrarily pick either #8 or #10 to stiff, as they are the most expensive to bribe. So it is not possible for the other pirates to completely predict #4's plan in advance. 7gp have to be paid out in bribes, #4 would keep 993.

In step 3, we need a 4-3 vote. #4 is voting against it; of the remaining six, four have to be bribed. Here's where the previous uncertainty allows #3 to be stingier. #8 and #10 aren't sure whether they're going to get 0 or 3 in #4's plan I don't know their risk aversion -- would they be willing to gamble?

Consider #8 and #10. If either of them knows in advance what #4 is going to do, they will respond accordingly. But then you can only predict #10's actions if you know what #4 is thinking!

Let's make a chart.

If it's step # 10 9 8 7 6 5 4 3 2 1
then the
ending for
pirate #
1 * * * * * * * *
2 * * * * * * * *
3 * * * * * * * 99?
4 * * * * * * 993 0
5 * * * * * 995 0 1
6 * * * * 996 0 1 2
7 * * * 999 0 1 2 0
8 * * X 0 1 2 0/3 2?
9 * 0 <--- 1 2 0 1 0
10 1000 1000 <--- 0 1 2 0/3 2?

* means died before this round,
X means died this round (no plan could possibly win),
numbers represent the amount of gold he takes home
when all is said and done (and show that he lived).

I'll say 2 and 2, assume that #4 will flip a coin, and #8 and #10 are risk neutral and can be induced to vote "yes" if they are bribed with more than 1.5 gold pieces.

Continuing the chart...

If it's step # 10 9 8 7 6 5 4 3 2 1
then the
ending for
pirate #
1 * * * * * * * * * 991
2 * * * * * * * * 992 0
3 * * * * * * * 993 0 1
4 * * * * * * 993 0 1 2
5 * * * * * 995 0 1 2 0
6 * * * * 996 0 1 2 3? 2
7 * * * 999 0 1 2 0 1 0
8 * * X 0 1 2 0/3 2 0? 2
9 * 0 <--- 1 2 0 1 0 1 0
10 1000 1000 <--- 0 1 2 0/3 2 0? 2
none 1-0 1-1 2-1 3-1 3-2 4-2 4-3 5-3 5-4

I don't like the ending of my argument because of all the unknowns and assumptions that are necessary, but I don't see how to get past the arbitrary choice in #4 and onward.

Think of "I stay alive" as being worth 1001 gold. "For what doth it profit a pirate if he gain all the gold but loseth his life".
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As has been said, #10 will vote against everything, because he might be the last alive and get everything.

Not necessarily. If it so happens that #n+1's plan will be accepted and leave #10 with zero, then #10 will vote in favor of #n's plan as long as it leaves #10 with some money.

Uhh...your logic is wrong. If there are 9 pirates there, there are plenty of plans that pirate 2 could use that would be accepted.
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jrr,

I think we mostly agree, but we've had the same problems trying to figure out how to predict the behavior of the higher numbers in the early rounds and come to different conclusions, especially about allocations of the gold.

I guess I figure if I'm #3, and #1 offers me 1, I would probably turn it down if I have even a 100:1 chance of getting that 993 if it comes to me. Same way down the line until we get to #7. So it seems like #7 should get a lot of dough.

I'm also having trouble with the judgment calls. If #9 is up and gives #10 all the gold, does #9 live? Why? Is there some assumption that given pareto-equal outcomes, a person will live rather than die? Because if the risk of having to deal with #10's whims is eliminated, I think it changes the answer.

Interesting problem, solasis.

dan
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Not necessarily. If it so happens that #n+1's plan will be accepted and leave #10 with zero, then #10 will vote in favor of #n's plan as long as it leaves #10 with some money.

I see what you're saying, but I think you have to know how risk averse the players are. Given the uncertainties, I'm not sure that you can ever say with certainty that n+1's plan will definitely be accepted. For instance, what if one of the high-number players decides to give up a certain but negligible payoff in the hope that it will come down to them and they'll get it all? They might be willing to come out a little less rich because they want the chance of the big score.

The payoffs are high enough that one's risk aversion plays a big role, unless you assume that everyone's risk neutral. I didn't see that assumption in the problem.

Another thing: if they collude, they can reduce some of these risks. But I'm not sure what the rigid solution to the collusion negotiations might be.

dan
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I guess I figure if I'm #3, and #1 offers me 1, I would probably turn it down if I have even a 100:1 chance of getting that 993 if it comes to me. Same way down the line until we get to #7. So it seems like #7 should get a lot of dough.
But it's not about "chance" or "probability" or "luck" at all. Each pirate is perfectly rational and acts so as to maximize his own payout. And all the pirates know that the other pirates are rational, too.

If you turn down an offer that gives you gold, the next player knows you are not acting rationally, and so won't bother to give you any the next time either.

I'm also having trouble with the judgment calls. If #9 is up and gives #10 all the gold, does #9 live? Why? Is there some assumption that given pareto-equal outcomes, a person will live rather than die? Because if the risk of having to deal with #10's whims is eliminated, I think it changes the answer.
This information is not available to us or the pirates, so we must reason without it. Suffice it to say that #9 will never get the chance to put forth his plan, because the pirates are rational. We're really just considering "What's the case if we started out with only 2 pirates in the boat" and then slowly increasing the number of pirates.

A pirate will never willingly put himself in a situation where his life or death is determined by someone else's whim. Whim or no whim does not change the answer.
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For instance, what if one of the high-number players decides to give up a certain but negligible payoff in the hope that it will come down to them and they'll get it all?

Because the pirates are not playing the lottery -- they are acting rationally. I think that idea is subsumed by "risk neutral".

Nothing forces a pirate to honor the results of the collusion negotiations.
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Because the pirates are not playing the lottery -- they are acting rationally. I think that idea is subsumed by "risk neutral".

As I said before, if you assume risk neutrality, a lot of my concerns go away.

But the original problem didn't say the pirates were risk neutral, and solasis often discuss concepts, especially on this board, that emphasize the point that making different decisions based on differing tolerances for risk is not inherently irrational.

I agree that failing to assume risk neutrality makes the problem much more complex. But then, I was always the one who tried to introduce private bank financing into Monopoly games as a kid.

dan
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jrr, I also have a question about one of your early steps:

So #7 can offer the plan: "999 for me, 0 for #8, 1 for #9, none for #10." #8 votes in favor of the plan because it lets him live. #9 will vote in favor of the plan because he at least gets some gold out of it. #10 will vote against it, but it won't amount to anything. So #7 hopes things will get to step 7! He can win if he gets there!

This situation comes up a lot in tax and estate planning. "If we (7 and 9) cooperate, then we can split $1000 in tax savings - if we don't cooperate, the IRS (10) gets all of it."

In my experience, the guy who has more leverage gets the bulk of the money. Here, 9 has more leverage because 7 is certain to die if 9 votes against, but 9 may not die if he and 10 are left.

So if 9 says to 7, "Give me $999 or I'm voting against you," wouldn't a rational 7 give it to 9 rather than dying to spite 9?

I'm probably making this way more complicated than it needs to be.

dan
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So if 9 says to 7, "Give me $999 or I'm voting against you," wouldn't a rational 7 give it to 9 rather than dying to spite 9?


9 *wouldn't* say that to 7, because 9 knows that if he says this, and #7 refuses, then 7 dies and 9 gets nothing. 9 will not spite himself; he knows that it's rational to only expect 1 and that he is liable to get nothing if he tries for anything more. Again, he's not playing the lottery here; he's not going to pass up a small payoff for a "chance" at a bigger payoff, because HE KNOWS THAT ALL THE OTHER PLAYERS ARE RATIONAL. He can deduce what everyone else must be thinking.

You're not comparing apples to apples when you're trying to measure leverage.
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HE KNOWS THAT ALL THE OTHER PLAYERS ARE RATIONAL.

Agreed. 9 knows that 7 is rational and will give 9 as much money as is necessary to buy 9's vote, because otherwise 7 will die.

Are you saying it would be rational in any circumstance for 7 to refuse 9's offer and die? If not, then how is it irrational for 9 to extract as much money as possible based on the absolute certainty that 7 won't refuse?
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I'm saying your hypothetical is outside the bounds of the problem.

All of the pirates are rational and make their decisions (what to offer, vote yes or no) without needing to communicate with the other players.

If it is rational for 9 to vote against 7, he will do so.

The problem is set up so that if at any time a player who is getting money (or who is not getting money but gets a chance to survive where he previously didn't have one) as a result of a deal, votes against the deal, if the other players are rational, the "defector" gets zilch (or killed).

7 doesn't "accept" or "refuse"; he just makes an offer.

Greater amounts are preferable to lesser amounts, but it is irrational to let your desire for a greater amount lead you vote in such a way as you could get zero.
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I'm saying your hypothetical is outside the bounds of the problem.

Fair enough.

In other words, there's no opportunity for negotiation. The proposing pirate has the advantage of defining the terms unilaterally, and once they're defined, the others either have to take it or leave it. So it makes sense to give the others as little as is necessary to induce the desired behavior.

I've been thinking like a lawyer too long. At least soon I'll get a break from that....

dan
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Solasis --

There are several versions of this puzzle roaming around the Internet. All of them say FIVE pirates.

Most of them specify that the pirates have three goals, in this order:
1. Greed
2. Throwing other pirates overboard
3. Self-preservation

which seems to be the OPPOSITE of the goals of the pirates in your problem:
1. Self-preservation
2. Greed

Your pirates' indifference to throwing people overboard makes the problem a little less well defined.
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you guys did a great job of it.

here is the solution, a discussion can follow later
pirate
1 2 3 4 5 6 7 8 9 10

1000 0 (and hope)
0 0 1000
1 1 0 998
0/2 2/0 1 0 997
0/2/2 2/0/2 2/2/0 1 0 995
two to any 2 of the 1st 4 1 0 995
two to any 3 of the 1st 5 1 0 993
two to any 3 of the 1st 6 1 0 993
two to any 4 of the first 7 1 0 991


i just find this a terrific deduction problem with respect to working the solution backwards to front.

tr
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sorry, here it is in table mode

      pirate
      1       2     3     4     5    6    7   8   9   10

      1000    0 (and hope)
      0       0  1000
      1       1     0   998
      0/2    2/0    1     0   997
      0/2/2 2/0/2 2/2/0   1     0  995
      two to any 2 of the 1st 4 1    0  995
      two to any 3 of the 1st 5      1    0 993
      two to any 3 of the 1st 6           1   0 993
      two to any 4 of the first 7             1   0  991
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Good job, jrr. As often happens with this type of puzzle, I got caught up in possibilities beyond the scope of the problem.

You said you found similar problems elsewhere on the internet - did the addition of the goal of wanting to drown other pirates change the answer?

dan
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Yep. It means that a pirate will vote against a proposal even if it's the only way he'd survive. So you have to increase the amount of the bribe you pay to the lower classes. The "cheapest pirate to bribe" changes.
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