## Pirate Democracy

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### Pirate Democracy

There is a ship of 100 perfectly logical pirates (who, nevertheless, continue to steal from people. Whatever). The pirates are ranked in order of fierceness. They divide their treasure of 1000 gold coins the following way:

The fiercest pirate proposes a scheme for dividing the treasure. The other pirates then vote on the scheme. If fewer than half* of the pirates approve the scheme, the fiercest pirate is thrown overboard, and the next fiercest pirate takes over.

You are the fiercest pirate of the 100. Propose a scheme where you keep as much of the treasure to yourself as possible without becoming shark bait.

* Your vote counts. So if 49 other pirates approve your scheme, you win. Also, you may only give out whole coins.

MrBawn

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This one is a classic

Gelsamel
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Just thinking aloud. Alright, not really aloud, because I'm typing my thoughts rather than causing pressure waves in the air that represent them, but you know what I mean.

Can we assume that each pirate will vote according to how much they get personally, with no regard for fairness to other pirates? I think we can. After all, they are logical, heartless pirates.

If we number the pirates in ascending order of fierceness (so I'm 100, and 1 is a total wimp), I think the best course of action is to come up with a plan that 1-49 will approve of by leaving 50-99 royally screwed. I'm probably about to do something really recursive. Should be interesting. I'll be back.
Teaspoon

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Location: Where you least expect me

Oh, this one. I don't particularly want to tackle it for 100, but I suspect the solution will look very similar to the one for 6.

It assumes very bizarre behavior on the part of the pirates in regard to loot preferences, but makes perfect sense in the context of robot pirates.
jwwells

Posts: 180
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I know it's not the best way, but I'm just writing my thoughts out...

My piratey plan:
Pirates of fiercenes 50-100 will recieve 20 gold pieces.

It's fair and you don't get much money but it absolutely guarantees the minimum vote of 50 for you to stay alive.

I think that saying "the pirates are all logical" is a bit vague. For example, my thinking is that if I gave pirate 50 19 gold pieces, 51-99 20, and then gave myself 21, I'd be sharkbait, because the logical pirates (especially number 50) would be pissed that all of them got less than me. On the other hand, you could argue that I'd still be good because the logical pirates would acknowledge my superior fierceness and permit my extra 1 gold. So thinking about it this way, what if I kept taking one more gold at a time and adding it to my cut? At what point would this become "illogical" to the pirates causing them to vote against me?

silverhammermba

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What is the pirates goal? To minimize the loss of life? Or to just get the most gold they can?

If they're all perfect logicians they'll see that the only outcome will be the second last pirate having all the gold and 98 pirates dead. They'd be best off just voting yes to the very first plan, no matter how the gold is distributed. Maybe though, the only real way to end it without someone dying is for the first pirate to give everything to the second last pirate.
I'm not lazy, I'm just getting in early for Christmas is all...

Ephphatha

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Location: Bathurst, NSW, Australia

### Re: Pirate Democracy

You get 951 coins, and every other even numbered person gets 1.
If you want to figure out why this works yourself, don't read the spoiler.
Note: This assumes that first the pirates want to stay alive, then given life, they want to maximize the amount of coins that they have, and finally, if the amount of coins you get is fixed, then you want to kill as many other pirates as possible

Spoiler:
Ok this is a hint if you're stuck. Try working it out for lower numbers of pirates, then use induction to see why it works for 100 pirates. If you still want to see the solution, look at the other spoiler. Also, I will call the fiercest pirate the captain and the 2nd fiercest pirate the 1st mate, so my proof is less clogged

Spoiler:
Ok this is the solution. I think we all will realize what will happen when there is only one pirate left (he obviously gets all the coins), if there are two pirates, then the captain will take all the coins because he still has half the votes. For three pirates this gets interesting, the captain can't count on the vote of the 1st mate (due to the fact that the 1st mate will get all the coins if the captain becomes shark bait anyway) so all the convince can do is convince (a.k.a. bribe) the lowest ranking pirate by giving him one coin and taking the other 999 (this is better for the lowest ranking pirate because if the captain gets thrown off then the lowest ranking pirate gets nothing. Now the captain has 2 votes for him, so he stays alive. For four people the best option for the captain is to give one coin to the 3rd ranking person and ensure that he gets two votes and stays alive. Similarly at each level the captain gives each pirate of the same fierceness (modulo 2)one coin, and give himself the rest. This ensures he gets at least half the votes and stays alive. Note: this does not work when the amount of people is more than twice the amount of coins there are, then it gets really interesting

I have a more interesting puzzle: After they found the 1000 coins (and decided the distribution), they now decide to downgrade the captain. The captain's position is decided by his generosity points. He gets +1 for every vote for him, and -1 for every vote against him. He is the lowest ranking person if he has 0 generosity points, and he gains a rank for every added generosity point. Now suppose that they find another 1000 coins, and the new captain decides the distribution according to the same rules as before. This cycle repeats infinitely many times (fine, the pirates are immortal, at least to the point of not dying by old age), but the value of the coins decreases by a factor of 0.9 every cycle. How can the captain maximize his coins value?
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
tomtom2357

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### Re:

@tomtom2357
Spoiler:
That pattern doesn't seem quite right though. Let's say a pirate of arbitrarily low fierceness is expecting to get 1 gold coin from almost any plan proposed before himself; why should he agree with it over any other proposed plan?
Scuttlemutt

Posts: 54
Joined: Wed Oct 27, 2010 9:08 pm UTC

### Re: Re:

Scuttlemutt wrote:@tomtom2357
Spoiler:
That pattern doesn't seem quite right though. Let's say a pirate of arbitrarily low fierceness is expecting to get 1 gold coin from almost any plan proposed before himself; why should he agree with it over any other proposed plan?

Spoiler:
The pattern is you give one coin to each pirate of the same fierceness-parity as yourself. They will vote in favor, because the next fiercest pirate is of the opposite parity and would give them nothing.
Goplat

Posts: 490
Joined: Sun Mar 04, 2007 11:41 pm UTC

### Re: Re:

Goplat wrote:
Scuttlemutt wrote:@tomtom2357
Spoiler:
That pattern doesn't seem quite right though. Let's say a pirate of arbitrarily low fierceness is expecting to get 1 gold coin from almost any plan proposed before himself; why should he agree with it over any other proposed plan?

Spoiler:
The pattern is you give one coin to each pirate of the same fierceness-parity as yourself. They will vote in favor, because the next fiercest pirate is of the opposite parity and would give them nothing.

Also: this is optimal because you can't hope to win by giving any fewer coins because you would get less votes and therefore become shark bait.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
tomtom2357

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### Re: Pirate Democracy

Tomtom2357 has the correct solution... For those interested, here's how I arrived at the solution:

Spoiler:
I started with the case of two pirates and worked upwards. Each pirate will support the current scheme if they receive more gold coins now then they would if there was one less pirate. The most ferocious pirate is the one with the highest number.
Code: Select all
Pirate Number         100  ...   6    5    4    3    2    1
Case:  2 pirates  (trivial)                        1000   0
Case:  3 pirates                               999   0    1
Case:  4 pirates                          999   0    1    0
Case:  5 pirates                     998   0    1    0    1
Case:  6 pirates                 998  0    1    0    1    0
Case:  100 pirates    951         1   0    1    0    1    0

tomtom2357 wrote:I have a more interesting puzzle: After they found the 1000 coins (and decided the distribution), they now decide to downgrade the captain. The captain's position is decided by his generosity points. He gets +1 for every vote for him, and -1 for every vote against him. He is the lowest ranking person if he has 0 generosity points, and he gains a rank for every added generosity point. Now suppose that they find another 1000 coins, and the new captain decides the distribution according to the same rules as before. This cycle repeats infinitely many times (fine, the pirates are immortal, at least to the point of not dying by old age), but the value of the coins decreases by a factor of 0.9 every cycle. How can the captain maximize his coins value?

Just clarifying the rules.

If there are 100 pirates and 50 vote for me, I become the lowest ranked pirate. With 100 pirates and 51 who vote for me, I become the 3rd lowest ranked pirate?

Does each pirate act knowing that they will continue voting and downgrading captains for eternity?

If so, (starting the problem off)
Spoiler:
Code: Select all
With two pirates, the lower ranked will always vote against the captain, ensuring his place as captain.  They will essentially split the wealth.

With three pirates, if the captain survives, they will end up rotating positions and splitting the wealth three ways.  Hence, the captain will always be killed, splitting the treasure 2 ways.

With 4 pirates, pirate 3 will always vote for the current scheme to stay alive.  If pirates 1 and 2 vote against the captain, they will all rotate and in the end, split the treasure 4 ways, 250 apeice.  Pirate 2 would be better off accepting 251 gold coins for his vote, and the gold is divided as you see below.

Pirate Number             4          3          2          1
Case:  2 pirates                            1000 (500)   0 (500)
Case:  3 pirates                     X      1000 (500)   0 (500)
Case:  4 pirates    749 (374.5)   0 (374.5)    251         0

Am I on the right track here?   Things get confusing pretty quickly.

Edit: I believe I have the reasoning for case 3 incorrect. Only the captain and pirate 2 would rotate, so for pirate 2 there is no monetary incentive to accept or decline the captains proposal. If, however, pirate 2 accepted the proposal, there is a chance that he will die when he becomes captain - if he declines the captains proposal (killing him), he is assured of his life.
TPressey

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### Re: Pirate Democracy

Yes, each pirate knows that this will go on for eternity, so I think your solution is valid. Also, Happy New Year!
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
tomtom2357

Posts: 429
Joined: Tue Jul 27, 2010 8:48 am UTC

### Re: Re:

Scuttlemutt wrote:@tomtom2357
Spoiler:
That pattern doesn't seem quite right though. Let's say a pirate of arbitrarily low fierceness is expecting to get 1 gold coin from almost any plan proposed before himself; why should he agree with it over any other proposed plan?

I sorry scuttle but I have to agree with tomtom.
because:
Spoiler:
the best plan for any pirate with n fierceness is: if n is odd offer 1 gold coin to every odd fierceness pirate and if n is even offer 1 gold coin to every even pirate, and then take the rest for him self.

if a pirate that is going to get a coin dose not vote for it then next vote he is going to be offered no coins
Tnarg

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### Re: Pirate Democracy

Next question, what happens if the number of pirates is sufficiently bigger than the number of coins such that this does not work?
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mike-l

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### Re: Pirate Democracy

Hmmm...
Spoiler:
Inductively, we have the following argument.

In the 2 pirate case, the second pirate keeps all the gold, and the first pirate can do nothing about it.
In the 3 pirate case, the 3rd pirate only needs one vote, and the 1st pirates vote is cheaper than the 2nd pirate's vote. By a lot.

Exactly how much depends on how the pirates deal with ties in personal offers, and weather or not they wanted to kill each other off. Historically, they didn't. However, that degenerates into the trivial case where the first pirate keeps 1000 gold, so we will presume that because of greed, you must buy a vote with more gold than they might otherwise get.

So in the 3 pirate case, the 3rd pirate must offer the first pirate 1 gold, to make his case better than the case the second pirate would offer in a 2 pirate case.

Now however, we must stop and examine things. If the second pirate's best strategy in the 3 pirate case is to simply hope it turns into a 2 pirate case and then keep all the gold, then in a 3 pirate case he will never get any gold in the 3 pirate case! That's not a very good strategy.

So we have to ask ourselves something... can a perfect logician realize that a perfect logician's strategy choosing mechanism is deterministic? Surely they can, even I can figure that out. That means that pirate 2 knows he cannot eventually act out the 2 pirate strategy with probability 1 in the 3 pirate starting case, and have anything other than a probability of 0 of obtaining gold. His ideal strategy must take into account the fact that his ideal strategy is known by all.

Suppose pirate 2's ideal strategy is to hope that pirate 3 dies and that he can claim all the gold. Pirate 3 can deduce this and offer pirate 1 1 gold, and pirate 2 will get no gold at all.
Now suppose that in the 3 pirate case, pirate 2's ideal strategy is to offer pirate 1 2 gold, to garner his vote over pirate 3's offer of 1 gold, and keep 998 gold. 998 gold is in fact, greater than 0 gold. You say that pirate 1 cannot trust pirate 2 not to change his mind.... but that is not the case! These are perfect logicians. They cannot be tricked. If it is not in pirate 2's ideal strategy to ever change his mind - keeping in mind that his entire strategy is known, by everyone, including any inclination to change his mind later - then why would changing his mind be part of his plan? A ship of perfect logicians and no communication is equivalent to a ship of perfect logicians and everyone saying their plans until no one has any more to say, and furthermore, being bound to those plans. They are bound not because they agreed to them, but because they were deduced by a perfectly logical, who cannot be wrong! Now, if they were not perfectly logical, but rather, extremely smart, and could request (once) any finite amount of time to do their plotting, and maximally greedy, then yes, betrayals are totally in. But that's not what was posed. There is nothing stopping a perfectly logical being from choosing a branch of their future action table that contains no betrayals. Note that for a perfectly logical being, your life's future action table is known, and cannot be modified, and you proceed into the future by pruning it with your actions. Once an action is gone, it's gone. The actions you took in the past were specifically those necessary to preclude it from ever coming back.

So even in the 3 pirate case, pirates 2 and 3 must enter into a mental bidding war for pirate 1's vote on pirate three. Right up until pirate two becomes the cheaper vote.

Suppose pirate 3's plan is to pay pirate 1 499 gold. Pirate 2 would then have to pay pirate 1 500 gold to beat him. That would leave 500 gold for pirate 2, meaning pirate 3 could buy his vote with 501 gold as easily as he could pirate 1's vote, and thus pirate 1 cannot ask pirate 2 for that much gold. Pirate 2 can ask pirate 3 for it however, as splitting 1000 gold with pirates 1 or 3 makes no difference to pirate 2, pirate 3 will have to pay to break the tie. I'm pretty sure it stops here.

Further analysis is getting just a little bit too recursive, so I'm going to stop here. I'm sure people will have enough to object with in this!

Next question, what happens if the number of pirates is sufficiently bigger than the number of coins such that this does not work?
Oh, that's easy, pirates get executed until there are just enough.
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WarDaft

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### Re: Pirate Democracy

WarDaft wrote:
Next question, what happens if the number of pirates is sufficiently bigger than the number of coins such that this does not work?
Oh, that's easy, pirates get executed until there are just enough.

Only if pirates would rather die instead of getting nothing. If your solution is correct, than every pirate before there is 'just' enough should vote yes to ANY proposal by the first pirate, otherwise they will die. In particular, by making N is at least double 'just' enough, then 'just' enough is actually N, and the lead pirate would get everything. Obviously not correct.
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
mike-l

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### Re: Pirate Democracy

Seeing as I ended up actually disagreeing with that solution... not sure why I left that in.

Hmm, suppose there was only 1 coin and 10 pirates.

Under the simplest inductive scheme...

2 pirates - 2 gets it
3 pirates - 1 gets it
4 pirates - 2 or 3 gets it
5 pirates - Now what?

Hmm.

Ahh, a mixed strategy. Give it to pirate 3 with 70% chance, and pirate 1 with 30% chance. This makes their expected payouts higher by keeping 5 alive. This is however, skirting very close to the 'no dividing coins rule, but without it... I'm pretty sure 5's dead.
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WarDaft

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### Re: Pirate Democracy

Guys, my strategy works until there are more than twice as many people as there are coins to go around. Let there be 100 gold coins. The 202nd pirate has to give 1 coin each to all of the pirates of the same fierceness parity as him/herself. That pattern ends with him, the 203rd person does not have enough coins to bribe everyone, so if it gets to him, he gets thrown overboard. however, that does not mean that he does not play a part in the voting. The 204th person gives 100 coins to everyone with the opposite parity to himself, except 203. Then he gets 101 votes, but 203 has to vote for 204, otherwise 204 gets thrown off and then 203 gets thrown off too, so 204 gets 102 votes. 205 does not have the extra person's vote so he gets thrown off. 206 gets 102 votes (including the one from 205) but still gets thrown off. 207 gets 103 votes (including 205 and 206's vote) but still gets thrown off. 208 has 104 votes (1 from himself, 100 from the people he gives gold to, and 205, 206, 207's vote). So if N is the number of people and N>200, the Nth pirate only stays on if N=200+2n, where n is an integer. This is true because he gets votes from the 2n-1 people (this includes himself) that get thrown off if pirate N gets thrown overboard, and the 100 people he gives gold to.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
tomtom2357

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### Re: Pirate Democracy

Just a random thought: what would happen in the original problem if the 2nd pirate were to defect and also disapproves of the scheme?

Then, if the 99th pirate would use his normal strategy (give everybody of his own parity 1 coin), it may not get accepted, as the other pirates will consider the option the 2nd pirate defects again. So, it's not clear at all what happens now, and the result may be favourable for the 2nd pirate.

In fact; with 99 pirates battling for 1000 gold pieces, and the 2nd pirate being sure of survival, he seems to have pretty good odds, so disapproving might be the better decision. Any thoughts?
finchy

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### Re: Pirate Democracy

The trouble with introducing irrational characters is that it's hard to anticipate future decisions, and therefore the steps of the logic break down quickly unless you know how P2 behaves. But regardless of the number of pirates, irrationality by a single pirate either does nothing, or loses that pirate a coin and kills one pirate.

Assume everyone but P2 votes as described above. If P2 votes down P100's scheme, then P99 gets to propose a division - 1 coin to the odd-numbered pirates and none to the evens. This proposal will pass based on the 50 votes of the odd-numbered pirates (including P99), so P2 gets nothing. Therefore, P2 should accept the initial offering of 1 coin if he believes his fellow pirates to be similarly rational. The same can be said of Pn where n is even, so P100's initial proposal is accepted with 50 votes. If P2 somehow clearly states his intention and is thought to be reliable, an alternative solution would be for P100 to offer 1 coin to all the evens except P2, and 2 coins to P1, ensuring his survival and taking a coin away from P2. Thus, he has no advantage by threatening the other pirates even if it seems to be a reliable threat.

I think.

Gwydion

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### Re: Pirate Democracy

One of the main problems in game theory is whether you are able to make credible threats. P2 cannot actually make his threat credible, as if the situation does arise, then the rational thing to co would be to go back on his word and vote for P100. Even if he somehow could make a credible threat (this in itself is changing the problem) he does not get anything extra, and ends up worse off than before. How about this: for every pirate that gets thrown off, he/she manages to steal 2 coins to take with him. This should make everything more interesting, and allow more pirates to survive.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
tomtom2357

Posts: 429
Joined: Tue Jul 27, 2010 8:48 am UTC

### Re: Pirate Democracy

There are no threats or betrayals between perfectly logical beings. There is no doubt that someone will go through with their best possible plan, because they didn't promise to do so but rather because their plan was deduced logically. As perfectly logical creatures, they cannot and will not act contrary to any proper logical deduction of their actions. If the most logical course of action is to commit to not betraying at some point in the future, then they will not betray at that point... otherwise they are acting erratically, not logically.

Remember, perfectly logical creatures do not have free will. They do not change their mind unless the situation changes in a way that cannot be predicted. Note that introducing the announcement of a random variable at the start is not enough, as a strategy can be chosen at the beginning to take into account any result of the random variable.
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WarDaft

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### Re: Pirate Democracy

WarDaft wrote:There are no threats or betrayals between perfectly logical beings. There is no doubt that someone will go through with their best possible plan, because they didn't promise to do so but rather because their plan was deduced logically. As perfectly logical creatures, they cannot and will not act contrary to any proper logical deduction of their actions. If the most logical course of action is to commit to not betraying at some point in the future, then they will not betray at that point... otherwise they are acting erratically, not logically.

Remember, perfectly logical creatures do not have free will. They do not change their mind unless the situation changes in a way that cannot be predicted. Note that introducing the announcement of a random variable at the start is not enough, as a strategy can be chosen at the beginning to take into account any result of the random variable.

This is... misleading. Perfectly logical creatures can make promises and threats all they want, and in fact, making these is an important aspect of game theory. The challenge is making these threats/promises credible, which is an impossibility in a single-instance game. However, when a game is to be repeated several times, players have both the ability to make threats/promises and the ability to see them through.

Take as an example, the case of a repeated prisoner's dilemma. A "perfectly logical" creature who defects every time never "loses", but certainly could do better by cooperating with the other player. By promising to cooperate, and simultaneously threaten to defect if the other player does so first, one never does strictly better than one's opponent, but ends up much better off.

Another, more straightforward example - take the example of global thermonuclear war between two players. The game is largely one of simultaneous moves - both sides choose a strategy/action dynamically in real-time. As both players feel that any nuclear strike is the worst payoff possible, it doesn't happen despite any perceived aggression, and there is plenty of room to stomp on your opponent's toes without fear of nuclear retaliation. If Russia threatens to fire nukes at any minimal aggression by the US, and the US feels this is a credible threat, it becomes a game of sequential moves - the US moves troops into a former Soviet state and risks destroying the world. Now, the US's strategy decision changes drastically. Of course, there's nothing to stop Russia from going back on it's word - which is why one requires an automatic missile launcher (and knowledge of the same). (See Strangelove, et al for a more thorough discussion of this situation )

tl;dr: Promises and threats, if they can be made credibly, can be (and are) used effectively to alter other players' decisions in games such as this.

Gwydion

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### Re: Pirate Democracy

This is... misleading. Perfectly logical creatures can make promises and threats all they want, and in fact, making these is an important aspect of game theory.
A perfectly logical beings entire life is a single instance of one gargantuan game. Another perfectly logical being that may eventually encounter them will deduce the entire scope of their possible future actions given their situation, and those deductions will never be wrong. Perhaps we can say that one perfectly logical being can 'threaten' another because it is optimal for A for B to choose C, so A ensures (as close as possible) that it is not optimal for B to choose not C, but that's not a threat as we would generally consider it.

And you're considering prisoner dilemmas where both players are choosing different options. That is impossible if they are both perfectly logical - It cannot be perfectly logical for player 1 to choose option A and player 2 to not choose option A in a symmetric game like PD. They must choose the same value, whichever it is. This makes the problem substantially easier actually.
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WarDaft

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### Re: Pirate Democracy

For an example of a symmetric game in which it is suboptimal for two perfectly rational players to play symmetrically, take the "Gift of the Magi" story - both players prefer to give the other a gift and not get one in return. The "optimal" outcome for either player is the same, but reflected, but when the wife sells her hair to buy a watch-chain and the husband sells his watch to buy a comb, they both lose.

If stating that you plan to choose an option which (on its face) is suboptimal for you and different from your expected rational behavior is "not a threat as we would generally consider it", please define "threat" in this context so that I can continue.

Gwydion

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### Re: Pirate Democracy

I didn't ask when it was optimal for two parties to act asymmetrically, I asked how it was possible for two perfectly logical parties to pick different choices in a symmetrical game. It's like meeting each other when starting on opposite sides of a circular hallway and communication is impossible.

I suppose we could say that they can employ a mixed strategy and thus may end up acting differently (if given a separate and unpredictable source of randomness, they can't do it on their own!) but they're going to choose the same mixed strategy.

If stating that you plan to choose an option which (on its face) is suboptimal for you and different from your expected rational behavior is "not a threat as we would generally consider it", please define "threat" in this context so that I can continue.
A threat is generally persuasion. I'm talking about one player eliminating a favorable outcome for another player if doing so is overall favorable to them. Player B reasons that (in the case of PD for example) there is no line of reasoning that allows player A to choose defect but not allow player B to choose defect - player B has eliminated the potential of (A defect, B co-op) from A's future outlook.
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WarDaft

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### Re: Pirate Democracy

First off, we've completely derailed this thread, so if I can't bring it back into line here, I'm happy to drag this discussion to PM.

A threat (or promise) is meant to be persuasive, but that's insufficient to call it a threat. A threat (or promise, switch to positive wording where needed) is when a player states that he plans to act in a way other than his expected behavior (which was presumably "rational" and therefore maximized his expected return for the expected play of his opponents). This threat is designed to cause the other players to act differently from their expected strategy in order to "make the best of a bad situation", which in the end should work out better for the player making the threat.

In the classic PD, the outcome (A defect, B co-op) never happens without some outside intervention. If A believes B to be rational (not super-rational), and absent any other information, the problem as stated gives B a strictly higher payoff by defecting for all potential decisions by A, therefore A acts to maximize his payoff in that scenario by defecting. Even if B promises to cooperate, A should still defect. Therefore, B telling A he plans to cooperate (a promise) doesn't change A's behavior, and is therefore wasted. B telling A he plans to defect is similarly unnecessary, and shouldn't even be thought of as a threat (perhaps an assurance or a warning?)

The real challenge with threats and promises is making them credible. While there are several ways to do this (I alluded to one earlier), the most important is going through with your threat/promise when the time arises, even if it's seemingly irrational. If you don't, your future threats and promises don't get taken seriously, and you end up even worse off than you started.

Any time a game is played once, between players who will not interact again in the future and have no long-term consequences other than the immediate payout, this is all irrelevant. The presumption with the pirates, however, is that the next time they find treasure, they'll split it in a similar way. If P2 threatens to vote down any proposal in which he's offered fewer than 2 coins, there are situations in which he can do better than what was described above (With 4 pirates, P4 has to give extra coins to P2 or risk dying -> does this change how P3 votes when P5 proposes?), but if the other pirates call his bluff and he takes his 1 coin, the threat isn't going to be believed the next time, and he ends up with 1 or 0 coins anyway.

Gwydion

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### Re: Pirate Democracy

There's no such thing as "super rational". If there is a best strategy, then there is obviously nothing better than it. Choosing a maximally good strategy is obviously the only logical choice.

Back into the pirate version, in the 3 pirate version, pirate 2 can "commit" to paying pirate 1 at least 2 gold should pirate 1 down vote pirate 3. He can do this merely by it being his only possible means of obtaining gold in the 3 pirate version. He cannot later betray pirate 1 by not giving him 2 gold after pirate 3 is tossed overboard, because that betrayal would have be part of his plan before pirate 1 down votes pirate 3. As pirate 2 is perfectly logical his plan is totally deducible by pirate 1, and so must not contain betrayal in order to obtain any gold in the 3 pirate version. Thus, 1 gold is not enough to guarantee pirate 3 gets pirate 1's vote.

If we presume no further deductive bartering, then both pirate 2 and pirate 1 have received more gold than if pirate 1 had voted for pirate 3, clearly it is a better choice for both of them. So pirate 1 voting for pirate 3 for the price of 1 gold cannot possibly be pirate 1's best choice.
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WarDaft

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### Re: Pirate Democracy

WarDaft wrote:There's no such thing as "super rational".

Yes, there is.

http://en.wikipedia.org/wiki/Superrationality
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jestingrabbit

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### Re: Pirate Democracy

How can we call a strategy the rational choice if there is a better strategy? The word loses its meaning...
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WarDaft

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### Re: Pirate Democracy

addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
mike-l

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### Re: Pirate Democracy

WarDaft wrote:How can we call a strategy the rational choice if there is a better strategy? The word loses its meaning...

Well, the superrational choice only works if everyone does it, whereas the rational choice works regardless of what everyone else does. Also, in the prisoners dilemma, the superrational choice in a single instance becomes the rational choice in the iterated game (essentially).

For our pirates dilemma, the rational strategy in a single run is known (it seems to me). Whether the rational strategy changes when its iterated is an interesting question imo.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

jestingrabbit

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### Re: Pirate Democracy

Well, the superrational choice only works if everyone does it, whereas the rational choice works regardless of what everyone else does. Also, in the prisoners dilemma, the superrational choice in a single instance becomes the rational choice in the iterated game (essentially).
Well, or if you know (as in know) who you're up against.

Let's consider the case where a SR PD player is up against an R PD player. Both of them know who their opponent is, but cannot communicate as usual. Is it still rational for R to defect? It doesn't make sense that if the R player was replaced with an SR player that they could do better, R is supposed to have the best possible strategy.

For our pirates dilemma, the rational strategy in a single run is known (it seems to me). Whether the rational strategy changes when its iterated is an interesting question imo.
Well, it has a sort of iteration built into it. We need to agree on whether or not perfectly logical pirates can change their minds, for one. I don't think they can, as should be obvious. To switch plans is to act erratically, which is directly opposed to logically.
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WarDaft

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### Re: Pirate Democracy

When the pirates play the game infinitely many times, there is an obvious strategy that is better than the original strategy. This exists because even though perfectly logical beings cannot change their minds if the situation is fixed, they can if the situation varies (even though we can predict exactly what the change in decision is). In the iterated pirate game, every pirate (except the captain) can vote against the strategy if they get less than two coins. Now, every pirate knows this is a credible threat, because the game will always be played again. However, this only works if the pirates come back to life after they get thrown overboard, this is necessary to have a complete reset of the game. If you do not agree to this then an entirely different strategy works.

Actually, we need utility values to work with infinite games, for example, getting a coin may have a utility of 1, but getting thrown overboard may have a utility of -1, -10, or even -100, depending on the situation and depending on whether the pirates stay alive when the game is reset.
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tomtom2357

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### Re: Pirate Democracy

WarDaft wrote:Let's consider the case where a SR PD player is up against an R PD player. Both of them know who their opponent is, but cannot communicate as usual. Is it still rational for R to defect? It doesn't make sense that if the R player was replaced with an SR player that they could do better, R is supposed to have the best possible strategy.

A SR PD player only cooperates if he knows he is up against another SR PD player. Otherwise he defects, just like the rational player. So in SR vs R, it's (Defect, Defect), which is clearly better for the second player than (Defect, Cooperate)

And yes, two super rational players can do better than two rational players, that's why it's super .
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mike-l

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### Re: Pirate Democracy

WarDaft wrote:...in the 3 pirate version, pirate 2 can "commit" to paying pirate 1 at least 2 gold should pirate 1 down vote pirate 3. He can do this merely by it being his only possible means of obtaining gold in the 3 pirate version. He cannot later betray pirate 1 by not giving him 2 gold after pirate 3 is tossed overboard, because that betrayal would have be part of his plan before pirate 1 down votes pirate 3. As pirate 2 is perfectly logical his plan is totally deducible by pirate 1, and so must not contain betrayal in order to obtain any gold in the 3 pirate version. Thus, 1 gold is not enough to guarantee pirate 3 gets pirate 1's vote.

If we presume no further deductive bartering, then both pirate 2 and pirate 1 have received more gold than if pirate 1 had voted for pirate 3, clearly it is a better choice for both of them. So pirate 1 voting for pirate 3 for the price of 1 gold cannot possibly be pirate 1's best choice.

The trouble with your solution is it begs the question throughout. You state a "logical strategy" then assume it was obtained logically to prove that it was obtained logically. P2 promising gold to P1 in exchange for voting down P3 is one possible strategy, but without some mechanism for enforcement there is nothing to stop P2 from taking all the gold once it is his turn to propose a division. In the 2-pirate game, P1 is shafted no matter what. Knowing this, P1 would rather get a minimal amount from P3 than nothing from P2, so one gold coin is in fact, enough.

In the repeated game, the amount P2 needs to promise in order to induce P1 to vote down P3 is a function of the likelihood of P2 actually doing it, the probability of future encounters, and the amount P1 values gold now over gold later. P2's likelihood of following through with his promise is further complicated by the number of outside observers, his valuation of his own reputation for honesty, the total number of gold coins found (and promised), and numerous other uncountable factors.

In my opinion, super-rationality comes from a desire for "rational" to mean "best". Frankly, game theory is rife with situations in which the truly "rational" players make horrible decisions that could've been better in theory, but break down quickly in practice. Real people (who are rarely rational in the game-theoretic sense) usually do much better than the "experts" would suggest, which makes some people very upset.

Gwydion

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### Re: Pirate Democracy

The 'enforcement' is that pirate 2 cannot act erratically, everything pirate 2 does is deducible logically. If we, somehow, logically deduce that pirate 2 will not betray pirate 1 after pirate 3 is thrown overboard... then how can pirate 2, a perfectly logical being, not act exactly as we have deducted? Pirate 2 cannot betray without that being deducible, being perfectly logical. So in order to get gold, pirate 2 must ensure that it is deducible that they will not betray. If it is deducible that they will not betray, I ask again, how does this perfectly logical pirate then betray?

This in and of itself does not prove that it is logical for pirate 1 to down vote pirate 3. That only becomes logical if, given the above, down voting pirate 3 can be made profitable by pirate 2, which is not necessarily the case (if the pirates take some gold with them when they jump, for example, it can easily not be.)
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WarDaft

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### Re: Pirate Democracy

WarDaft wrote:The 'enforcement' is that pirate 2 cannot act erratically, everything pirate 2 does is deducible logically. If we, somehow, logically deduce that pirate 2 will not betray pirate 1 after pirate 3 is thrown overboard... then how can pirate 2, a perfectly logical being, not act exactly as we have deducted? Pirate 2 cannot betray without that being deducible, being perfectly logical. So in order to get gold, pirate 2 must ensure that it is deducible that they will not betray. If it is deducible that they will not betray, I ask again, how does this perfectly logical pirate then betray?

Well, if P2 believes that by lying to P1 and saying he'll give coins in exchange for voting off P3, he'll end up better off, then it is rational to lie.

Gwydion

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### Re: Pirate Democracy

Except he doesn't have to say anything. It's being deduced that it is more profitable for him to commit absolutely to not betraying. If we deduce that he would only lie, and always betray, then he will always receive 0 gold in the three pirate situation, yet if we can deduce that he will commit to not betraying, then there is an avenue for pirate 2 to make 998 gold and give 2 to pirate 1. 998 > 0 and 2 > 1. Surely, if possible, pirate 2 will commit (not say he will commit, but actually do so, so that we may deduce a desirable course of action about him committing) to anything he must to make this possible.
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WarDaft

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### Re: Pirate Democracy

WarDaft wrote:Except he doesn't have to say anything. It's being deduced that it is more profitable for him to commit absolutely to not betraying. If we deduce that he would only lie, and always betray, then he will always receive 0 gold in the three pirate situation, yet if we can deduce that he will commit to not betraying, then there is an avenue for pirate 2 to make 998 gold and give 2 to pirate 1. 998 > 0 and 2 > 1. Surely, if possible, pirate 2 will commit (not say he will commit, but actually do so, so that we may deduce a desirable course of action about him committing) to anything he must to make this possible.

We can't deduce that, since there's no avenue to enforce that commitment. Surely he'd like there to be one, but the way the problem is set up, there's nothing stopping him from 'betraying' and just taking 1000 gold.
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