A easy problem

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Amicitia
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A easy problem

Postby Amicitia » Mon Oct 29, 2007 7:40 am UTC

Two economics post graduates convicted for crimes against humanity are trapped in a prisoner's dilemma. What does each do in order to maximise their payouts, knowing their opponent is doing the same?
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Macbi
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Re: A easy problem

Postby Macbi » Mon Oct 29, 2007 9:22 am UTC

Spoiler:
If their opponent has to do the same thing then they should stay silent.
I often think that (in the normal dilemma) as both are perfectly logical, and they know this, and their sitiuations are identical, they know that they must do the same thing.
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Re: A easy problem

Postby Ari » Mon Oct 29, 2007 9:48 am UTC

A single-shot dilemma?

Well, that depends if they're rational or superrational. ;)
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Re: A easy problem

Postby Robin S » Mon Oct 29, 2007 12:58 pm UTC

By "knowing their opponent is doing the same", do you mean "knowing their opponent will choose the same action as them" or "knowing their opponent will follow the same lines of reasoning as them"?
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Re: A easy problem

Postby Macbi » Tue Oct 30, 2007 11:19 am UTC

Robin S wrote:By "knowing their opponent is doing the same", do you mean "knowing their opponent will choose the same action as them" or "knowing their opponent will follow the same lines of reasoning as them"?

Their situations are identical, so I think the two are equivalent.
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Re: A easy problem

Postby EdgarJPublius » Tue Oct 30, 2007 3:48 pm UTC

No, that's the essence of the Prisoners Dilemma, that tho two aren't equivalent.
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Re: A easy problem

Postby Token » Tue Oct 30, 2007 4:40 pm UTC

EdgarJPublius wrote:No, that's the essence of the Prisoners Dilemma, that tho two aren't equivalent.

Not really. If you are reasoning the same as someone, you'll pick the same answer given the same information. The essence of the Prisoner's Dilemma is that the Nash equilibrium is not the "optimal" choice.
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Re: A easy problem

Postby zingmaster » Tue Oct 30, 2007 7:21 pm UTC

And so since they're economists assumedly working together, they know the payout matrix and have agreed beforehand to avoid Nash Equilibrium.
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Re: A easy problem

Postby Ralp » Tue Oct 30, 2007 9:29 pm UTC

Token wrote:
EdgarJPublius wrote:No, that's the essence of the Prisoners Dilemma, that tho two aren't equivalent.

Not really. If you are reasoning the same as someone, you'll pick the same answer given the same information. The essence of the Prisoner's Dilemma is that the Nash equilibrium is not the "optimal" choice.

If you reason the same as your opponent, and your situations are identical, then you'll both play by the same strategy, but that doesn't mean every or even any move will be identical. We might both (independently) decide to cooperate with probability p initially, and thereafter cooperate with probability q or match our opponent's last move with probability 1-q.

Incidentally I think what I just described is a pretty good general strategy for the iterated game (for appropriately chosen p and q), but if they both know the other is a post-grad econ student (as Common Knowledge), I don't see how the optimum strategy wouldn't just be "always cooperate".

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Re: A easy problem

Postby Token » Tue Oct 30, 2007 10:24 pm UTC

Ralp wrote:Incidentally I think what I just described is a pretty good general strategy for the iterated game (for appropriately chosen p and q), but if they both know the other is a post-grad econ student (as Common Knowledge), I don't see how the optimum strategy wouldn't just be "always cooperate".

It's the optimum strategy no matter what each person knows about the other. However, there is a difference between an optimum strategy and an equilibrium strategy. Always cooperate is the former, always betray is the latter.
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Re: A easy problem

Postby Ralp » Wed Oct 31, 2007 6:13 am UTC

Token wrote:
Ralp wrote:I don't see how the optimum strategy wouldn't just be "always cooperate".

It's the optimum strategy no matter what each person knows about the other. However, there is a difference between an optimum strategy and an equilibrium strategy. Always cooperate is the former, always betray is the latter.

It's certainly not the optimum strategy for me if I happen to know you're playing "always betray", and it might not be the optimum strategy if I know there's a good chance you will betray me sometimes (depending on that probability and the specific payoffs). I think it's also pretty important whether I know anything about whether my potential cooperation or betrayal might affect your future decisions to cooperate or betray me.

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Re: A easy problem

Postby Macbi » Wed Oct 31, 2007 1:20 pm UTC

Ralp wrote:
Token wrote:
EdgarJPublius wrote:No, that's the essence of the Prisoners Dilemma, that tho two aren't equivalent.

Not really. If you are reasoning the same as someone, you'll pick the same answer given the same information. The essence of the Prisoner's Dilemma is that the Nash equilibrium is not the "optimal" choice.

If you reason the same as your opponent, and your situations are identical, then you'll both play by the same strategy, but that doesn't mean every or even any move will be identical. We might both (independently) decide to cooperate with probability p initially, and thereafter cooperate with probability q or match our opponent's last move with probability 1-q.

Incidentally I think what I just described is a pretty good general strategy for the iterated game (for appropriately chosen p and q), but if they both know the other is a post-grad econ student (as Common Knowledge), I don't see how the optimum strategy wouldn't just be "always cooperate".

They only play one game so multiple game strategies don't work (although the one you suggested is good).
Actually, looking at the first post that might be false, but if they both cooperate in the first game (which they should) then your strategy is equal to mine.
In one game: If they both cooperate with probability p then their expected gain is highest when p=1.
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Re: A easy problem

Postby Indon » Wed Oct 31, 2007 4:07 pm UTC

If I know my opponent would use the exact same reasoning as I will, thus leading to the same result, I will choose to stay silent, regardless of if my opponent betrays me... which I know he won't, because he's following the same line of reasoning.

More, uh, reasonably, though, I would play the game as I would in any other circumstance, because knowing what action someone will choose (which is what knowing their strategy essentially means) seems to trivialize the scenario.
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Re: A easy problem

Postby morjax » Thu Nov 01, 2007 4:50 pm UTC

Token wrote:
EdgarJPublius wrote:No, that's the essence of the Prisoners Dilemma, that tho two aren't equivalent.

Not really. If you are reasoning the same as someone, you'll pick the same answer given the same information. The essence of the Prisoner's Dilemma is that the Nash equilibrium is not the "optimal" choice.


"What does each do in order to maximise their payouts, knowing their opponent is doing the same?"

They aren't necesarily reasoning the same, They are only both trying to maximize thier payouts, knowing that their opponent is also trying to maximize her payouts. Also, How can we be sure that our opponent is reasoning soundly?

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Re: A easy problem

Postby Robin S » Thu Nov 01, 2007 5:24 pm UTC

I posted here about Iterated Prisoner's Dilemma strategies. Ignore the first paragraph.
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