## Rethinking the Prisoner's Dilemma

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### Rethinking the Prisoner's Dilemma

Reference, just in case.
What if the two prisoners were perfect logicians and each also knows that they are both perfect logicians?

Spoiler:
They would both cooperate!
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lordatog
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### Re: Rethinking the Prisoner's Dilemma

You're gonna have to explain your reasoning. I have a feeling that I'll disagree with it.

brenok
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### Re: Rethinking the Prisoner's Dilemma

I guess something like:

"If we are both perfect logicians and know we are perfect logicians, we will make always the same decision faced with the same data. So we either both cooperate or both defect, and it's better to both cooperate. I know it is better to cooperate and he knows I know it's better to cooperate so we both cooperate"

quintopia
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### Re: Rethinking the Prisoner's Dilemma

I think the hypothesis is insufficient to achieve superrationality. I think you may have to go as far as specifying the fact that they are both perfect logicians is common knowledge between them.

SPACKlick
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### Re: Rethinking the Prisoner's Dilemma

Yes without common knowledge the problem devolves back to "He knows that whatever I do He is better of betraying so he is guaranteed to betray so I must betray to protect myself."

With common knowledge I'm still not sure you get there. I see the argument and it's compelling but I'm not sure it's rigorously valid which it would need to be for a logician to get there.

jestingrabbit
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### Re: Rethinking the Prisoner's Dilemma

So, just to lay out what is going on here. There are two kinds of game solvers that are being discussed here. One is called a rational agent whereas the other is known as a a superrational agent.

Normally when we talk about a game player being "logical" or what have you, we mean that it is a rational agent ie "always chooses to perform the action with the optimal expected outcome for itself from among all feasible actions." That's very different from the superrational perspective.

I also want to point out that we have done this to death in this subforum. Like, there are few things that have wasted more pages with more futility. So, I want to set ground rules.

1) Say things that are cognisant of the distinction that I have just drawn.
2) Read carefully what others have written.
3) Try to give the best possible interpretation to the statements of others.

-jr
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

quintopia
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### Re: Rethinking the Prisoner's Dilemma

I'm not sure there is anything left to say.

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### Re: Rethinking the Prisoner's Dilemma

quintopia wrote: I think you may have to go as far as specifying the fact that they are both perfect logicians is common knowledge between them.

That was what I was trying to get at when I said:
each also knows that they are both perfect logicians

Sorry if it wasn't clear.

I was actually not aware of superrationality. Thanks for bringing it up! Interesting read.
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quintopia
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### Re: Rethinking the Prisoner's Dilemma

Ah. My bad. I thought you were trying to say that "each knowing the other is a perfect logician" and no more than that was sufficient to achieve superrationality.

Vytron
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### Re: Rethinking the Prisoner's Dilemma

I don't think Superrationality is achievable in practice even with common knowledge. Suppose we had this variant of the problem:

-What if the two prisoners were perfect logicians and each also knows that they are both perfect logicians? And one of them is told the other has already made a decision on how to act?

This breaks the symmetry and now the second logician has an advantage. Thinks "if the other is a perfect logician and I'm a perfect logician he'd have known that we were going to cooperate and that that was our best possible outcome, so he just cooperated." At this point you can't call her a perfect logician if she makes at this point the mistake to cooperate, because betraying is better for her at this point!

Note however, that at no point were they told what the other prisoner chose, only that they had already made a decision. Since defecting when the other defects is better for you than cooperating when the other defects, the first agent had already concluded that defecting is the optimal move for the second agent, so it's better to defect first. And we end with the original problem when both defect.

Now, how close in time does it need to be the first agent's decision to the second to cause both of them to defect? My claim is that the distance doesn't matter, because once it goes to the negative the second agent picks first and the first agent is told the other has already picked, and both defect.

There's no magic simultaneous moment at which both cooperate, so both always defect.

For short, being a perfect logician, and knowing your opponent is a perfect logician, and knowing that this is common knowledge, isn't enough to achieve superrationality, because at the first point in your thought chain where you have concluded the opponent is going to cooperate, your perfectly logical move is to defect.

So it should look like this:

What if the two prisoners were superrational?
Spoiler:
They would both cooperate!

But we already knew that, and I don't think there's a way to force conditions to make agents superrational, because rationally it's better to defect regardless of the opponent's choice.

quintopia
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### Re: Rethinking the Prisoner's Dilemma

Vytron wrote:For short, being a perfect logician, and knowing your opponent is a perfect logician, and knowing that this is common knowledge, isn't enough to achieve superrationality, because at the first point in your thought chain where you have concluded the opponent is going to cooperate, your perfectly logical move is to defect.

And this is exactly why your problem is different from the original one. You've handed one of them a piece of information that could change their mind. Without that piece of information, there is never any point in your thought chain where you can validly conclude the opponent is going to cooperate. I will admit that any amount of asymmetry could throw off superrationality. This is why superrational agents who also care about pareto efficiency would request to be put in separate rooms and be given no information about the other's decision process. AND that they be guaranteed that every piece of information they are given about the other player, the other player also receives about them. AND that their final decisions be "locked in" simultaneously.

Cauchy
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### Re: Rethinking the Prisoner's Dilemma

Superrationality is a specific type of irrationality that may occasionally lead to better outcomes than similar games with all rational players, in the case that every player is superrational.

Basically, superrationality is deductively begging the question. The argument for why cooperate is the logical choice hinges upon the fact that the actor has made the logical choice, and hence the opponent will also make the same choice. But you can't use the fact that cooperate is the logical choice to prove that cooperate is the logical choice!

Alternately, superrationality relies on purposefully stopping your logical train at some point. You say "If I and the other person both choose cooperate, we get a lot of money.". Then, you willfully stop yourself from realizing that you'd personally get even more money by defecting instead at this point, and in doing so, you expect everyone else to stop themselves from realizing this as well.
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quintopia
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### Re: Rethinking the Prisoner's Dilemma

The reasoning usually looks sort of like this:

-the other person is exactly as intelligent as me and is in an exactly symmetric situation.
-therefore we will make exactly the same decision.
-so I am choosing between CC and DD
-CC gets me more reward than DD
-so I will choose cooperate

At this point, there is no reason for the superrational actor not to consider that defecting would get him even more reward. But then he knows the other person will think of it too, and defecting would move them back to DD.

If you are going to argue that something is begging the question, it should be that first step. Why would an equally intelligent person in a symmetric situation always make the same decision? Or: why, once we know we'll make the same decision, would we not also know what that decision will be (DD)?
But this scenario is often brought up in situations for which the players can prove the other player is like them in some way. For instance, two artificial intelligences capable of advanced reasoning who both know that the other is running on exactly the same underlying algorithms.

douglasm
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### Re: Rethinking the Prisoner's Dilemma

Rationality vs superrationality, and various permutations of "perfect logician" stipulations, are all beside the real point, I think. Discussion of this nature is usually prompted by an intuitive feeling that the DD solution has to be wrong, somehow, because of how clearly inferior it is to CC and how CC tends to dominate over DD in analogous real world situations.

The true resolution to this bit of cognitive dissonance is to recognize that true pure prisoner's dilemma situations are extremely rare in real life. Analogous real life situations are far more likely to actually be iterated prisoner's dilemma with an unknown and indefinite number of iterations. In that variation of the problem, the Defect strategy's payout is altered by the ability of the other player to punish you in future iterations, and the lack of knowledge about the number of iterations means that potential for punishment exists (as far as the players know) on every iteration without exception. This results in variations on Tit-For-Tat being the dominant strategies, with the precise optimal variant depending on what strategies the other players use.

quintopia
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### Re: Rethinking the Prisoner's Dilemma

I have no such intuitive feeling. DD is obviously right in a situation where it really and truly is a classical PD. I would most likely defect in such a situation. But the fact remains that DD is not Pareto-optimal, and that's not just an intuition.

Vytron
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### Re: Rethinking the Prisoner's Dilemma

quintopia wrote:At this point, there is no reason for the superrational actor not to consider that defecting would get him even more reward.

Yes, but that only applies to superrational actors. Perfect logicians with perfect knowledge and common knowledge about each other's perfect knowledge will conclude that if the other being were to cooperate then defecting is better, and if the other being were to defect then defecting is better, so they would defect.

If superrational actors have to truncate their logic to achieve the better outcome, then their logic is truncated, and therefore not perfect.

And, anyway, I don't think the original PD is about logic, it is about trust. Even if we were only on the single stance scenario, my mother and me, or my sister and me, or a close friend and me, would be certain to cooperate, because we'd trust that we'd not betray each other. Superrational beings have the trust that whatever they decide to do will also be decided by the other being, this is something perfect logicians with common knowledge don't have.

It's like in the blue eyes puzzle: superrational beings can leave the island in n days, where n is the number of possible eye colors, while perfect logicians leave the island in n days, where n is the number of people with your eye color, because by day n if they didn't leave you know all they share the same color with you.

There's nothing that can make perfect logicians superrational.

quintopia
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### Re: Rethinking the Prisoner's Dilemma

I misspoke. That should have read "the perfect logicians has no reason not to consider that defecting would get him even more reward" but you are right. There is one more thing they need to have as common knowledge to be superrational: a shared desire to exploit the poser of the dilemma for as much total utility as possible.

The argument was that he would consider defecting, and then consider that the other player would also consider it. Then they would both realize that the other player would also realize that both were going to defect and that that would fail to exploit the dilemma organizer. So then both players consider cooperating again . This would go around forever, unless both players truncate their consideration somewhere, so they both then think "we're both going to do the same thing, what thing would accomplish our shared goals the best?"

Vytron
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### Re: Rethinking the Prisoner's Dilemma

Yeah, if the Prisoner Dilemma turns into a competition to defeat the dilemma organizer, I don't think it's necessary for them to be perfect logicians or have common knowledge to both cooperate. Common logicians would do it.

SirGabriel
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### Re: Rethinking the Prisoner's Dilemma

Vytron wrote:It's like in the blue eyes puzzle: superrational beings can leave the island in n days, where n is the number of possible eye colors

How could superrational beings know their eye color so quickly?

Vytron
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### Re: Rethinking the Prisoner's Dilemma

Offtopic
Spoiler:
(since this isn't the blue eyed puzzle thread)

I recall seeing many solutions that would only have worked with superrational beings, and one was really quick, but it was pointed out to the person that proposed it that it wouldn't work because even though everyone was a perfect logician, they weren't guaranteed to all agree to the same strategy (the fallacy being that they believed that if there was a strategy that allowed them to escape this quick, it would be used by all perfect logicians.)

What the problem asks is this: There's many copies of you in the island, and after the guru says she sees a person with blue eyes, you know all other people will use whatever strategy you come up with.

So, suppose there's 10 blue eyed yous on the island. If you weren't superrational then this follows:

I see 9 blue eyed people, so it's them who are waiting to leave, and I don't have blue eyes, or we all have blue eyes. What would person 9 be thinking if my eyes were brown? Well, they'd be thinking that they see 8 people with blue eyes, so either those are the ones waiting to leave, or everyone except me (the 10th) have blue eyes. And then they'd wonder what person 8 would be thinking if person 9 and 10 didn't have blue eyes (7 people have blue eyes, so they will leave without me unless I have blue eyes, etc.)

And so on, until the last person, which wouldn't see any other person with blue eyes and will know they have blue eyes and will leave today. So tomorrow, if person 1 is still here, it becomes common knowledge that person 1 and person 2 would leave tomorrow. The day after tomorrow, it becomes common knowledge that since 1 and 2 are still here, person 3 has blue eyes and they'll leave the day after tomorrow.

And so on, so person 10 concludes that if there's 9 people with blue eyes they'll leave the 9th day, but if they don't it means their own eyes are blue, so they'll all leave the 10th.

Perfect logicians don't have a way around that (if they did I'd be the hero of the other thread, and a statue would be built after me.) However, with superrationality you know you can skip days!

Yes, because whenever you see 9 people with blue eyes, you know that they all see at least 8 blue eyed people, though they don't know that. But they know all of them see at least 7. This means you can skip at least 6 days!

Ah, so the problem is that you know this, but if your eyes are brown, then the 9th guys doesn't know this. They only know they can skip at least 5 days. But if your own eyes are brown, then the 9th guy goes through all this and realizes only 4 days can be skipped.

In the original problem, this kind of reasoning is a bottomless pit, and you end not being able to skip any days. When you're superrational, you can choose to arbitrarily truncate the chain at, say, skipping 5 days is fine, because I know that everyone knows that everyone is seeing at least 6 blue eyed people in reality.

So, just make the rules:

If you see 9 blue eyed people, skip 5 days.
If you see 8 blue eyed people, skip 7 days.

And that's it. Everyone will follow that strategy, so that, on the original problem, everyone sees 9 blue eyed people and skips to day 5. If you believe that you're on that case but it turns out your eyes aren't blue, everyone else is seeing 8 blue eyed people, so they're skipping 7 days. They'll leave without you and that will tell you your eyes weren't blue.

This isn't optimal, but if it turns out there's 100 people on the island, they can use similar concepts and set similar rules to skip the first 90 days if they see 98 blue eyed people, 92 days if they see 99, and 94 if they see 100. Or something. If this is optimized then it's possible superrational beings of the original puzzle would leave the island in 3 days (but it's impossible to leave in 2.)

douglasm
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### Re: Rethinking the Prisoner's Dilemma

Vytron wrote:Offtopic
Spoiler:
So, just make the rules:

If you see 9 blue eyed people, skip 5 days.
If you see 8 blue eyed people, skip 7 days.

Spoiler:
Suppose there are, in fact, 8 people with blue eyes. The people with brown eyes will see 8, will therefore skip 7 days, and will conclude on day 2 (which they're calling day 9) that their eyes are blue. And they will be wrong.

Superrationality doesn't matter for blue eyes, it is not possible no matter how clever your strategy to leave early without having a case where your strategy will fail. I proved this in the blue eyes thread once, with induction from the day of departure back to the Guru's announcement, opposite the usual direction of induction for the puzzle.

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### Re: Rethinking the Prisoner's Dilemma

Offtopic
Spoiler:
Superrationality doesn't matter for blue eyes

It should matter.

Suppose there are 1000 blue eyed people, and 1000 brown eyed people.

With superrationality you can do this:

Rule - if I see at least 750 blue eyed people, then I pretend I start at day 500.

Everything else remains the same, but everyone leaves in half the time.

SirGabriel
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### Re: Rethinking the Prisoner's Dilemma

Vytron wrote:Offtopic
Spoiler:
Superrationality doesn't matter for blue eyes

It should matter.

Suppose there are 1000 blue eyed people, and 1000 brown eyed people.

With superrationality you can do this:

Rule - if I see at least 750 blue eyed people, then I pretend I start at day 500.

Everything else remains the same, but everyone leaves in half the time.

Spoiler:
But any such rule would be arbitrary. I don't think superrationality means that everyone else will make the same arbitrary choice as you. And there is no obvious best choice, since not everyone sees the same number of blue-eyed people as you do. So I don't think superrationality would actually help you leave any earlier.

PeteP
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### Re: Rethinking the Prisoner's Dilemma

SirGabriel wrote:
Vytron wrote:Offtopic
Spoiler:
Superrationality doesn't matter for blue eyes

It should matter.

Suppose there are 1000 blue eyed people, and 1000 brown eyed people.

With superrationality you can do this:

Rule - if I see at least 750 blue eyed people, then I pretend I start at day 500.

Everything else remains the same, but everyone leaves in half the time.

Spoiler:
But any such rule would be arbitrary. I don't think superrationality means that everyone else will make the same arbitrary choice as you. And there is no obvious best choice, since not everyone sees the same number of blue-eyed people as you do. So I don't think superrationality would actually help you leave any earlier.

Spoiler:
Also let's say you generalize that rule to say "jump back to the last full hundred lower than my" or any rule like that. The two groups need to jump back to the same number. If you can see X eyes either you are blue eyed and there are other who see X+1 or you aren't and there are other who see X-1. To keep synchronized X+1,X and X-1 must all lead to the same jump. However if not all numbers are projected to the same jump then there has to be an X so that X and X+1 don't result in the same jump. Therefore you won't find a general projection from number of eyes you see to jump length that always work.

quintopia
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### Re: Rethinking the Prisoner's Dilemma

There is one type of superrational agent that can shortcut the blue eyes puzzle:

Spoiler:
If they all know each other's mental algorithms perfectly and know they will pick the same random number independent of the number of blue eyes they can see. Perhaps because the Guru said 'there is at least one person with blue eyes', they all agree to skip the first day, and immediately assume all the reasoning they would otherwise have taken the first day to do has already been done. Because this number does not depend on the actual number of blue eyes (except that there are any, which is common knowledge) and therefore doesn't depend on what they can observe (which may differ between them), and because they all know each other will think of this number because they all have identical mental algorithms, they finish one day sooner. (The number "one" here is arbitrary. It could be any random number that they did not use their knowledge of how many eyes they can see to arrive at. They will fail if the random number they all think of is greater than or equal to the actual number of blue eyes, but this does not contradict their perfect logicianhood, since a perfect logician is described as being able to correctly derive any derivable conclusions instantly, not that they are incapable of deriving wrong conclusions from additional magical information that happens to be wrong.)

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### Re: Rethinking the Prisoner's Dilemma

Offtopic
Spoiler:
(The only reason I haven't moved this to the Blue Eyed Puzzle thread is because I still haven't finished reading it, so any discussion there would break my unread pages)

@SirGabriel: Well, that only shows that my examples suck. Actual superrational beings would never actually use the rules I'm depicting them using, they would come up with something that works, a kind of rule that isn't arbitrary but that each superrational being knows all other superrational beings arrived to. More on that below.

--------

So I'll use quintopia's superrationality example that seems to work and ask the question.

If all superrational beings carried a die that always falls on 1 and they could use it to skip 1 day, can they logically deduce that it's safe to skip at least one day?

I claim that the answer is yes, so, for any given situation it's safe to skip 1 day, unless there's too few people.

Now, the idea is superrational beings would be able to skip the maximum number of days that can be skipped, as if they had a die with infinite sides, and was fixed to fall on the highest number so that all superrational beings skip to this number.

If you wake up on the island and you see 99 blue eyed people, how many days are safe to skip?

Is it safe to skip 1 day?

Yes

Example:

All superrational beings just act as if one day had already happened, because they know nobody would have left that day anyway. Everything follows as if they weren't superrational, and they leave one day earlier.

Is it safe to skip 2 days?

Yes

Example:

All superrational beings just act as if two days had already happened, because they know nobody would have left the first 2 days anyway. Everything follows as if they weren't superrational, and they leave two days earlier.

Is it safe to skip 3 days?

Yes

...

So, there should be a point at which you get:

Is it safe to skip n days?

No, because then it's impossible to synchronize X+1,X and X-1 as they all don't lead to the same jump.

Whenever the superrational beings realize this they jump the last day that was safe, and skip that many days. Perfect logicians can't do this, so that's what makes different perfect logicians with common knowledge from superrational beings in Rethinking the Prisoner's Dilemma.

SPACKlick
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### Re: Rethinking the Prisoner's Dilemma

Offtopic
Spoiler:
The question vytron is whether n is the same for the group that see 99 blue eyes and the groups that see 98 or 100.
If the answer to that is yes then we also need the same n for 97 and 101
" "96 and 102
" "95 and 103
...
" "1 and 197
" "0 and 198

If n is the same for 0 it must be 0 or everyone leaves night 1.

lordatog
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### Re: Rethinking the Prisoner's Dilemma

Back on topic:

-the other person is exactly as intelligent as me and is in an exactly symmetric situation.
-therefore we will make exactly the same decision.
-so I am choosing between CC and DD
-CC gets me more reward than DD
-so I will choose cooperate

The problem with superrationality in this instance is that this logic is not complete. Our agent is solving half of the problem, then just guessing rather than solving the other half. Here's what this argument should actually look like:

-The other person is exactly as intelligent as me and is in an exactly symmetric situation.
-Therefore we will make exactly the same decision.
-So I have ruled out CD and DC as possible outcomes, leaving CC and DD as the only remaining options. Now I have to determine which of them is correct.
-I cannot actually influence the other person's actions. Intentionally acting illogically would not magically force the other person to do so.
-So choosing C is strictly inferior to choosing D, as it always has inferior rewards.
-So I have also now ruled out CC, leaving DD as the only option.
-So I will choose to defect.

(and really, we can only actually rule out CD and DC in this way if we've already proven that the optimal strategy is not a mixed strategy)

Cauchy
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### Re: Rethinking the Prisoner's Dilemma

Offtopic:

Spoiler:
Superrational agents can't leave early, because leaving isn't volitional. You leave exactly when you determine the color of your eyes, and not a moment before or after. Since everything is deterministic, agents can't change their behaviors, so there's no way to coordinate an effort. If everyone agrees "let's start counting at 1 instead of 0", then everyone is going to count to 100 and still not know the color of their eyes, so they can't leave on day 99. You can't shortcut a day, because each day the islanders all learn actual new information during the public reveal that no one left.

Superrational agents also can't glean extra information off of knowing everyone else is superrational. That's because there's not a known symmetry in situation (beyond the "is a perfect logician" thing that everyone already knows). If one islander knew he were in the same situation as every other blue eyed islander, then he'd know he had blue eyes, so he should just leave immediately. Otherwise, he doesn't know that he's the same, so he can't conclude that everyone else knows the things he knows.
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SPACKlick
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### Re: Rethinking the Prisoner's Dilemma

lordatog wrote:Back on topic:

-the other person is exactly as intelligent as me and is in an exactly symmetric situation.
-therefore we will make exactly the same decision.
-so I am choosing between CC and DD
-CC gets me more reward than DD
-so I will choose cooperate

The problem with superrationality in this instance is that this logic is not complete. Our agent is solving half of the problem, then just guessing rather than solving the other half. Here's what this argument should actually look like:

-The other person is exactly as intelligent as me and is in an exactly symmetric situation.
-Therefore we will make exactly the same decision.
-So I have ruled out CD and DC as possible outcomes, leaving CC and DD as the only remaining options. Now I have to determine which of them is correct.
-I cannot actually influence the other person's actions. Intentionally acting illogically would not magically force the other person to do so.
-So choosing C is strictly inferior to choosing D, as it always has inferior rewards.
-So I have also now ruled out CC, leaving DD as the only option.
-So I will choose to defect.

(and really, we can only actually rule out CD and DC in this way if we've already proven that the optimal strategy is not a mixed strategy)

I don't believe the agent is solving only half the problem.

If we accept the fact that super-rational agents will always use the same strategy and know that as common knowledge. Of the two discrete strategies CC has higher expected rewards than DD. If we allow a probablistic strategy where p is the probability of cooperating C is the reward for double co-operate, D is the reward for double defect B is the reward for betraying and 0 is the reward for being betrayed and B>C>D>0 Then there is a formula for the tactic the players should take whcih I can't quite work out but it turns out

if 2C<(D+B) then there is a non co-operate tactic the superrational agents will use.

At B=101, C=55, D=1 E(x)max = 55 at p=1
At B=101, C=45, D=1 E(x)max ~= 45.55 at p=0.9

Edit to add maths
E(x) = (C+D-B)p^2 + (B-2D)p +D

Completing the square for the above polynomial gives
E(x) = (C+D-B) * [ (B-2D) / (2 {C+D-B} ) +p]^2 - [ (B-2D)^2 / (4 {C+D-B} ) ] +D

The relevant component of that is [ (B-2D) / (2 {C+D-B} ) +p]^2 because E(x) reaches a limit when that = 0
That limit is a maxima when B>C+D (limit is maxima when coefficient of second order is negative)
Solving that for p gives

p = (B-2D)/(2B-2C-2D) (where B<>C+D {Which is fine because as established above B>C+D in all relevant cases})

0<=p<=1
@ p=1 gives 2B-2C-2D = B - 2D
B-2C=0
B=2C

@ p=0 0 = B-2D
B=2D

@ P=0.5 B-C-D=B-2D
-C=-D
C=D

so E(x)max is found at 0.5<p<1 when B>C+D, B>2C

douglasm
Posts: 630
Joined: Mon Apr 21, 2008 4:53 am UTC

### Re: Rethinking the Prisoner's Dilemma

Vytron wrote:Offtopic
Spoiler:
Superrationality doesn't matter for blue eyes

It should matter.

Suppose there are 1000 blue eyed people, and 1000 brown eyed people.

With superrationality you can do this:

Rule - if I see at least 750 blue eyed people, then I pretend I start at day 500.

Everything else remains the same, but everyone leaves in half the time.

Spoiler:
That rule fails for 750 blue eyed people.

I think the principle I described here is sufficient, but I'll take a different, more direct, approach.

Suppose that you have some superrational rule that, given a number N of blue-eyed people that you see, you skip D = f(N) days. For this rule to be worth anything, there has to be at least one value for N where f(N) > 0. For this rule to work for the single blue-eyed person case, f(0) and f(1) must be 0. Therefore, there are at least two distinct output values for f(N). Therefore, there exists an N where f(N) != f(N+1).

Consider the case of N where f(N) != f(N+1). This can be true in two different ways, either f(N) < f(N+1) or f(N) > f(N+1).

Consider f(N) < f(N+1). Suppose that there are, in fact, N+1 blue-eyed people on the island. The people with blue eyes will see N blue-eyed people and will skip f(N) days. Everyone else (with brown eyes) will see N+1 blue-eyed people and will skip f(N+1) days. The brown-eyed people are waiting for day N+1 - f(N+1). The blue-eyed people are waiting for day N - f(N). Because f(N) < f(N+1), the former will be at or before the latter, and all the brown-eyed people will conclude that they have blue eyes and attempt to leave. Puzzle failed.

So, there cannot be any N where f(N) < f(N+1). This means f(N) cannot increase. Combine this with the known value f(1) = 0, which is required for the single blue-eyes case, and f(N) cannot exceed 0 for any positive N.

It doesn't matter how clever your rule is, even if everyone's superrational or discussed it ahead of time, if there is any case where your rule has you skipping a positive number of days then there is a case where your rule causes the solution to fail.

Vytron
Posts: 429
Joined: Mon Oct 19, 2009 10:11 am UTC
Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

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