Other prespective on resolving the Prisoner's dilemma

I had a thought related to this (apologies if this has been bought up on LW before, but I haven't seen it exactly).

Let's say you're playing a game theory game, and you have access to an oracle that either answers correctly or refrains from answering.

Obviously in a lot of games it makes sense to ask what move your opponent will make, so you can then optimise your own move. But in the Prisoner's Dilemma, the opponent's move is useless information - whatever you learn, your optimal move is to defect.

What you want to ask is not "what move will he make ?", but "will he make the same move as me ?". Learning which row or column you're in doesn't help; learning whether you're on-diagonal or off-diagonal does.

They don't have to be known to be impossible. Just unlikely. If you're facing someone similar to yourself, it might be that choosing to defect makes it more likely that they defect, and enough so to counter out any gain you'd have, but you still don't know they'll defect.

If the off-diagonals are impossible, then it's not the Prisoners' Dilemma, it's just Cake or Death. If you're facing an identical copy of yourself, then it's really the Newcomb Paradox. Open code (or, at least, unilateral open code) only works in the Ultimatum Game. One study found that knowing how the other person chose actually increases defection; the probability that Player Two defects given that Player One defects > probability given Player One cooperates > probability given Player One's action is unknown. Furthermore, open code only makes sense if you have a way of committing to following the code, and part of the Prisoners' Dilemma is that there is no way for either player to commit to a course of action. And another part of the whole concept of the one-off Prisoners' dilemma is that there is no way to retaliate. If the players can sue each other for reneging on agreements, then it's basically an iterated Prisoners' Dilemma (and if there's a defined endpoint, then you have the Backwards Induction Paradox).

Why not put some figures on 'identicality' of the players and see what comes out ?

A simple way is to consider the probability P that both players will play the same move. That's a simple mesure of how similar both players are.

Remember I am not stating that there is any causal dependency between players (it's forbidden by the rules):

• A and B could be twins raised in a tight familly

• A and B could be one unique person asked to play against several unknown opponents and not knowing he is playing agaisnt herself (experimental psychologist can be quite perverse)

• A and B could be two instances of one computer program

• A and B could even be not so similar persons, but merely play alike two times out of three. It's already correlation enough.

• A and B could be imagined to be so different as to always play the opposite move for one another, given the save initial conditions (but I guess in this case I can't imagine how they could both be rational)...

• etc.

That gives use an inequation of this parameter and a result depending of the values inside the PD matrix.

Notations:

Player A move is x, move can be: x=C (cooperation) or x=D (defection)

Player B move is y, move can be: y=C (cooperation) or y=D (defection)

P(E) denotes probability of event E

G(E) denotes the expected (probabilist) payoff if event E occurs.

We also assume a stable definition of rationality. That means something like what physicians calls gauge Invariance : you should be able to exchange the rôle of x and y without changing equations. Gauge invariance gives use some basic properties:

We can assume P(y=C) = P(x=C) = P(C) ; P(y=D) = P(x=D) = P(D).

It follows:

P(x=C and y=D) = P(x=D and y=C) = P(x!=y)

P(x=C and y=C) = 1 - P(x=C and y=D) = P(x=y)

P(x=D and y=D) = 1 - P(x=C and y=D) = P(x=y)

Now, keeping in mind these properties, let's find the payoff for x=C G(x=C), and the payoff for x=D G(x=D).

Gx(C) = Gx(x=C and y=C) P(x=C and y=C) + Gx(x=C and y=D) P(x=C and y=D)

Gy(D) = Gx(x=D and y=D) P(x=D and y=D) + Gx(x=D and y=C) P(x=D and y=C)

Gx(C) = Gx(x=C and y=C) p(X=Y) + Gx(x=C and y=D) P(x!=y)

Gy(D) = Gx(x=D and y=D) P(x=y) + Gx(x=D and y=C) P(x!=y)

Gx(C) = (Gx(x=C and y=C) - Gx(x=C and y=D)) * P(x=y) + Gx(x=C and y=D)

Gx(D) = (Gx(x=D and y=D) - Gx(x=D and y=C)) * P(x=y) + Gx(x=D and y=C)

The rational choise will be C for x if Gx(C) > Gx(D)

on the contrary the reasonable choise will be D for x if Gx(C) < Gx(D)

if Gx(C) = Gx(D) there is no obvious reason to choose one behavior or the other (random choice ?).

The above inequations are quite simple to understand if we consider P(x=y) as a variable in a geometric equation. We get equations for too lines. The line that is above the other should be considered as the rational move.

The mirror argument match the case where P(x=y) = 1,

Then we have Gx(C) = Gx(x=C and y=C), Gx(D) = Gx(x=D and y=D),

with usual parameters where Gx(x=C and y=C) > Gx(x=D and y=D),

C is rational for identical players.

The most interesting point is were the two lines meet.

At that point Gx(C) = Gx(D)

It yields :

P(x=y) = (Gx(x=D and y=C) - Gx(x=C and y=D))/(Gx(x=C and y=C) - Gx(x=C and y=D) - Gx(x=D and y=D) + Gx(x=D and y=C))

PD criterium is such that this is always a positive value.

With the usual values we get :

P(x=y) = (5-0)/(3-0+5-1) = 5/7 = 0.71

It simply means that if probability of same behavior is 71% or above it is rational to cooperate in a Non iterated Prisonner Dilemma.

My point is really that if both players are warned that the other one is a (mostly) rationale being it is enough for me to believe he is identical to me (he will behave the same) with a probability above 71%.

You should understand that a probability of 50% of identical behavior is what you get when the other player is random. As I understand it defectors are just evaluating the probability of identical behavior of the other between 50% and 71%. It is a bit too random for my taste.

What I also find interresting is that my small figures match results I remember having seen in real life experiences (3 on 5 cooperating, 2 on 5 defecting). [I remember a paper about "Quasi-magical reasonnning" from the 90's but I lost pointers to it]. It does not imply ant more that some of these people are rational and others are mislead, just divergence on raw evaluation of probability for other human players to do the same as them.

As an afterword, I should also say something about Dominance Argument, because this argument is the basis for the current belief of most academics that 'D = rational'.

It goes like this:

What should A play ?

if B choose C, A should choose D because  Gx(x=D and y=C) > Gx(x=C and y=C)

if B choose D, A should also choose D because Gx(x=D and y=D) > Gx(x=C and y=D)

Hencefoth A should choose D whathever B plays. Right ?

Wrong.

The above is only true if x and y are idependant variables. Basically that is what you get when P = 50%

The equations are above, easy to check.

Mathematically x and y are independant variable means the behavior of y is random relating to x.

This is a much stronger property than just stating there is no causal relationship between x and y. And not exactly a realistic one...

That is like stating that because two traders do not communicate/agree between each other, they won't choose to buy or sell the same actions on marketplace. Or that phone operators pricing won't converge if operators do not communicate between each others before publishing new package pricings ? I'm not pretending they will perfectly agree, or that convergence can not be improved through communication, but just that usually the same cause/environment/education give the same effects and that some correlation is to be expected. True random independance between variables only exist in the mathematical world.

Suppose that you were playing Omega in single PD; Omega is presumed to be able to accurately predict your moves in advance, as in Neucomb's problem.

Omega's strategy will maximize its score; your goal is to maximize your score.

What would Omega do? Should its strategy be dependent on yours?

         Player 1\Player 2: {c:C;d:C} {c:C;d:D} {c:D;d:C} {c:D;d:D}
         c....................c:C 3/3...c:C 3/3...c:D 0/5...c:D 0/5
         d....................d:C 5/0...d:D 1/1...d:C 5/0...d:D 1/1