So, what is wrong believing in probabilities ?

To ask that question is already to presuppose the one-boxing answer, and to miss the problem that the problem itself may be problematic. I don't take simple two-boxing any more seriously than Amanojack does, but the third possibility, of disputing that the problem is well-posed, is worth exploring. On LW, self-professed two-boxers are usually taking that alternative. (Elsewhere, I see two-boxing philosophers actually saying that two-boxing loses, but is still the rational thing to do.)

The problem is best disputed not by simply asserting, as some have, that no such Omega can exist, but by thinking in detail about what it would take for someone to predict the decisions of a decision-maker who knows you're trying to predict their decisions. What that sort of thinking looks like is this. That paper is about Prisoners Dilemma, but similar investigations could be made of Newcomb, Parfit's Hitchhiker, etc.

That is what fighting the hypothesis looks like, done right.

Newcomb's Problem is silly. It's only controversial because it's dressed up in wooey vagueness. In the end it's just a simple probability question and I'm surprised it's even taken seriously here. To see why, keep your eyes on the bolded text:

The problem is, such emphatic declarations of confidence in the right answer can just as easily be followed by one-boxing, two-boxing, or declaring the hypotheses self-contradictory. That is, in fact, what makes it a Problem, even if, to any individual, it is not a problem.

If you find yourself in a hole, stop digging.

-- Denis Healey

I'm with you. You have to look at the outcomes, otherwise you end up running into the same logical blinders that make Quantum Mechanics hard to accept.

After reading some of the Quantum Mechanics sequence, I am more willing to believe in Omega's omniscience. Quantum mechanics allows for multiple timelines leading to the same outcome to interfere and simply never happen, even if they would have been probable in classical mechanics. Perhaps all timelines leading to the outcome where one-boxing does not yield money happen to interfere. Even if you take a more literal interpretation of the problem statement, where it is your own mind that determines the box's content, your mind is made of particles which could conceivably affect the universe's configuration.

Sometimes the most remarkable things seem commonplace. I mean, when you think about it, jet travel is pretty freaking remarkable. You get in a plane, it defies the gravity of an entire planet by exploiting a loophole with air pressure, and it flies across distances that would take months or years to cross by any means of travel that has been significant for more than a century or three. You hurtle above the earth at enough speed to kill you instantly should you bump into something, and you can only breathe because someone built you a really good tin can that has seams tight enough to hold in a decent amount of air. Hundreds of millions of man-hours of work and struggle and research, blood, sweat, tears, and lives have gone into the history of air travel, and it has totally revolutionized the face of our planet and societies.

But get on any flight in the country, and I absolutely promise you that you will find someone who, in the face of all that incredible achievement, will be willing to complain about the drinks.

The drinks, people.

--Harry Dresden, Summer Knight, Jim Butcher

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.