The Nash Equilibrium and the traveller's dilemma

[CallMeIshmael]

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Then the common use of the "assumption of common rationality" is different than the "assumption of infinite rationality" that you used in setting up this problem.

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Nope.

http://en.wikipedia.org/wiki/Common_knowledge_(logic)

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Then the accepted answer is wrong. Come up with a couterargument that isn't an argument from authority.

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We are talking about the definition of "the assumption of common/infinite rationality."

Your definition is wrong. I read through a book here to get a better understanding of it, and posted a wiki link (wiki isnt the greatest, but Im sure there are better readings online available).


If we disagree on a definition, then what means do I have beyond referencing authority to give credence to the claim that my definition is superior?


I mean, imagine the following dialogue:

A: That movie was brilliant!

B: No way. It was full of plot holes, had poor acting, the camera work was terrible and the ending made no sense.

A: While all that is true, the movie's name began with the letter 'Q', which means that it must be brilliant!

B: Well, thats not really the definition of 'brilliant.' Let me get a dictionary to show you.

A: OMG. APPEAL TO AUTHORITY.

[GMontag]

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Then the common use of the "assumption of common rationality" is different than the "assumption of infinite rationality" that you used in setting up this problem.

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Nope.

http://en.wikipedia.org/wiki/Common_knowledge_(logic)

[/ QUOTE ]

Then the accepted answer is wrong. Come up with a couterargument that isn't an argument from authority.

[/ QUOTE ]



We are talking about the definition of "the assumption of common/infinite rationality."

Your definition is wrong. I read through a book here to get a better understanding of it, and posted a wiki link (wiki isnt the greatest, but Im sure there are better readings online available).


If we disagree on a definition, then what means do I have beyond referencing authority to give credence to the claim that my definition is superior?


I mean, imagine the following dialogue:

A: That movie was brilliant!

B: No way. It was full of plot holes, had poor acting, the camera work was terrible and the ending made no sense.

A: While all that is true, the movie's name began with the letter 'Q', which means that it must be brilliant!

B: Well, thats not really the definition of 'brilliant.' Let me get a dictionary to show you.

A: OMG. APPEAL TO AUTHORITY.

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I wasn't disagreeing with the definition. I was disagreeing with the "accepted answer" to the PD given that definition. The appeal to authority I was talking about was your reference to the accepted answer.

[CallMeIshmael]

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Then the common use of the "assumption of common rationality" is different than the "assumption of infinite rationality" that you used in setting up this problem.

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Nope.

http://en.wikipedia.org/wiki/Common_knowledge_(logic)

[/ QUOTE ]

Then the accepted answer is wrong. Come up with a couterargument that isn't an argument from authority.

[/ QUOTE ]



We are talking about the definition of "the assumption of common/infinite rationality."

Your definition is wrong. I read through a book here to get a better understanding of it, and posted a wiki link (wiki isnt the greatest, but Im sure there are better readings online available).


If we disagree on a definition, then what means do I have beyond referencing authority to give credence to the claim that my definition is superior?


I mean, imagine the following dialogue:

A: That movie was brilliant!

B: No way. It was full of plot holes, had poor acting, the camera work was terrible and the ending made no sense.

A: While all that is true, the movie's name began with the letter 'Q', which means that it must be brilliant!

B: Well, thats not really the definition of 'brilliant.' Let me get a dictionary to show you.

A: OMG. APPEAL TO AUTHORITY.

[/ QUOTE ]

I wasn't disagreeing with the definition. I was disagreeing with the "accepted answer" to the PD given that definition. The appeal to authority I was talking about was your reference to the accepted answer.

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It really does just boil down to a definitional thing

You believe that "common assumption of rationality" means that players are deciding between symetrical payoffs, and are choosing the best option therein.

This is not part of that definition.

[Nicholasp27]

in callme's oot example, u have to go $2, because otherwise u get nothing since ur opponent is john nash; u can go 100, but he'll go 2 and he'll get 4 and u'll get 0

the only way u get money playing against nash is to go 2, in which case u get 2

if u were playing some random person off the street, u could go in the high 90s, but against nash, being rational and knowing u [img]/images/graemlins/heart.gif[/img] game theory, u have to go 2

[Nicholasp27]

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I say again, 99 is optimal.

The problem here is that game theory is butting its head in where it's not needed. This is a simple maths/logic question.

Every number is dominated by the number 1 lower. However, it dominates every choice 2 numbers lower

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apparently, u have not properly read the question

if u pick 100 and i pick 98, i'll get paid 100 and u'll get paid 96


if u pick 100 and i pick 2, i'll get paid 4 and u'll get paid 0


u seem to think that u will just lose 2 off of YOUR bid...but u lose 2 off of the lowest bid


that's why u HAVE to go 2 against another rational like nash if he knows u are rational, as he will go 2, so ur choices in this game are to get $2 or to get $0...take ur pick

[Nicholasp27]

i agree that people are biased by the fact that it's 'only' $2

if u were playing against nash and it was a number between $2 billion and $100 billion, and u were wearing a "i heart nash equilibrium" tshirt, so he knew u were rational and knew game theory, then would u REALLY not pick $2 billion?

i would and i'd go home with $2 billion; whomever doesn't goes home with $0

[Nicholasp27]

$100 is dominated by 99, thus should be eliminated from consideration


if opp goes: u go

100: 100->100, 99->101
99: 100->97, 99->99
98: 100->96, 99->96
... they are equal

thus 100 should be eliminated

we now look at the game as 2->99

well 99 is dominated by 98...

opp goes: u go
99: 99->99, 98->100
98: 99->96, 98->98
97: 99->95, 98->95
...

so remove 99 from your choices

play game with 2->98

and keep doing that all the way down to NE of 2,2

that's how game theory works; u shouldn't pick a strategy that is dominated by another; so eliminate that strategy and re-run the game without that as an option...do this until u find either a dominant strategy (over all other strategies) or u eliminate all dominated strategies

if u still have more than 1 left, then u may need to rand options; if u have only 1 strategy left, then u have equilibrium

[Nicholasp27]

the traveler's dilemma experiment with game theorists also built in a bias that changed the results of the experiment, thus changing the game

they said they'd give a rand() player $20 times their average payout from playing the game

that gives people the incentive to go for higher numbers to try to get some meaningful amount of money; again, the low $ amount biases players...they could go $2 every time and average between 2-4, so $40-80 ev divided by 151 players...let's say 3 on average...that's 60/151=40 cents expected value

instead, players went for high amounts, and the rand winner got $1700 by averaging $85...

also, 10/51 'game theorists' chose 100, which is dominated by 99, for every round

so while interesting, that experiment was not the same game

[wazz]

apparently you haven't read anything i've said

[Siegmund]

In trying to resolve alleged paradoxes like these, people tend to forget the details behind how the equilibria are defined: the minimax solution of a game is its value against malevolent opponents, a Nash equilibrium is a position where you would play the same even if you had been told your opponent's strategy in advance, or, in other words, the best that a completely non-reactive player who ignores your plays even after they are revealed in an iterative game, can guarantee for himself.

There is a large class of games, including the traveller's dilemma, the prisoner's dilemma, the multi-person prisoner's dilemma, and the new 100-person game in the other thread, in which all players in the game face symmetric situations.

It's easy to prove, for any of the accepted definitions of optima or equilibria, that the solution of a symmetric game (if it exists) must be symmetric.

A very natural question, when faced with a symmetric game, is to find an optimal strategy over the sub-game where we know all players in the game will play the same (possibly mixed) strategy - that is, a 1-D optimization along the main diagonal of the payoff matrix, instead of the usual simultaneous and opposing row and column optimizations. Finding the solution is usually computationally very easy, and very often leads to what a lot of people believe to be the "right answer" to games like these.

I'm sure someone has studied this way of solving symmetric games before, and discussed solutions found in this manner before, but I don't recall reading about it. In the meantime, by analogy with the Cauchy Principal Value of an improper integral(*), I propose to call this the Diagonal Principal Value of a game, and the corresponding stragetics the Diagonal Optimal Strategies.

(*) - Quick calculus refresher: the improper integral from -infty to +infty only exists if the limit of the definite integral from a to b exists as a goes to -infty and b goes to +infty. Even when the improper integral is not defined (for instance, the integral of 1/x), the Cauchy Principal Value, defined as the limit of the integral from -a to a as a goes to infty, may exist (for 1/x it is zero.)