The Tao of Gaming

Boardgames and lesser pursuits

Game Theorist — “Dumb is the new Smart”

An interesting article points out that even game theoreticians don’t believe their own results (when money is on the line).

a research team repeated the experiment using professional game theorists playing for real money. But even among game theorists, game theory failed

One hypothesis is that you can get good results by playing dumb. If your opponent knows you are totally rational, then they have to give up a lot to keep from getting screwed. (This particular example deals with the Traveler’s Dilemma, but it applies to the Prisoner’s Dilemma, as well).

I remember a book that dealt with various puzzle aspects of Game Theory as told by Sherlock Holmes, et al. One passage discussed the prisoner’s dilemma, after a clever person tries to use it with real prisoners. When it doesn’t work, he goes to Holmes, who then sighs and calls forth one of the prisoners.

“So you know that it’s always better [to defect].”
“Yes, guv’ner.”
“Then pray explain to [this doofus] why you don’t.”
“Me mates would beat me senseless.”

Glad to see the theoreticians catching up.

Now to just figure out how this relates to unconvincing cylons, and the applications will be endless!

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Written by taogaming

December 9, 2008 at 5:38 pm

Posted in Game Theory

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3 Responses

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  1. I think the author kind of screwed the article up (either that or she was trying to give herself a provocative angle). The subheadline got the concept right–the “solution” to the Traveler’s Dilemma is a paradox–but the article refers to the $2 value as the game theoretic solution to the problem, which it clearly is not. The issue is the same as with the Unexpected Hanging paradox: repeated use of prior valuation assumptions will lead to faulty results. The game theorists didn’t set a $2 value because they know it’s not the proper answer to the problem.

    Larry Levy

    December 9, 2008 at 8:58 pm

  2. The idea is that if the other player bids $100, then you can bid $99 to increase your winnings to $101. But if he knows that you are bidding $99, he can bid $98 (to win $100), and so forth, iterated down to $2. One problem is that I doubt anyone in real life iterates more than two or three times. Also, as Alex says, you need to look at the big picture. There is only ONE situation in which defecting gains you anything, which is if you manage to bid exactly $1 less than the other guy. Thus defecting will rarely be worth the risk. It’s a case of missing the wood for the trees (which seems to be a common malady of game theory actors).

    Kester Jarvis

    December 12, 2008 at 5:22 pm

  3. Yes, $99 dominates $100, since it always does better or at least as well, no matter what the other fellow does (if he bids $100, I gain a buck; if he bids $99, I gain $2; and if he bids anything else, it’s a wash). So an answer of $99 is justifiable. However, $98 doesn’t dominate $99, unless you assume that your opponent will never bid $100 because it is dominated. However, that leads you on the spiral toward the clearly suboptimal answer of $2. They even have a lable for that kind of reasoning–leaky induction (cool name, huh?). So while it’s not the only answer, I think $99 is the best and probably the most likely value for the game.

    Nash Equilibrium can get you into trouble. The ($99,$99) solution for the game isn’t stable, in that one of the players can benefit by choosing $98. The only stable solution is ($2,$2). But in non-zero sum games, unlike zero-sum ones, stable doesn’t mean optimal! It could just be a local maxima, where if both players stray from the solution, they can both do better. Some people (and maybe even some Game Theoreticians) cling to the belief that Nash Equilibrium = Game Solution, but then again, some people still believe in Santa Claus and the Eternal Bull Market. I just can’t believe that the majority of academics in Game Theory would say that the value of this game is $2.

    Larry Levy

    December 12, 2008 at 7:43 pm


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