The Tao of Gaming

If I played the Travelling Problem, I would absolutely say either $100 or $99, and would consider it stupid to do otherwise.

Its a probabilistic risk problem to me.


If I say $2, I am guaranteed to get either $2 or $4.

If I say $99 or $100, then I will get between $99-101 if the other play also does the same, Or I will get $X-2 if they say X (some lower number).

Lets say that even if I assume there is a 90% chance they will be "rational" and say $2, and a 10% chance of them saying 90-100. My expectation value is still higher by saying 99 or 100.

If you believe that most people will follow a probabilistic model, instead of using the nash equilibirum answer, then only saying 99-100 makes sense.

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).

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.