I’ve been busy at FEW the past few days, but thanks to everyone who has responded to my previous post. Anyway, in the airport on the way back from Pittsburgh, I saw that the current issue of Scientific American has several philosophically interesting articles, including ones about the origin of life (did it start with a single replicating molecule, or a process involving several simple ones?) and anesthesia (apparently, the operational definition of general anesthesia isn’t quite what you’d expect, focusing on memory blockage more than we might have expected). (It looks like you’ll have to pay to get either of those.)

But I want to discuss an interesting article by economist Kaushik Basu on the Traveler’s Dilemma (available free). This game is a generalization of the Prisoner’s Dilemma, but with some more philosophically interesting structure to it. Each player names an integer from 2 to n. If they both name the same number, then that is their payoff. If they name different numbers, then they both receive the smaller amount, with the person who named the smaller number getting an additional 2 as a bonus, and the one with the larger number getting 2 less as a penalty. If n=3, then this is the standard Prisoner’s Dilemma, where naming 2 is the dominant strategy. But if n≥4, then there is no dominant strategy. However, every standard equilibrium concept still points to 2 as the “rational” choice. We can generalize this game further by letting the plays range from k to n, with k also being the bonus or penalty for naming different numbers.

Unsurprisingly, in actual play, people tend not to actually name k. Interestingly, this is even the case when economics students play, and even when game theorists at an economics conference played! Among untrained players, most play n, which interestingly enough is the only strategy that is dominated by another (namely, by n-1). Among the trained players, most named numbers between n-k and n-1.

In the article, this game was used to suggest that a better concept of rationality is needed than Nash equilibrium play, or any of the alternatives that have been proposed by economists. I think this is fairly clear. The author also uses this game to suggest that the assumption of common knowledge of rationality does a lot of the work in pushing us towards the choice of k.

I think the proper account of this game may bear some relation to Tim Williamson’s treatment of the Surprise Exam Paradox in *Knowledge and its Limits.* If we don’t assume common knowledge of rationality, but just some sort of bounded iteration of the knowledge operator, then the backwards induction is limited.

Say that an agent is rational^{0} only if she will not choose an act that is dominated, based on what she knows about the game and her opponent’s options. Say that an agent is rational^{i+1} iff she is rational^{i} and knows that her opponent is rational^{i}. (Basically, being rational^{i} means that there are i iterations of the knowledge operator available to her.) I will also assume that players are reflective enough that there is common knowledge of all theorems, even if not of rationality.

Now I claim that for i^{i}, then when she plays the Traveler’s Dilemma, she will pick a number less than n-i.

Proof: By induction on i. For i=0, we know that the agent will not choose any dominated strategy. However, the strategy of picking n is dominated by n-1, so she will not pick n=n-i, as claimed. Now, assume that it is a theorem that if an agent is rational^{i}, then when she plays the Traveler’s Dilemma, she will pick a number less than n-i. Then the agent knows this theorem. In addition, if an agent is rational^{i+1}, then she knows her opponent is rational^{i}, and by knowing this theorem, she knows that her opponent will pick a number less than n-i. Since she is also rational^{i}, she will pick a number less than n-i. But given these two facts, picking n-(i+2) dominates picking n-(i+1), so by rationality^{0}, she will not pick n-(i+1) either, proving the theorem, so the induction step goes through, QED.

Thus, if an agent picks a number n-i, then she must be at most rational^{i-1}. But based on what Williamson says, iterations of the knowledge operator are generally hard to come by, so it should not be a surprise that even game theorists playing with common knowledge that they are game theorists will not have very high iterations of rationality. I wonder if it might be possible to use the Traveler’s Dilemma to estimate the number of iterations of knowledge that do obtain in these cases.

Since so many people play n, can we conclude that there are 0 iterations of knowledge?

Well, we can conclude that in these cases there are 0 iterations of “rationality”. With the game-theorists the results were as follows:

This might suggest somewhere between 1 and 5 iterations of “rationality”.

Or of course, it might suggest that something else is at work here, like altruism or a sense of community or cooperation or something, as I think you were suggesting.

Yeah – I think this game is too similar, in some crude respects, to ultimatum games to really be a test of rationality.

I’m pleased though that a lot of people picked between 95 and 99. I think there are some decent norms according to which 99 is the right choice, but I have to think that through. Certainly anything lower than 95 seems like a mistake.

Here is one quick thought. There is a sense in which 100 is a better choice than 2. If I was faced with only those two choices, and the other player had the choices they actually have, then 100 would be the uniquely rational choice. Now one might argue that although 100 beats 2 in a pairwise comparison, it is worse in a group setting. I’m a little dubious of that move actually, but I need to think it through more carefully.

Why would 100 be the uniquely rational choice if you had only the two choices and the opponent had all of the regular ones? If the opponent picks either 2 or 3, you would have been better off picking 2.

But it would definitely be weird if adding “irrelevant alternatives” switched the ordering of two choices.

I would say that for the game theorists, those who picked 100 had 0 iterations, and those who picked 99 had 1 iteration. Those who picked 98 did not have 2 iterations, since that requires knowing that your opponent is rational. But one usually does not “know” a false fact, and 10 of their opponents were not rational.

98 might still be a better choice than 99 in this context though. 98 is superior to 99 by 1 point when facing 99, and by 2 when facing 98; but inferior by 1 when facing 100. So it depends on how many people picked 98, 99, and 100.

I was thinking if A’s choices were {2, 100}, and B’s choices were all of 2 through 100, then the only equilibria are <2, 2> and <100, 99>. But the second equilibria is better by a lot – and rather obviously so – so that’s what they should settle on. Hence A should play 100. (In that restricted setting.)