Prisoner's Dilemma and Centipede Game - what's wrong with this analysis?

Question

This is an analysis which to me seems trivial, but which I very rarely see brought up in any discussion of games like The Prisoner's Dilemma or The Centipede Game which are well known for having 'counterintuitive' 'rational' strategies (the only similar thing I have seen is here). I apologize if any of the below is unclear- I'm not especially familiar with the terminology of game theory.

In each of these games, as they are typically formulated, it is assumed that the players are perfectly rational and that they understand the game. It seems to me as though a rational player would assume that any other rational player would make exactly the same choice they would, since they are themselves rational, and therefore in symmetric games like the Prisoner's Dilemma the only rational choice is to behave cooperatively, as both players choosing to behave cooperatively gives better results for each of them than both of them choosing to behave traitorously.

The centipede game is more complicated, but in the formulation I typically see (two pots, one k * the size of the other, two players, choose to pass and double the pot sizes or take one pot and give the other player the other pot, at most N steps) the only rational strategy is for the players to pass the pot between each other until the number of steps remaining is small enough that the small pot can no longer exceed the large pot's current value and then pass it between them an even number of times. The 'winner' and 'loser' are decided at the game's start by N and k; the game then is about maximizing everyone's profit. The typical argument against this strategy is that at the $n$th step, the choosing player has no incentive not to take the pot, but it ignores that for a rational player to take the pot, taking the pot must be the rational choice, and if it were rational to take the pot now it would have been rational to take the pot at the first opportunity, and if it were rational to take the pot at the first opportunity then they would have taken the pot at the first opportunity, and gotten a much smaller pot, which is clearly not the rational choice. The incentive against taking the pot before the point at which the small pot can no longer reach the size of the large pot is that taking it sooner guarantees a smaller payout than the above strategy. After that point taking the pot out of turn would invalidate the earlier incentive, and so cannot be the rational choice.

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Edit in response to Shane's comment, since it seems like this will come up a lot.

It seems like my argument regarding the prisoner's dilemma is being misconstrued. I will restate it here using hopefully more precise language.

The game is set up so that there are two, rational players, each of whom choose between A (cooperation) or B (betrayal). If both players choose A they each get a large reward. If both choose B they each get a small reward. If one chooses A and the other chooses B, the one choosing A gets no reward and the one choosing B gets a very large reward. Both players have complete knowledge of the game, and can take as long as they'd like to choose between A and B, though they cannot communicate with each other.

The game is completely symmetrical, and there is no way to distinguish between the players, and the players know this. Therefore, each player independently can reason that, if there is a better choice for themselves between A and B, then that same choice is also better for the other player to make. Therefore, if it is better for them to choose B, then it is also better for the other player to choose B, resulting in the small reward for each of them. However, if it is instead better to choose A, then it is also better for the other player to choose A, resulting in the large reward for each of them. Since both players getting a large reward is better for both players than both players getting a small reward, it is better to choose A, and therefore both players choose A.

Answer 1

I'm going to focus on your argument applied to the prisoner's dilemma. For concreteness, take the following bimatrix game:

$$ \begin{matrix} 1\backslash 2 & C & D \\ C & 3,3 & -6,10 \\ D & 10,-6 & 0,0 \end{matrix} $$

Let's get the standard approach straight first before I directly answer your question. The classic solution concept for games such as the one above is called a Nash Equilibrium. For us to be able to define what that is, we first need a notion of a strategy.

In the class of simultaneous move games, such as the one above, a strategy is simply a (possibly degenerate) probability distribution over (a single player's) actions. A collection of strategies, one for each player, is called a strategy profile. (These strategies must be independent of each other.)

Call a strategy a best response if it maximises a player's payoffs given the strategy of the other player(s). Notice that in the prisoner's dilemma, the strategy that plays $D$ with probability $1$ is a best response to any strategy of the other player.

What, then, is a Nash Equilibrium? A Nash Equilibrium is a strategy profile such that each player's strategy is a best response to the other players' strategies specified by the given strategy profile. An alternative characterisation of a Nash Equilibrium strategy profile (that should be straightforward to derive from the previous one) is one where no player can unilaterally change their strategy and strictly increase their payoff by doing so. That is, a strategy profile is a Nash Equilibrium if there does not exist a profitable unilateral deviation for each player.

It is easy to see that the unique Nash Equilibrium of the prisoner's dilemma given above is $(D,D)$ -- the strategy profile where each player plays $D$ with probability $1$.

Ok, now that I have this out of the way, let's move on to your question. What is wrong with the argument you just gave?

One answer might be that your argument is not wrong per se, it just invokes a solution concept different from the one most commonly used in game theory -- Nash Equilibrium. Why is this the most commonly used solution concept? One reason might be is that the structure of Nash Equilibria is well-understood in large classes of games. In particular, for finite games (i.e. finite number of players and actions), a Nash Equilibrium always exists. There are also stronger existence theorems, but the aforementioned one is the one most likely to appear in any given game theory course. Another reason is that, in most simple games, Nash equilibrium is fairly straightforward to compute. Another argument I hear often is that Nash Equilibrium gives us a persuasive and somewhat accurate account of actual behaviour, but I'm personally not sure if this is the case -- I'm not an expert on empirical approaches to game theory. However, I will have something to say about the persuasiveness of this account next.

One objection to your argument (that is admittedly somewhat similar to my first answer, and I imagine you're familiar with) is that your solution requires players to play strategies that are not best responses. If a player believes that the other player is going to choose $C$, then the best thing for them to do is to choose $D$. If they choose $C$, they only get a payoff of $3$, whereas if they choose $D$, they secure a payoff of $10$. Thus, the player should choose $D$.

In fact, your solution requires players to play strategies that are strictly dominated. That is, regardless of the strategy of the other player, there exists an alternative strategy that is strictly better than the one currently being employed. It is a well-known result in epistemic game theory that the common knowledge of rationality implies that players will never play strictly dominated strategies. Hence, for players to play the strategies you propose, the players cannot possess common knowledge of their rationality. Admittedly, this assumption is difficult to adjudicate either way, but personally I think the common knowledge of rationality is a natural assumption when it comes to thinking about games, and it certainly seems natural in the setting you describe. At the very least, it makes certain complicated games much more tractable than they would be otherwise.

Moreover, choosing a dominated strategy is not robust to misspecification of beliefs about the other player, so at minimum, a solution concept that assumes players do not play dominated strategy seems to be desirable.

A key part of your reasoning seems to rely on the claim that rationality and symmetry guarantees that player $1$ chooses $C$ if and only if player $2$ does. As mentioned earlier, this possibility is excluded when we invoke Nash Equilibrium as a solution concept. Under Nash Equilibrium, players' actions cannot be correlated. Moreover, your reasoning seems to rely on a Newcomb-type paradox: it seems to require that each player be able to perfectly predict what the other player does. However, allowing players to do that seems to lead to results that are either paradoxical or not well-defined, which seems like a good reason to assume that players cannot do so.

Answer 2

Note that both players being perfectly rational does not imply both players knowing that the other player is perfectly rational as well. Indeed, since the setup is that the two players cannot communicate, there is no way for one player to figure out the other player's strategy before doing the move.

Note that there is the variant of the repeated prisoner's dilemma, and there the situation is different, exactly because you can learn about the other player's strategy through past actions. And it turns out that in that case, cooperating by default, but answering betrayal from the other side with your own betrayal in the next move, is a much better strategy than just betraying consistently. And if both sides follow that strategy, they will indeed always cooperate.