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I'll phrase my confusion in the context of the Prisoner's Dilemma (although it is applicable to other situations).

There are two "rational" players $A$ and $B$, each faced with a decision between options $X$ and $Y$. The payoff matrix is,

$$ \begin{array}{|c|c|c|} \hline (A\text{ payoff},B\text{ payoff}) & A\text{ chooses }X & A\text{ chooses }Y\\ \hline B\text{ chooses }X & (a,a) & (b,c) \\ \hline B\text{ chooses }Y & (c,b) & (d,d) \\ \hline \end{array} $$

Where $b > a > d > c$. As the usual reasoning goes, if player $A$ chooses $X$, then player $B$ should choose $Y$ since $b > a$. If player $A$ chooses $Y$, then player $B$ should again choose $Y$, since $d > c$. Thus, player $B$ should always choose $Y$; by similar reasoning, player $A$ should always choose $Y$.

My confusion comes from the fact that the situation is symmetric. Since $A$ and $B$ are both rational players in a symmetric situation, if there exists a single best choice, that choice must be the best choice for both players. From what I understand, in game theory it is assumed that all players have shared information and all players know that the other players are rational. But this would imply that $A$ and $B$ have the information necessary to conclude that they will both make the same choice. Thus, they are now faced with two outcomes: choose $X$ and receive payoff $a$, or choose $Y$ and receive payoff $d$. Since $a > d$, they would both choose $X$.

What is wrong with this reasoning? Are they not given the knowledge that all other players are rational? Are they somehow forbidden from "meta" reasoning like this; and if so, what constitutes "meta"?

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  • $\begingroup$ If i'm not mistaken, is this the Nash equilibrium? $\endgroup$
    – Bumblebee
    Nov 14, 2021 at 7:54
  • $\begingroup$ The Nash equilibrium is both players choosing $Y$; I'm confused why, if they can deduce that they will both make the same choice, they would still choose $Y$. $\endgroup$
    – Vedvart1
    Nov 14, 2021 at 7:55
  • $\begingroup$ Because both will lose then. Think of it as a game of split and steal. If they both choose to split they each get money but if the both steal they get nothing. They choose to both steal so the other get's nothing, vice-versa. The question to ask is why let the other win if I can't win? $\endgroup$
    – Bumblebee
    Nov 14, 2021 at 7:57
  • $\begingroup$ Yes, if they both split, they get money; if they both steal, they get nothing. As far as I can tell, they can deduce that these are the only two options - it is not possible for one person to split while the other steals. Thus, in this scenario, would they not both choose split, since that is the option with better payoff for them? $\endgroup$
    – Vedvart1
    Nov 14, 2021 at 7:59
  • $\begingroup$ That depends entirely if they know each other, if they are friends they will split but if they are not friends but strangers how do you guarantee a split? The safest option would be to steal. $\endgroup$
    – Bumblebee
    Nov 14, 2021 at 8:01

1 Answer 1

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A rational player is seeking to maximize their own play, without concern for how it affects others. Indeed, rational play pushes the players toward $(d,d)$, because there's no reason not to -- whether I'm guessing what player A will do, or player A acts first and explicitly reveals their action, I'm better off choosing Y.

Rational players don't see a reason to cooperate unless the incentive to do so is greater (ie, if $a>c$ and $b>d$).

What you're thinking about, with this "meta reasoning" discussion, is what's known as a hyper-rational player. Hyper-rational players can look at outcomes based on how they affect everyone at the table, and can consider whether looking out for everyone is a better strategy. This act of "looking at everyone" is needed to escape the mutually assured destruction of $(d,d)$, and allows the hyper-rational player to consider that if everyone plays X, everyone is better off in the long run.

The reason that rational players won't do the same thing is because the hyper-rational move isn't a stable one. Once you know that everyone will agree to play X, the rational player realizes that if he's the only one that plays Y, he will cash in big... (and so on, and so on.)

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  • $\begingroup$ Is "looking out for everyone" really required though? If a player in a perfectly symmetric game deduces that everyone will make the same choice because they are using the same decision algorithm, then couldn't they simply eliminate all of the non-diagonal entries of the payoff matrix, since they are unreachable? In my thinking, any given player wouldn't switch to $Y$ for the greater payoff, because that greater payoff is a mirage of sorts - if they switch, then by the nature of the decision, all other players switch, and the player knows this will happen. $\endgroup$
    – Vedvart1
    Nov 14, 2021 at 8:22
  • $\begingroup$ The problem in your logic is that you can't make other players switch by virtue of your own decision to switch. Rather, because everyone is using the same algorithm, they have already made a decision, and you are forced to comply with them to follow the same algorithm. I'm a rational player. I know that no matter what you play, I'm better off playing $d$. You are also a rational player; how will you respond? Deciding to play $c$ doesn't change my mind, and doesn't benefit you; you'll end up playing $d$ as well. $\endgroup$
    – Vaekor
    Nov 29, 2021 at 14:06

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