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This question is an extension of the Centipede Game.

My prof. posed this to me in class and I can't figure out how to approach this problem.

Imagine in this game, there is an alternative possibility (with very low probability, 0.0001, 0.0000001, etc.) that the first player is irrational, and always goes right instead of down. If this were true, it would make sense for player 2 to always go right (except at the last node) to achieve the highest payoff.

The tricky part for me is: now assume that P1 and P2 are both rational, and that they know that each other are rational, but P1 does not know that P2 knows that P1 is rational — how does that affect the game (with regards to the equilibrium outcome)?

Backwards induction is pretty straightforward in the original form of the game, is there an alternative method that can be better used for this incomplete information version?

(I was thinking of assigning probabilities — starting from the second-to-last node — that would make the payouts equivalent for P2's response to an irrational/rational version of P1, but that seems quite tedious for the ~100 nodes, and I feel there must be come generalized approach to this.)

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What does rational mean in this context? What is the equilibrium notion? –  Michael Greinecker May 12 '12 at 18:24

2 Answers 2

There is no easy way to answer. But consider this strategy profile $s$: pass at round $t$ if the probability the other is irrational is greater than $p_t$.

  1. If players follow the the strategy profile $s$ then by using Bayes' rule you can compute the probability that that a player is rational conditional round $t$ was reached when they play accordingly to $s$, call it $\pi_t$.
  2. Given $\pi_t$ you can find the player best response, call it $\hat{s}$.
  3. You want to solve the "fixed-point" problem where $s=\hat{s}$.

It is a good idea to read the gang of four paper: Kreps, David, Paul Milgrom, John Roberts, and Robert Wilson (1982). "Rational Cooperation in the Finitely‑Repeated Prisoners’ Dilemma". Journal of Economic Theory 27 (2): 245‑252. doi:10.1016/0022-0531(82)90029-1.

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We know that if it is common knowledge that both players are rational then player one will always play down at the start.

Dropping common knowledge, if with probability p>0 player 1 irrationally always plays right and player 2 is rational and knows this, then on the occasion that player 1 is irrational they reveal this information to player 2 on the first move at which point it is rational for player 2 to continue playing right until the right most terminal node is reached.

I had something for this last part following from above but it wasn't quite right.

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That is not true. It is consistent with rationality of both players that people stay in for quite some time in the absence of common knowledge. So this will not signal that player $1$ is irrational. –  Michael Greinecker May 12 '12 at 18:49

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