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Entangled Equilibria and the Twin Prisoners' Dilemma

0Vanessa Kosoy

0Scott Garrabrant

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I think you meant to divide by in the equation for the utility differential?

A quick thought (haven't thought much about this): It seems natural to require that the counterfactual system is involutive (of course for two strategy games the condition is trivial). I don't know what the game theoretic implications would be.

In this post, I present a generalization of Nash equilibria to non-CDT agents. I will use this formulation to model mutual cooperation in a twin prisoners' dilemma, caused by the belief that the other player is similar to you, and not by mutual prediction. (This post came mostly out of a conversation with Sam Eisenstat, as well as contributions from Tsvi Benson-Tilsen and Jessica Taylor)

This post is sketchy. If someone would like to go through the work of making it more formal/correct, let me know. Also, let me know if this concept already exists.

Let A1,…,An be a finite collection of players. Let Mi be a finite collection of moves available to Ai. Let Ui:∏iMi→R be a utility function for player Ai.

Let Δi be the simplex of all probability distributions over Mi. Let Vi denote the vector space of functions v:Mi→R, with ∑m∈Miv(m)=0.

A vector given a distribution in p∈Δi and a vector v∈Vi we say that v is unblocked from p if there exists an ε>0, such that p+εv∈Δi. i.e., it is possible to move in the direction of v from p, and stay in Δi.

Given a strategy profile P=(p1,…,pn) in ∏j≤nΔj, and a vector V=(v1,…,vn) in ∏j≤nVj, we say that V improves P for Ai if

limε→0∑m1∈M1…∑mn∈MnUi(m1,…,mn)(∏j≤npj(mj)+εvj(mj)−∏j≤npj(mj))ε>0.

We call this limit the utility differential for Ai at P in the direction of V.

i.e., V improves P for Ai if Ui is increased when P is moved an infinitesimal amount in the direction of V. (not that this is defined even when the vectors are not unblocked from the distributions)

A "counterfactual gradient" for player Ai is a linear function from Vi to ∏j≤nVj, such that the function into the Vi component is the identity. This represents how much player Ai expects the probabilities of the other players to move when she moves her probabilities.

A "counterfactual system" for Ai is a continuous function which takes is a strategy profile P in ∏j≤nΔj, and outputs a counterfactual gradient. This represents the fact that a player's counterfactual gradient could be different depending on the strategy profile the game ends up in. We fix a counterfactual system Ci for each player.

Claim: There exists a strategy profile P=(p1,…,pn) in ∏j≤nΔj such that for all players Ai, if vi∈Vi is unblocked from pi, then Ci(P)(vi) does not improve P.I will call such a point an entangled equilibrium.

Proof Sketch: (Very sketchy, and I have not verified this. There is probably a better way. It might not be true.)

We will construct a continuous function from ∏j≤nΔj to itself. We do this by moving pi by adding the gradient of function from pi to to Ui, assuming all other players probabilities change according to the linear function Ci(P). If adding this gradient would take the point out of the simplex Δi, you hit a boundary after moving some proportion α of the gradient. Then, now that you are on the boundary, you use move by 1−α times the gradient of the same function restricted to the boundary, and repeat.

This function has a fixed point, by Brouwer. For each player, this fixed point must have 0 gradient, or be on the boundary, pointing outward. If it is on a boundary, it has the same property when restricted to that boundary.

If there was an unblocked direction a player could move that would improve its utility, then the gradient would be nonzero. If the player is on a boundary, the gradient is pointing outward, and there is an unblocked direction that would be an improvement, there will be such a direction that stays on that boundary, and the gradient restricted to that boundary would be nonzero. □

Example 1Consider a twin prisoners' dilemma game. Two players can either cooperate or defect. They get 0 utility for being exploited, 3 for exploiting, 2 for mutual cooperation, and 1 for mutual defection.

Both players believe that the other is using a decision procedure that is entangled with their own, but they are not completely sure.

Formally, both players think that when they increase their probability by ε, the other player increases by 2ε/3, regardless of where the probabilities start.

Thus C1(p,q)(v1)=(v1,2v1)/3, where v1 is the vector (A1 cooperates)-(A1 defects) in V1, and v2 is the analogous vector for player 2. Similarly, C2(p,q)(v2)=(2v2/3,v2).

Since both players always think that increasing their probability of cooperation increases utility, the only equilibrium is when both players cooperate with probability 1.

Example 2Now let look at a case where the gradients are a function of the distributions. Again, both players believe that the other is using a decision procedure that is entangled with their own, but they are not completely sure.

Formally, player 1 thinks that when they increase (or decrease) their probably of cooperation by ε, the other player will increase (or decrease) their probability by δε, where δ is one minus the square of the difference between their two probabilities.

Player 2 on the other hand, thinks the other player will increase (or decrease) their probability by δε, where δ is one minus the absolute value of the difference between their two probabilities.

Thus C1(p,q)(v1)=(v1,(1−(p−q)2)⋅v2), where v1 is the vector (A1 cooperates)-(A1 defects) in V1, and v2 is the analogous vector for player 2. Similarly, C2(p,q)(v2)=((1−|p−q|)⋅v1,v2).

Thus, if player 1 cooperates with probability p, and player 2 cooperates with probability q, then player 1 expects to lose gain (2(1−(p−q)2)−1)ε utility by increasing his probability by ε. This function is only 0 if |p−q|=1/√2. Thus the only entangled equilibria with mixed strategies for player 1 have (p−q)=1/√2. Similarly, the only entangled equilibria with mixed strategies for player 2 have |p−q|=1/2. Thus, there are no mixed strategies for both players in equilibria.

The only pure equilibrium is p=q=1

If p=0, then (2(1−(p−q)2)−1) must be nonpositive, so q must be at least 1/√2. None of these points are equilibria for player 2.

If p=1, then (2(1−(p−q)2)−1) must be nonnegative, so q must be at least 1−1/√2. This is in equilibrium if q=1/2.

If q=0, then p must be at least 1/2, This in in equilibrium if p=1/√2.

If q=1, then p must be at least 1/2, and there is no equilibrium.

Thus, there are three entangled equilibria (1,1), (1,1/2), (1/√2,0).