Lecture 2 - 22/01

Prisoner's Dilemma

If agents do not cooperate, the best (global) outcome possible is missed.

COP 21 Game

• N governments

• 2 actions per states

  • Do not pollute (Cost = 3)

  • Pollute (cost=1 and +1 for everyone)

What is the equilibrium?

Multi-Player Game

• Siimultaneous move games

• n players, each player pick a strat and occurs a loss.

$$l_k(s_1, ...s_n) = l_k(s_k, s_{-k})$$

Goal of the player: Minimize their loss

Zero-Sum Two-player Games

Zero-sum: $\sum_{k=1}^n l_k = 0$

n=2

Action for each players: $i \in [n] = {1,....,n}$ and $j \in [m]$

Game

$$\min_{i\in[n]} \max_{j \in [m]} l_{ij}$$

Mix strategies

For example in the game of rock-paper-scissor

We have probabilities over actions of each player as $p=[p1, p2, .... p_n] \in \Delta_n$and $q=[q1,q2, ..., q_m] \in \Delta_m$

$\Delta_n := {p \in R^n: p_1+...p_n=1, p_i >= 0}$

Payoff: $l(p,q):= E_{i\sim p, j \sim q} [l_{ij}] = p^TLq$

Game: $\min_{p\in \Delta_n} \max_{q \in \Delta_m} p^TLq$

Nash Equilibrium of a Game

Best worst-case move

$$s^* \in \text{NASH} \implies l_k(s^*_k, s^*_{-k}) \leq l_k(s_k, s^*_{-k}) \forall s$$

Theorem

Any game with a finite set of players and a finite set of strategies has a Nash equilibrium of mixed strategies.

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