Mixed-Strategy Anti-Coordination Game

Setup

Definition:

Mixed-Strategy Anti-Coordination Game

• Players: $n$ identical players.
• Strategies: Each player chooses Blonde or Brunette.

Rules

• Start with a group choosing between Blonde and Brunette; Players simultaneously choose one target.
• The player who is alone on Blonde, or avoids crowding on Blonde, gets the preferred outcome.
• Choices are made simultaneously.
• Payoffs satisfy $a>b>c>d$, with the notes using $a=3$, $b=2$, $c=1$, $d=0$; A player choosing Blonde gets $a$ if alone on Blonde and $d$ if at least one other player also chooses Blonde.
• A player choosing Brunette gets $c$ if exactly one other player chooses Blonde and $b$ otherwise.

Payoff Matrix

• For the representative two-player case $(n=2)$, the payoff matrix is:

Mixed Strategy Derivation

• Let $p$ be the probability that each of the other $n-1$ players chooses Blonde.
• Expected payoff from choosing Blonde:

• Expected payoff from choosing Brunette:

• Simplifying:

Derivation (Best Response Analysis)

• If the probability that another player chooses Blonde is low, being the unique player on Blonde is attractive.
• If the probability that another player chooses Blonde is high, congestion on Blonde makes Brunette more attractive.

• In a symmetric mixed equilibrium, each player must be indifferent:

• Rearranging:

Derivation (Nash Equilibrium)

• Asymmetric pure equilibria exist when exactly one player chooses Blonde and all others choose Brunette.
• In that profile:
  • the Blonde player does not want to switch because $3>1$,
  • each Brunette player does not want to switch because switching would create congestion and reduce payoff from $1$ to $0$.
• A symmetric mixed equilibrium is obtained by solving the indifference condition above.
• For $n=2$:

• For $n=3$:

Nash Equilibrium

Result:

The game has:

• asymmetric pure Nash equilibria where exactly one player chooses Blonde,
• a symmetric mixed equilibrium where, in the strategy order (Blonde, Brunette), each player uses the probability vector $(p, 1-p)$, so the equilibrium profile is

$$\{(p,1-p), (p,1-p), \dots, (p,1-p)\},$$

with one vector for each of the $n$ players and with $p$ solving

$$(1-p)^{n-2}\bigl(3 + (n-4)p\bigr) = 2.$$

For the cases highlighted in the notes:

• $n=2$: $\{(\tfrac{1}{2},\tfrac{1}{2}),(\tfrac{1}{2},\tfrac{1}{2})\}$,
• $n=3$: $\{(2-\sqrt{3},\sqrt{3}-1),(2-\sqrt{3},\sqrt{3}-1),(2-\sqrt{3},\sqrt{3}-1)\}$.

Social Optimum

• If nobody chooses Blonde, each player gets $2$, so total welfare is $2n$.
• If exactly one player chooses Blonde, total welfare is $3+(n-1)\cdot 1 = n+2$.
• For $n\geq 3$, $2n>n+2$, so the welfare-maximising outcome is that everyone chooses Brunette.

Insights

Insight:

• The equilibrium problem is driven by congestion on the salient action.
• The film intuition does not imply that one player approaching Blonde is socially optimal.
• Mixed strategies capture uncertainty about who will try for the salient option.