Symmetric $3 imes 3$ simultaneous-move game with full-support and support-restricted mixing

Content map: SMU H3 Game Theory Map

Setup

Definition:

Symmetric $3\times 3$ simultaneous-move game with full-support and support-restricted mixing

Players: Two players, Player 1 and Player 2.

Strategies: Each player chooses $A$ , $B$ , or $C$ .

Rules

Start with a symmetric $3\times 3$ payoff matrix; Players simultaneously choose pure strategies or mix over supports.
The player who makes the opponent indifferent over the relevant support reaches mixed equilibrium.
Players move simultaneously.
Diagonal outcomes give payoffs $1$ , $2$ , and $3$ at $(A,A)$ , $(B,B)$ , and $(C,C)$ ; The off-diagonal entries below are a compact formalisation consistent with the visible support-analysis results in the lecture notes.

Payoff Matrix

	A	B	C
A	1, 1	1, 0	0, 0
B	0, 1	2, 2	1, 0
C	0, 0	0, 1	3, 3

Derivation (Best Response Analysis)

Against $A$ , Player 1 compares $1,0,0$ , so $BR_1(A)=A$ .
Against $B$ , Player 1 compares $1,2,0$ , so $BR_1(B)=B$ .
Against $C$ , Player 1 compares $0,1,3$ , so $BR_1(C)=C$ .
By symmetry, $BR_2(A)=A$ , $BR_2(B)=B$ , and $BR_2(C)=C$ .
Let Player 2 mix with probabilities $(p,q,1-p-q)$ on $(A,B,C)$ . Then Player 1’s expected payoffs are:

E[\pi_1(A)]=p+q,

E[\pi_1(B)]=0\cdot p + 2q + (1-p-q)=1-p+q,

E[\pi_1(C)]=3(1-p-q)=3-3p-3q.

Pairwise indifference boundaries are:

E[\pi_1(A)]=E[\pi_1(B)] \iff p=\frac{1}{2},

E[\pi_1(A)]=E[\pi_1(C)] \iff p+q=\frac{3}{4},

E[\pi_1(B)]=E[\pi_1(C)] \iff p+2q=1.

Hence, for $0\le p\le 1$ and $0\le q\le 1-p$ ,

BR_1(p,q)= \begin{cases} A & \text{if } p>\frac{1}{2} \text{ and } p+q>\frac{3}{4}, \\ B & \text{if } p<\frac{1}{2} \text{ and } p+2q>1, \\ C & \text{if } p+q<\frac{3}{4} \text{ and } p+2q<1, \\ \{A,B\} & \text{if } p=\frac{1}{2},\ q>\frac{1}{4}, \\ \{A,C\} & \text{if } p+q=\frac{3}{4},\ p>\frac{1}{2}, \\ \{B,C\} & \text{if } p+2q=1,\ p<\frac{1}{2}, \\ \{A,B,C\} & \text{if } \left(p,q\right)=\left(\frac{1}{2},\frac{1}{4}\right). \end{cases}

By symmetry, if Player 1 mixes with $(r,s,1-r-s)$ , then Player 2’s expected payoffs are:

E[\pi_2(A)]=r+s,

E[\pi_2(B)]=1-r+s,

E[\pi_2(C)]=3-3r-3s.

Hence, for $0\le r\le 1$ and $0\le s\le 1-r$ ,

BR_2(r,s)= \begin{cases} A & \text{if } r>\frac{1}{2} \text{ and } r+s>\frac{3}{4}, \\ B & \text{if } r<\frac{1}{2} \text{ and } r+2s>1, \\ C & \text{if } r+s<\frac{3}{4} \text{ and } r+2s<1, \\ \{A,B\} & \text{if } r=\frac{1}{2},\ s>\frac{1}{4}, \\ \{A,C\} & \text{if } r+s=\frac{3}{4},\ r>\frac{1}{2}, \\ \{B,C\} & \text{if } r+2s=1,\ r<\frac{1}{2}, \\ \{A,B,C\} & \text{if } \left(r,s\right)=\left(\frac{1}{2},\frac{1}{4}\right). \end{cases}

Diagram (Best Response Regions)

Each graph uses the first two probabilities on the axes. The probability on $C$ is the residual term.

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diagram

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diagram

Derivation (Nash Equilibrium)

Pure mutual best responses give

(A,A), \qquad (B,B), \qquad (C,C).

For support $\{A,B\}$ , set the probability on $C$ equal to $0$ and impose

p+q = 1-p+q.

This gives

p=\frac{1}{2}.

Since $q=1-p$ on this support,

p=\frac{1}{2}, \qquad q=\frac{1}{2}.

Check the omitted action:

E[\pi_1(C)]=0<1.

For support $\{A,C\}$ , set the probability on $B$ equal to $0$ and impose

p+q = 3(1-p-q).

Since $q=0$ on this support, this gives

p=\frac{3}{4}, \qquad q=0, \qquad 1-p-q=\frac{1}{4}.

Check the omitted action:

E[\pi_1(B)]=1-\frac{3}{4}=\frac{1}{4}<\frac{3}{4}.

For support $\{B,C\}$ , set the probability on $A$ equal to $0$ and impose

1-p+q = 3(1-p-q).

Since $p=0$ on this support, this gives

q=\frac{1}{2}, \qquad 1-p-q=\frac{1}{2}.

Check the omitted action:

E[\pi_1(A)]=\frac{1}{2}<\frac{3}{2}.

For a fully mixed equilibrium, impose

p+q = 1-p+q = 3-3p-3q.

Solving gives

p=\frac{1}{2}, \qquad q=\frac{1}{4}, \qquad 1-p-q=\frac{1}{4}.

By symmetry, Player 1 uses the same probability vector in each mixed equilibrium.

Nash Equilibrium

Result:

The game has seven symmetric Nash equilibria:

pure equilibria:

(A,A), \qquad (B,B), \qquad (C,C),

two-action mixed equilibria:

\left\{\left(\tfrac{1}{2},\tfrac{1}{2},0\right),\left(\tfrac{1}{2},\tfrac{1}{2},0\right)\right\},

\left\{\left(\tfrac{3}{4},0,\tfrac{1}{4}\right),\left(\tfrac{3}{4},0,\tfrac{1}{4}\right)\right\},

\left\{\left(0,\tfrac{1}{2},\tfrac{1}{2}\right),\left(0,\tfrac{1}{2},\tfrac{1}{2}\right)\right\},

a fully mixed equilibrium:

\left\{\left(\tfrac{1}{2},\tfrac{1}{4},\tfrac{1}{4}\right),\left(\tfrac{1}{2},\tfrac{1}{4},\tfrac{1}{4}\right)\right\},

with strategy order $(A,B,C)$ for both players.

Total payoff at $(A,A)$ is $2$ .
Total payoff at $(B,B)$ is $4$ .
Total payoff at $(C,C)$ is $6$ .
Every off-diagonal outcome gives total payoff $0$ or $1$ .
Therefore $(C,C)$ is the unique social optimum.

Insights

Insight:

The new matrix supports one symmetric equilibrium for every nonempty support subset of $\{A,B,C\}$ .

Support restrictions change the indifference equations, so each two-action support generates a different mixed equilibrium.

The Pareto-dominant equilibrium $(C,C)$ is also the unique social optimum.

Symmetric $3 imes 3$ simultaneous-move game with full-support and support-restricted mixing

Setup

Rules

Payoff Matrix

Derivation (Best Response Analysis)

Diagram (Best Response Regions)

Derivation (Nash Equilibrium)

Nash Equilibrium

Social Optimum

Insights