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Mixed Strategy Nash Equilibrium

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Mixed Strategy Nash Equilibrium

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Mixed Strategy Nash Equilibrium

🎯 Learning Objectives

Define a mixed strategy as a probability distribution over pure strategies
Apply the indifference condition to find mixed-strategy NE
Solve for mixed NE in 2×2 games
Interpret mixed strategies in real-world contexts
Understand why randomisation can be optimal

3.1 Intuition: Why Mix?

In Matching Pennies, there's no pure-strategy Nash equilibrium — whatever you choose, your opponent can exploit you. But if you randomise (play Heads with 50% probability), your opponent cannot predict your move.

🔑 Key Insight: The purpose of mixing is to keep your opponent indifferent — making them unable to exploit any pattern in your play. Real-world examples:

Soccer penalty kicks: kickers mix left/right/centre
Poker: players bluff with a certain probability
Tennis serves: mixing direction keeps the opponent guessing

3.2 Formal Definition

A mixed strategy for player

i

is a probability distribution

\sigma_i

over their pure strategy set

S_i

. If

S_i = \{s_{i1}, s_{i2}, \dots, s_{ik}\}

, then

\sigma_i = (p_1, p_2, \dots, p_k)

where

p_j \geq 0

and

\sum p_j = 1

. A mixed-strategy Nash equilibrium is a profile

\sigma^* = (\sigma_1^*, \dots, \sigma_n^*)

such that for every player

i

\mathbb{E}[u_i(\sigma_i^*, \sigma_{-i}^*)] \geq \mathbb{E}[u_i(\sigma_i, \sigma_{-i}^*)] \quad \forall \sigma_i \in \Delta(S_i)

3.3 The Indifference Condition

Key result: In a mixed-strategy NE, each player must be indifferent between any pure strategy they play with positive probability. If player

i

plays

s_i

and

s_i'

both with positive probability in equilibrium, then:

\mathbb{E}[u_i(s_i, \sigma_{-i}^*)] = \mathbb{E}[u_i(s_i', \sigma_{-i}^*)]

Why?

If one strategy gave strictly higher expected payoff, the player would play it exclusively — contradicting that they mix.

3.4 Solving 2×2 Games

Example: Matching Pennies

\begin{array}{c|cc} & H & T \\ \hline H & (1, -1) & (-1, 1) \\ T & (-1, 1) & (1, -1) \end{array}

Let Player 1 play

H

with probability

p

T

with

1-p

. Let Player 2 play

H

with probability

q

T

with

1-q

. Step 1: Find Player 2's equilibrium mix that makes Player 1 indifferent. Player 1's expected payoff from

H

q(1) + (1-q)(-1) = 2q - 1

Player 1's expected payoff from

T

q(-1) + (1-q)(1) = 1 - 2q

Set equal:

2q - 1 = 1 - 2q \implies 4q = 2 \implies q = \frac{1}{2}

Step 2: Find Player 1's equilibrium mix that makes Player 2 indifferent. Player 2's expected payoff from

H

p(-1) + (1-p)(1) = 1 - 2p

Player 2's expected payoff from

T

p(1) + (1-p)(-1) = 2p - 1

Set equal:

1 - 2p = 2p - 1 \implies 4p = 2 \implies p = \frac{1}{2}

Mixed-strategy NE:

(\sigma_1 = (\frac{1}{2}, \frac{1}{2}), \sigma_2 = (\frac{1}{2}, \frac{1}{2}))

General Formula for 2×2 Games

\begin{array}{c|cc} & L & R \\ \hline U & (a, e) & (b, f) \\ D & (c, g) & (d, h) \end{array}

Player 1 plays

U

with

p

D

with

1-p

. Player 2 plays

L

with

q

R

with

1-q

. Player 1's indifference:

\mathbb{E}[U] = qa + (1-q)b = qc + (1-q)d = \mathbb{E}[D]

q(a - c) + (1-q)(b - d) = 0

q(a - c - b + d) = d - b

q^* = \frac{d - b}{a - c - b + d}

Player 2's indifference:

\mathbb{E}[L] = pe + (1-p)g = pf + (1-p)h = \mathbb{E}[R]

p^* = \frac{h - g}{e - f - g + h}

Example: Battle of the Sexes

\begin{array}{c|cc} & \text{B} & \text{F} \\ \hline \text{B} & (2, 1) & (0, 0) \\ \text{F} & (0, 0) & (1, 2) \end{array}

Player 1 (row) plays B with

p

\mathbb{E}[B] = q(2) + (1-q)(0) = 2q

\mathbb{E}[F] = q(0) + (1-q)(1) = 1 - q

Set:

2q = 1 - q \implies 3q = 1 \implies q = \frac{1}{3}

Player 2 (column) plays B with

q

\mathbb{E}[B] = p(1) + (1-p)(0) = p

\mathbb{E}[F] = p(0) + (1-p)(2) = 2 - 2p

Set: $p = 2 - 2p \implies 3p = 2 \implies p = \frac{2}{3}

**Mixed NE:** $p = \frac{2}{3}, q = \frac{1}{3}$ Expected payoffs: - P1: $\frac{2}{3} \cdot \frac{1}{3} \cdot 2 + \frac{1}{3} \cdot \frac{2}{3} \cdot 1 = \frac{4}{9} + \frac{2}{9} = \frac{2}{3}$ - P2: $\frac{2}{3} \cdot \frac{1}{3} \cdot 1 + \frac{1}{3} \cdot \frac{2}{3} \cdot 2 = \frac{2}{9} + \frac{4}{9} = \frac{2}{3}$ ## 3.5 Support Identification The **support** of a mixed strategy is the set of pure strategies played with positive probability. To find a mixed NE: 1. **Guess the support** (which strategies are played with positive probability) 2. Apply the **indifference condition** on the support 3. Check that the resulting mix makes **other players indifferent** 4. Verify **no profitable deviation** to strategies outside the support *(Diagram)* ## 3.6 Interpretation of Mixed Strategies | Interpretation | Description | | --- | --- | | **Randomisation** | Player literally flips a coin (penalty kicks) | | **Population** | Proportion of a population playing each strategy (evolutionary) | | **Conjecture** | Opponent's uncertainty about what the player will do | | **Purification** | Mixed strategy represents small unobserved payoff variations | * * * ## 📊 Formula Summary | Concept | Formula | | --- | --- | | Mixed strategy | $\sigma_i \in \Delta(S_i)$ , a probability distribution | | Indifference condition | $\mathbb{E}[u_i(s_i, \sigma_{-i})] = \mathbb{E}[u_i(s_i', \sigma_{-i})]$ for $s_i, s_i'$ in support | | Mixing probability (P1, 2×2) | $q^* = \frac{d - b}{a - c - b + d}$ | | Mixing probability (P2, 2×2) | $p^* = \frac{h - g}{e - f - g + h}$ | * * * ## ✅ Practice Questions **Q1:** Find the mixed-strategy NE:

\begin{array}{c|cc} & L & R \ \hline U & (2, 1) & (0, 0) \ D & (0, 0) & (1, 2) \end{array}

> **Solution** > > This is Battle of the Sexes as above. P1 plays U with $p$, P2 plays L with $q$. > > P1 indifferent: $2q = 1 - q \implies q = \frac{1}{3}$ P2 indifferent: $p = 2 - 2p \implies p = \frac{2}{3}$ > > **Mixed NE:** $(\frac{2}{3}U + \frac{1}{3}D, \frac{1}{3}L + \frac{2}{3}R)$ Expected payoff for each: $\frac{2}{3}$ **Q2:** Find the mixed-strategy NE for Chicken:

\begin{array}{c|cc} & S & D \ \hline S & (0, 0) & (-1, 1) \ D & (1, -1) & (-5, -5) \end{array}

> **Solution** > > P1 plays S (Swerve) with $p$, D (Dare) with $1-p$. P2 plays S with $q$, D with $1-q$. > > P1 indifferent: $\mathbb{E}[S] = q(0) + (1-q)(-1) = -1 + q$ $\mathbb{E}[D] = q(1) + (1-q)(-5) = q - 5 + 5q = 6q - 5$ Set: $-1 + q = 6q - 5 \implies 4 = 5q \implies q = \frac{4}{5}$ > > P2 indifferent (by symmetry): $p = \frac{4}{5}$ > > **Mixed NE:** $(\frac{4}{5}S + \frac{1}{5}D, \frac{4}{5}S + \frac{1}{5}D)$ **Q3:** Find all NE (pure and mixed):

\begin{array}{c|cc} & L & R \ \hline U & (3, 2) & (0, 0) \ D & (0, 0) & (2, 3) \end{array}

> **Solution** > > Pure NE: (U, L) and (D, R) — check by underlining. > > Mixed: P1 indifferent: $3q = 2(1-q) \implies 3q = 2 - 2q \implies 5q = 2 \implies q = \frac{2}{5}$ P2 indifferent: $2p = 3(1-p) \implies 2p = 3 - 3p \implies 5p = 3 \implies p = \frac{3}{5}$ > > **Three NE:** (U, L), (D, R), and $(\frac{3}{5}U + \frac{2}{5}D, \frac{2}{5}L + \frac{3}{5}R)$ [Join Discord](https://discord.gg/gE2m4Qrdqv) [Previous**Nash Existence**](/notes/04-degree-electives-bsms4023-game-theory-week02-02b-nash-existence)[Next**Extensive-Form Games**](/notes/04-degree-electives-bsms4023-game-theory-week04-04-extensive-form-games)