Why do we set the two expected payoffs equal when solving for the opponent's mix?

Conditions

• The game is in a mixed-strategy Nash equilibrium.
• The player is randomizing over multiple pure strategies.
• The expected payoffs for the player's pure strategies are calculated based on the opponent's mixed strategy.

Reasoning, step by step

1. Identify the player who is randomizing over their pure strategies.
2. Calculate the expected payoff for each of that player's pure strategies given the opponent's mixed strategy.
3. Set the expected payoffs of all pure strategies being mixed equal to each other.
4. Solve the resulting equations for the opponent's mixing probabilities.

Example

To find Player 1's mixing probability $p$, we set Player 2's expected payoffs equal: $\pi_2(R) = \pi_2(L)$. This ensures Player 2 is indifferent between R and L, which is required for Player 2 to be willing to mix their strategies in equilibrium.

Common misconceptions

• Students might intuitively think a player should choose probabilities to maximize their own expected payoff directly, rather than to make the opponent indifferent.
• Thinking that each player's own payoff is used to solve that player's mixing probability. Actually, Player 1's mix is chosen to make Player 2 indifferent.

Watch the explanation

Explore next

• How do you calculate the probability p for Player 1's mixed strategy in a Nash equilibrium?
• How do you find Player 2's mixing probability q in a mixed-strategy Nash equilibrium?