When solving for Player 2's q, whose payoffs are made equal?
When solving for Player 2's mixing probability , Player 1's expected payoffs are made equal. Specifically, you equalize Player 1's expected payoff from choosing their pure strategy R with Player 1's expected payoff from choosing their pure strategy L. This is because Player 2's mixture must make Player 1 indifferent between their own actions in order for Player 1 to be willing to randomize.
Conditions
- The context is finding a mixed-strategy Nash equilibrium.
- Player 2 is the one whose mixing probability is being solved for.
- Player 1 is the opponent who must be made indifferent.
Reasoning, step by step
- Identify the player whose mixing probability is being solved for (Player 2).
- Identify the opponent (Player 1).
- Calculate the opponent's (Player 1's) expected payoff for each of their pure strategies.
- Set the opponent's expected payoffs equal to each other.
- Solve the equation for the mixing probability .
Example
To find , the board computes and , and then sets to get . Player 1's payoffs are equalized, not Player 2's.
Common misconceptions
- A common error is to use Player 2's own payoffs when solving for Player 2's . The indifference principle requires making the *other* player indifferent.
- Confusing which player's payoffs are used to solve for which player's mixing probability.
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