How do you calculate the probability p for Player 1's mixed strategy in a Nash equilibrium?

Conditions

• The game is a finite normal-form game with a 2x2 payoff matrix.
• Player 1 chooses R with probability $p$ and L with probability $1-p$.
• Player 2's expected payoffs from playing R and L must be calculated using Player 1's mixed strategy.
• The solution for $p$ must lie in the interval $[0, 1]$ to represent a valid mixed strategy.

Reasoning, step by step

1. Identify Player 2's payoffs for each outcome in the payoff matrix.
2. Calculate Player 2's expected payoff if they choose strategy R, denoted as $\pi_2(R)$, by weighting Player 2's payoffs in the R column by Player 1's probabilities $p$ and $1-p$.
3. Calculate Player 2's expected payoff if they choose strategy L, denoted as $\pi_2(L)$, by weighting Player 2's payoffs in the L column by Player 1's probabilities $p$ and $1-p$.
4. Set the two expected payoffs equal to each other: $\pi_2(R) = \pi_2(L)$.
5. Solve the resulting algebraic equation for $p$.

Example

In the displayed game, Player 2's expected payoff from choosing R is $\pi_2(R) = p(1) + (1-p)(0) = p$. Player 2's expected payoff from choosing L is $\pi_2(L) = p(0) + (1-p)(2) = 2 - 2p$. Setting them equal gives $p = 2 - 2p$. Adding $2p$ to both sides yields $3p = 2$, and dividing by 3 gives $p = 2/3$.

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.
• A common error is to use Player 1's own payoffs when solving for Player 1's mixing probability $p$. The mixing probability $p$ is chosen to make Player 2 indifferent, so Player 2's payoffs must be used.

Watch the explanation

Explore next

• What is the formula for expected payoff when one player uses a mixed strategy?
• How do you find Player 2's mixing probability q in a mixed-strategy Nash equilibrium?
• Why do we set the two expected payoffs equal when solving for the opponent's mix?