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How to Calculate a Mixed Strategy Nash Equilibrium

Set one player's expected payoffs from each move equal and solve. That gives the probability mix the other player uses, because a mix is stable only when the rival is indifferent.

The Mixed Strategy Equilibrium formula

Let q = the chance the column player picks Left. Set EV(Up) = EV(Down) and solve for q | Let p = the chance the row player picks Up. Set EV(Left) = EV(Right) and solve for p | q = (Down-Right payoff − Up-Right payoff) ÷ (Up-Left − Up-Right − Down-Left + Down-Right), all read off the row player's payoffs

Calculator

Enter both players' payoffs in a 2x2 game to get each side's equilibrium mix and the payoff it delivers.

The row player's number in the top left cell.

The row player's number in the top right cell.

The row player's number in the bottom left cell.

The row player's number in the bottom right cell.

The column player's number in the top left cell.

The column player's number in the top right cell.

The column player's number in the bottom left cell.

The column player's number in the bottom right cell.

Column player plays Left
50%

Playing Left 50% of the time leaves the row player with nothing to gain by favoring either row.

Row player plays Up
58.33%

Playing Up 58.33% of the time is what leaves the column player indifferent between Left and Right.

Row player's expected payoff
55

Both Up and Down are worth 55 against this mix, which is the check that the equilibrium is right.

Column player's expected payoff
45

Left and Right are both worth 45 against the row player's mix.

Equilibrium check
Both players mix

Both probabilities sit between 0 and 1, so randomizing is stable for each side.

How to calculate Mixed Strategy Equilibrium, step by step

  1. 1
    Draw the payoff matrix. Give the row player two moves, Up and Down, and the column player two moves, Left and Right. Write both numbers in each of the four cells, the row player's payoff first.
  2. 2
    Check that no pure strategy settles it. Look for a cell neither player wants to leave. If one player has a move that beats the other whatever the rival does, that move is dominant and there is nothing to mix over.
  3. 3
    Write the row player's two expected payoffs. Let q be the chance the column player picks Left. Then EV(Up) = q × (Up, Left) + (1 − q) × (Up, Right), and EV(Down) is built the same way from the Down row.
  4. 4
    Set them equal and solve for q. The row player is only willing to mix when both rows pay the same, so EV(Up) = EV(Down) pins down q. Notice that q depends on the row player's payoffs but describes the column player's behavior.
  5. 5
    Repeat from the other side for p. Let p be the chance the row player picks Up, set the column player's EV(Left) equal to EV(Right) using the column player's own numbers, and solve for p.
  6. 6
    Read the equilibrium payoffs. Put q back into either row to get the row player's expected payoff, and p into either column for the column player's. Both rows must give the same answer, which is the check on your arithmetic.

Worked example: Mixed Strategy Equilibrium

A kicker aims Left or Right while the keeper dives Left or Right at the same instant. Payoffs are scoring chances in percent, written (kicker, keeper): kick Left against a Left dive (30, 70), kick Left against Right (80, 20), kick Right against Left (90, 10), kick Right against Right (20, 80). Setting the kicker's expected payoffs equal gives 30q + 80(1 − q) = 90q + 20(1 − q), so 80 − 50q = 20 + 70q, then 120q = 60 and q = 0.5. The keeper dives Left half the time. Doing the same with the keeper's numbers gives 70p + 10(1 − p) = 20p + 80(1 − p), so 60p + 10 = 80 − 60p, then 120p = 70 and p = 0.5833, so the kicker aims Left 58.33 percent of the time. The kicker's expected payoff is 0.5 × 30 + 0.5 × 80 = 55 and the keeper's is 0.5833 × 70 + 0.4167 × 10 = 45.

Mixed Strategy Equilibrium questions

When does a game need a mixed strategy?

When no pair of fixed moves is stable, so every cell has at least one player who wants to switch. Games of pure conflict such as matching pennies or a penalty kick are the standard cases, and John Nash proved every finite game has an equilibrium once mixing is allowed.

Why do my own payoffs set the other player's probabilities?

Because the equilibrium mix has to leave the rival with no better reply. Your payoffs decide which mix by the rival would make you indifferent, and indifference is exactly the condition that stops you from settling on one predictable move.

What if the formula gives a probability above 1 or below 0?

There is no interior mixed equilibrium, which almost always means one player has a dominant move. Go back and check the matrix for a row or column that wins whatever the rival does, then solve the game in pure strategies instead.

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