How to find a Nash equilibrium in a payoff matrix

Reading a payoff matrix

Rows are one player's strategies, columns the other's, and every cell holds what each player earns when that pair meets. Textbooks write both payoffs in one cell, such as (28, 28). In a spreadsheet it is easier to keep two tables with the same layout, one per player, so each can be checked with ordinary formulas. Here Crema chooses the row and Bean the column, and the profits are in $ thousands per week.

Step 1: best responses

A best response is the strategy with the highest payoff given what the other player does.

• For the row player (Crema), the rival's choice is the column, so its best response is the maximum of each column of Crema's table: =IF(B2=MAX(B$2:B$4),1,0). The mixed reference B$2:B$4 keeps the rows fixed and lets the column change as you copy right.
• For the column player (Bean), the rival's choice is the row, so its best response is the maximum of each row of Bean's table: =IF(B6=MAX($B6:$D6),1,0).

In the example, Crema's best reply to Bean Low is Low (28), to Bean Mid also Low (46) and to Bean High it is Mid (58). Bean's best replies mirror these, because the game is symmetric.

Step 2: cells where both best responses meet

Multiply the two flag tables cell by cell: =B10*B14. A 1 marks a cell where both players are already playing a best response, which is exactly a pure Nash equilibrium. =SUM(B18:D20) counts them. Here there is one, Crema Low and Bean Low. You can also write it in one step with =IF(AND(B2=MAX(B$2:B$4),B6=MAX($B6:$D6)),1,0), see nested IF vs IFS for more on combining conditions. A formula rule in conditional formatting can colour the equilibrium cells like the circles in a textbook.

Worked example: a 3×3 pricing game between two coffee chains

Rows 2 to 8 are the two payoff tables (yellow). Rows 10 to 16 flag each player's best responses, rows 18 to 20 the cells where both meet, and rows 21 to 24 count the equilibria and read their payoffs.

Worked example: a 3×3 pricing game between two coffee chains

	A	B	C	D
1	Crema's profit ($k)	Bean Low	Bean Mid	Bean High
2	Crema Low	28	46	54
3	Crema Mid	22	40	58
4	Crema High	12	30	50
5	Bean's profit ($k)	Bean Low	Bean Mid	Bean High
6	Crema Low	28	22	12
7	Crema Mid	46	40	30
8	Crema High	54	58	50
9	Crema's best response? (1 = yes)	Bean Low	Bean Mid	Bean High
10	Crema Low	1	1	0
11	Crema Mid	0	0	1
12	Crema High	0	0	0
13	Bean's best response? (1 = yes)	Bean Low	Bean Mid	Bean High
14	Crema Low	1	0	0
15	Crema Mid	1	0	0
16	Crema High	0	1	0
17	Nash equilibrium? (1 = yes)	Bean Low	Bean Mid	Bean High
18	Crema Low	1	0	0
19	Crema Mid	0	0	0
20	Crema High	0	0	0
21	Number of pure Nash equilibria			1
22	Crema's payoff in equilibrium ($k)			28
23	Bean's payoff in equilibrium ($k)			28
24	Highest joint profit in the matrix ($k)			100

Crema and Bean are fictional and every payoff is illustrative. The payoff readings in rows 22 and 23 assume a single equilibrium. Educational only, not business or investment advice.

Why the equilibrium is not the best outcome

Both chains would earn $50k at High, High, the highest joint profit in the matrix ($100k). But it is not stable: if Bean charges High, Crema earns more by switching to Mid (58 instead of 50), and so does Bean. The undercutting continues until both sit at Low and earn $28k. This is the logic of the prisoner's dilemma: individually rational choices lead to an outcome both would like to avoid. A Nash equilibrium is a prediction of stable behaviour, not a recommendation of what is best for the players together. In repeated games, the threat of future price wars can sustain higher prices, which is why the discount factor matters.

Dominated strategies: a shortcut that agrees

A strategy is strictly dominated when another one pays more against every choice of the rival. Mid dominates High for both chains (22 > 12, 40 > 30, 58 > 50 for Crema). Once High is removed, Low beats Mid in both remaining columns (28 > 22, 46 > 40), so iterated elimination of dominated strategies also lands on Low, Low. Elimination does not always shrink a game to one cell, but the best-response method always finds every pure equilibrium.

No pure equilibrium, or several

Some games have no cell marked twice, like matching pennies or the timing of promotions where each player wants to surprise the other. Then the equilibrium is in mixed strategies: each player randomises with probabilities that make the other indifferent between its options. Every finite game has at least one Nash equilibrium once mixed strategies are allowed. Other games, such as choosing locations or technical standards, have several pure equilibria, and which one is reached depends on who moves first, history or convention.

Common mistakes

• Looking at the wrong direction. The row player's best responses are column maxima, the column player's are row maxima. Swapping them finds the wrong cells.
• Comparing payoffs across players. Each player compares only its own payoffs. Crema's 58 and Bean's 30 in the same cell are never compared with each other.
• Confusing equilibrium with the best joint outcome. High, High pays most in total but is not an equilibrium because each player gains by deviating.
• Stopping at the first equilibrium. Check every cell: a game can have several pure equilibria, or none.
• Wrong anchors in the MAX range. Use B$2:B$4 for column maxima and $B6:$D6 for row maxima, or the flags shift when copied.
• Ignoring ties. If two strategies give the same maximum, both are best responses, and both cells can be part of an equilibrium.