Row player’s expected payoff as the column player’s chance of Left rises
Add your numbers to see the visual breakdown.
Best responses, cell by cell
Each cell’s payoffs, whether each is a best response to the other player’s strategy, and whether the cell is a Nash equilibrium (it is when both are).
Cell
Row payoff
Column payoff
Row: best response?
Column: best response?
Nash equilibrium?
Up–Left
1,000
1,000
No
No
No
Up–Right
200
1,500
No
Yes
No
Down–Left
1,500
200
Yes
No
No
Down–Right
400
400
Yes
Yes
Yes
The row player chooses Up or Down and the column player Left or Right; each cell takes both players’ payoffs. Choose costs for years or losses, where less is better. The default is the OpenStax duopoly: two firms deciding whether to hold output down or increase it.
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What it calculates: Pure-strategy Nash equilibria, Row player’s dominant strategy, Column player’s dominant strategy, Mixed equilibrium: row plays Up.
Two firms that could each earn $1,000 settle at $400, and neither can do better alone
The default game is the duopoly from OpenStax Principles of Economics (Table 10.3). Each firm either holds output down or increases it. If both hold down, each earns $1,000; if both increase, each earns $400; a firm that increases while the other holds down earns $1,500 and leaves the other with $200.
Nash equilibrium: no player gains by changing strategy while the other keeps theirs
Increasing output is the better reply to either choice by the rival: $1,500 beats $1,000 when the rival holds down, and $400 beats $200 when it increases. That makes it a dominant strategy for both firms, and Down–Right, both increasing, is the only Nash equilibrium. A firm that switches alone from there drops from $400 to $200.
Yet both firms would rather be in Up–Left at $1,000 each, and the result says so: another cell is better for both players. That is the prisoner’s dilemma, and it is why a cartel is hard to hold together without a way to punish the firm that cheats.
Duopoly, OpenStax Table 10.3
Rival holds: 1,500 > 1,000, so increase
Rival increases: 400 > 200, so increase
Equilibrium: both increase, $400 each
How to find a Nash equilibrium in a 2 × 2 game: mark each best response, then find the cell with two marks
Take the column player’s choices one at a time. Against Left, compare the row player’s two payoffs in that column and mark the larger; against Right, do the same. Then take the row player’s choices: against Up, compare the column player’s two payoffs in that row and mark the larger; against Down, the same.
A cell with both payoffs marked is a Nash equilibrium: each player is already playing a best response to the other, so neither gains by switching alone. The table under the calculator does this marking for your numbers, cell by cell. Textbooks often call it the underline method.
A tie means both strategies are best responses, so both get the mark. Ties are also where a game can have more equilibria than this page lists: a player who is indifferent can mix in any proportion, which can produce a whole range of partly mixed equilibria along one edge of the table. The result says when your numbers contain a tie.
FigureIncreasing is each firm’s best reply whatever the other does, so both end at $400, though holding down together would pay each $1,000.
A dominant strategy is best against everything; a Nash equilibrium only has to be best against what the other player does
A strategy is dominant when it pays more than the alternative whatever the other player chooses. When both players have one, the game has exactly one Nash equilibrium, where both play it, as in the duopoly. Most games have no dominant strategy and still have equilibria.
Take two drivers choosing a side of the road, scoring 1 each for passing safely and 0 for a collision (enter 1 in Up–Left and Down–Right for both players, 0 elsewhere). Neither side is dominant, yet Up–Left and Down–Right are both Nash equilibria, because once everyone keeps to one side no single driver wants to switch. There is also a mixed equilibrium at 50% each, where both drivers are indifferent and expect 0.5.
A weakly dominant strategy never pays less and sometimes pays more. The result labels it as weak, because it can sit alongside other equilibria in which a player uses the weakly dominated strategy.
When no cell is an equilibrium, each player mixes to leave the other indifferent
In matching pennies the row player wins 1 if the coins match and loses 1 if they differ, and the column player wins when they differ. Every cell has a player who wants to switch, so there is no pure equilibrium. John Nash proved in 1950 that every game with finitely many players and strategies still has at least one equilibrium once players may randomize.
The mix comes from indifference. The column player plays Left with the probability q that makes the row player’s two strategies pay the same, and the row player plays Up with the probability p that does the same for the column player. In matching pennies both are 50%, and each player expects 0.
Each player’s mix is set by the other player’s payoffs, not their own. In the battle of the sexes, two people prefer different events but most of all want to be together: 2 at your favourite, 1 at theirs, 0 apart. The row player goes to their own favourite with probability 66.67%, the column player to the row player’s favourite with 33.33%, and each expects 0.67, less than the 1 that even the worse of the two pure equilibria pays.
Battle of the sexes
q = (1 − 0) ÷ ((2 − 0) + (1 − 0)) = 1/3
p = (2 − 0) ÷ ((1 − 0) + (2 − 0)) = 2/3
Each expects 2/3
The prisoner’s dilemma in years: set payoffs to costs, and confessing is dominant for both
OpenStax tells the story in years in prison (Table 10.2): both silent, 2 years each; both confess, 5 years each; one confesses while the other stays silent, 1 year for the one who confesses and 8 for the one who does not. Years are something each prisoner wants fewer of, so switch “Payoffs are” to costs and enter them as they are.
With Up and Left as “remain silent” and Down and Right as “confess”: against a silent partner, confessing means 1 year instead of 2; against a confessing partner, 5 instead of 8. Confessing is dominant for both, so both confess and serve 10 years between them, where silence would have cost 4.
Entering the years as negative numbers under gains gives the same answer. What does not work is entering them as positive gains: the calculator would then treat the 8-year sentence as the best outcome.
OpenStax Table 10.2, as costs
Partner silent: 1 < 2, so confess
Partner confesses: 5 < 8, so confess
Equilibrium: both confess, 10 years in total
What a 2 × 2 table leaves out: timing, repetition and more than two choices
This page solves simultaneous one-shot games, where both players choose at once and only once. When one player moves first and the second sees the move, the game is solved by working backwards from the end, which can rule out some of the equilibria found here.
Repetition changes the prisoner’s dilemma. Firms that meet every quarter can sustain cooperation with strategies such as matching whatever the rival did last time, because cheating today costs cooperation tomorrow. A one-shot table cannot show that.
A game with three or more strategies needs a larger matrix. The best-response marking works the same way; the mixed-equilibrium formulas here are specific to two strategies each. And an equilibrium is a prediction about stable play, not a recommendation: it does not say which of several equilibria people will coordinate on.
Frequently Asked Questions
What is a Nash equilibrium?
A combination of strategies in which no player can do better by changing their own strategy while the others keep theirs. Each player’s choice is a best response to the others’.
How do you find the Nash equilibrium of a 2×2 game?
Mark the row player’s better payoff in each column and the column player’s better payoff in each row. Any cell with both payoffs marked is a pure-strategy Nash equilibrium. If there is none, find the mixed equilibrium by making each player indifferent.
Can a game have no Nash equilibrium?
It can have none in pure strategies, as matching pennies does. But Nash proved in 1950 that every game with a finite number of players and strategies has at least one equilibrium once mixed strategies are allowed.
Can there be more than one Nash equilibrium?
Yes. A coordination game, such as choosing which side of the road to drive on, has two pure equilibria and a third, mixed one. The theory does not say which one players will reach.
Is a Nash equilibrium always the best outcome?
No. In the prisoner’s dilemma the only equilibrium leaves both players worse off than another cell. Equilibrium means no one gains by switching alone, not that the outcome is efficient.
What is the difference between a dominant strategy and a Nash equilibrium?
A dominant strategy is best whatever the other player does. A Nash equilibrium only requires each strategy to be best against the other player’s actual choice. Every dominant-strategy outcome is a Nash equilibrium, but most equilibria involve no dominant strategy.
How do I enter years in prison or losses?
Set “Payoffs are” to costs and enter the numbers as they are, so fewer years counts as better. Entered as gains, the longest sentence would be treated as the best outcome.
Sources & References
Figures on this page are checked against primary, authoritative sources. Links open in a new tab.
Both players choose at the same time, once, and both know the whole payoff table.
Each player cares only about their own payoff, and the numbers already include everything that matters to them.
Read the mixed equilibrium only with cardinal payoffs: an expected value means something only if a payoff of 2 is twice as good as 1.
Where it stops
Two strategies per player only; a larger game needs a larger matrix.
Moves in sequence and repeated play are not modelled, and both can remove or add equilibria.
With tied payoffs there can be a continuum of partly mixed equilibria, which the result mentions but does not list.
An equilibrium predicts stable play; it does not say which of several equilibria players will reach.
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Published the Nash equilibrium calculator for 2×2 games: every pure-strategy equilibrium by best responses, each player’s dominant strategy, the mixed-strategy equilibrium from the indifference conditions, and a flag when another cell is better for both players (the prisoner’s dilemma).
Checked against OpenStax Principles of Economics Tables 10.2 and 10.3, and against the definition of equilibrium itself in 4,000 random games.
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