Nash Equilibrium in a Two-by-Two Game

Type the eight payoffs. The tool marks each player's best reply on the cells themselves — a cell marked twice is an equilibrium — then solves the indifference conditions for the mixed one.

Worked example — Nash Equilibrium in a Two-by-Two Game

The game	Chicken
Equilibria	3
Row plays Top	33.3%
Column plays Left	33.3%
Mixed equilibrium pays	1.67 / 1.67
Both play Bottom	44.4%

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Lesson

The theory — Nash Equilibrium in a Two-by-Two Game

A Nash equilibrium is a pair of strategies, one for each player, where neither can do better by changing alone. It is a statement about unilateral moves only: it does not say the outcome is good, fair, or the best either player could hope for. A mixed strategy is a probability over the two moves, and a mixed equilibrium is the case where each player is deliberately unpredictable — which they can only be at one exact frequency.

What each symbol means

p: how often the row player chooses their top row
q: how often the column player chooses their left column

Assumes: That both players know the whole payoff matrix, choose at the same moment, and cannot promise each other anything. Every one of those can be relaxed and each relaxation is a different branch of game theory — sequential games, incomplete information, cooperative games — with different answers to the same eight numbers.

Your own probabilities are none of your business 🖖

Look at the rows for Row plays Top and Column plays Left. The equation that fixes how often the row player plays Top contains only the column player's four numbers, and the one that fixes the column player's frequency contains only the row player's. That is what mixing is for: you choose your frequencies so that your opponent has no reason to prefer either of their options, so the equation you are solving is theirs and not yours. Type over one of your own payoffs, leaving your opponent's alone, and watch your own percentage sit there unmoved — right up to the point where your change destroys the mixed equilibrium altogether and the row reads none. On the penalty kick the two matrices are locked together, because every point the kicker gains is one the keeper loses, so a kicker who really does get better changes both of them and his own mix moves after all: take his unguarded shot from 95 to 99 and drop the keeper's 5 to 1 to match, and he goes from shooting Top 36.4% of the time to 33.9%.

Getting better at something can leave you worse off 🖖

Press Chicken, then Chicken, rewarded for nerve. One number changed, and it belongs to the row player: holding their nerve against a swerver now pays 12 instead of 5. Their own behaviour does not budge — they still swerve 33.3% of the time — but their equilibrium payoff falls from 1.67 to 1.20, a drop of 28%. What moved is the other driver, down from swerving 33.3% of the time to 10.0%. With a prize that big, a one-in-ten chance of the other car yielding is already enough to make driving straight worth it, and that is therefore all the yielding the equilibrium leaves room for. Both cars now drive straight 60.0% of the time where it used to be 44.4%. The reward for being better at brinkmanship is an opponent who calls your bluff more often, more crashes, and a smaller payoff — while the column player's own payoff does not move at all.

One or three, never two 🖖

Set the eight payoffs so that no player is left exactly indifferent, then count the equilibria: you get one or three. Sweep all 65,536 two-by-two games whose payoffs are whole numbers from 0 to 3 and it comes out 31,968 with one and 7,968 with three. There are 25,600 with an even count, and every single one of them contains a tie — two payoffs a player is equally happy with, so a whole edge of the square becomes a best reply and the paths in the diagram overlap instead of crossing. Type 2 into all eight boxes and you will see the extreme case: every cell an equilibrium, four of them. Robert Wilson proved in 1971 that almost every finite game has an odd number of equilibria, and the diagram shows you why. Two paths that each run from one side of a square to the other have to meet, and they meet an odd number of times unless they lie on top of one another.

Problem solved in full

What does mixing actually buy you? The tool prints the equilibrium frequencies, and it is easy to read them as a way of exploiting the other player. They are not. Work out what the row player earns when the column player abandons the equilibrium entirely, in each of two games.

Open this problem in the tool
1. Take the penalty kick. Write the kicker's expected scoring chance as a function of both mixes, with p for shooting left and q for the keeper diving left.
  
  $E_{kick} = p (60 q + 95 (1 - q)) + (1 - p) (90 q + 70 (1 - q))$
2. Collect the q terms. The kicker's chance depends on the keeper only through the bracket multiplying q, and at p = 20/55 that bracket is exactly zero.
  
  $E_{kick} = 70 + 25 p + q (20 - 55 p)$
3. So at his equilibrium mix the kicker scores 79.09% whatever the keeper does — dive left every time, dive right every time, flip a coin, it makes no difference. That is a guarantee, and it is what he is buying.
  
  $p = \frac{20}{55} ⟹ 20 - 55 p = 0, E_{kick} = 70 + 25 \cdot \frac{20}{55} = 79.09%$
4. Now the same calculation for Chicken, where the two payoff matrices are not tied together.
  
  $E_{row} = p (3 q + 1 (1 - q)) + (1 - p) (5 q + 0) = p + q (5 - 3 p)$
5. At p = 1/3 the bracket is 4, not 0. The row player's payoff still depends entirely on the other driver: 0.33 if they never swerve, 4.33 if they always do.
  
  $p = \frac{1}{3} ⟹ 5 - 3 p = 4, E_{row} = \frac{1}{3} + 4 q \in [\frac{1}{3}, 4 \frac{1}{3}]$
6. The reason is that a penalty kick is constant-sum: every point the kicker gains is one the keeper loses, so making the keeper indifferent and making your own payoff safe are the same act. Chicken's two matrices are independent, its cells sum to 6, 6, 6 and 0, and the two come apart.
  
  $A_{r c} + B_{r c} = 100 \forall r, c vs A_{r c} + B_{r c} \in {6, 6, 6, 0}$
Answer

In a constant-sum game the equilibrium mix is a guarantee. The kicker's 79.09% holds against every keeper, including one who has read this page, which is von Neumann's minimax theorem and the reason professional footballers became a research subject: Ignacio Palacios-Huerta estimated the scoring chances from league penalty kicks, worked out the minimax frequencies they imply, and could not reject the hypothesis that the players were using them. Chicken has no such guarantee. The row player who mixes 1/3 earns anywhere from 0.33 to 4.33 depending on a driver they do not control, and 1.67 is merely what arrives if that driver plays their part too. Constant-sum is enough for the guarantee but it is not what the guarantee requires — set the row player's payoffs to 4, 0, 0, 4 against the column player's 7, 2, 3, 8 and the cells sum to 11, 2, 3 and 12 while the row player is still pinned at exactly 2.00 whatever the column player does. What the guarantee actually needs is for the mix that makes your opponent indifferent to be the same mix that cancels their influence on you, and constant-sum forces those two to coincide.

$\forall q 79.09% vs 0.33 \to 4.33 \frac{1}{3} + 4 q$

Learning path

The best reply, and what happens when nobody computes one

References (3)

Two pages, and the reason every game on this tool has an equilibrium at all: J. F. Nash, "Equilibrium points in n-person games." Proceedings of the National Academy of Sciences 36(1), 1950, pp. 48–49.
Where insight block 3's "one or three, never two" comes from: R. Wilson, "Computing Equilibria of N-Person Games." SIAM Journal on Applied Mathematics 21(1), 1971, pp. 80–87.
The empirical test the worked problem ends on — whether people actually mix at the equilibrium frequencies: I. Palacios-Huerta, "Professionals Play Minimax." Review of Economic Studies 70(2), 2003, pp. 395–415.

By the 42Math team · Reviewed against published sources. Corrections welcome.