Solving 2×2 Games by Hand: Best Responses, Pure Equilibria, and Mixing

You already know the Prisoner's Dilemma as a story. Our goal is to turn that kind of story into a calculation you can carry out by hand. Two people choose actions, their choices select one outcome, and we want every outcome where neither person benefits by changing alone. The method has four stages. Read the payoff pairs, mark each player's best replies, keep every cell with both marks, and if no cell survives, ask whether randomized choices can create an equilibrium. Start with the object the whole method uses: a payoff matrix. Row has two actions, Top and Bottom. Column has two actions, Left and Right. One action from each player selects exactly one of the four inner cells. Each inner cell contains an ordered pair of numbers. The first number is Row's payoff, and the second number is Column's payoff. In the upper-left cell, Row receives four and Column receives three. A payoff can mean money, points, votes, years of freedom, or simply a ranking. The scale depends on the story. For solving the game, each player compares only that player's own numbers, and a larger number means a preferred outcome. The players choose without first observing the other's current choice. Still, to analyse incentives we ask a conditional question. If I knew which action the other player had chosen, which of my actions would give me the largest payoff? That conditional answer is called a best response. Begin with Row and suppose Column chooses Left. Compare Row's first numbers in the Left column: four from Top and two from Bottom. Four is larger, so Top is Row's best response to Left. Now suppose Column chooses Right. Row compares zero from Top with three from Bottom. Three is larger, so Bottom is Row's best response to Right. We mark that first payoff in red as well. Now change viewpoints. Hold Row fixed on Top and compare Column's second numbers across that row. Column gets three from Left and one from Right. Left is better, so the second payoff three receives Column's blue mark. Hold Row fixed on Bottom. Column compares zero from Left with four from Right. Right is better, so the four receives the second blue mark. Row's comparisons ran down columns; Column's comparisons ran across rows. Now inspect whole cells. The upper-left cell carries Row's red mark and Column's blue mark. The lower-right cell also carries both. At either outcome, each player's chosen action is a best response to the other player's chosen action. A cell with both marks is a pure-strategy Nash equilibrium. Pure means that each player chooses one action with certainty. Nash equilibrium means that, holding the other player's action fixed, neither player gains by changing alone. Check the upper-left outcome directly. If Column stays Left, Row falls from four to two by switching. If Row stays Top, Column falls from three to one by switching. Neither unilateral change helps. The same check works at the lower-right outcome. Row would fall from three to zero by switching, and Column would fall from four to zero. A game can therefore have more than one pure equilibrium, and we must keep every double-marked cell. There is one small rule about ties. If two available actions give the same maximum payoff against an opponent's action, both are best responses. Mark both. Never break a payoff tie merely to force one answer. Now put the familiar Prisoner's Dilemma into this formal language. Each prisoner chooses Cooperate or Defect. Mutual cooperation gives two each. A lone defector receives three while the cooperator receives zero, and mutual defection gives one each. Mark Row's best responses first. Against Column's cooperation, Row prefers three from defecting to two from cooperating. Against Column's defection, Row prefers one from defecting to zero from cooperating. Both red marks land in the Defect row. Now mark Column's best responses using the second numbers. Against Row's cooperation, Column receives three by defecting. Against Row's defection, Column receives one by defecting. Both blue marks land in the Defect column. Only the lower-right cell has both marks, so mutual defection is the unique pure Nash equilibrium. Mutual cooperation gives both players more, but either player can gain individually by defecting while the other cooperates. That distinction matters. Equilibrium is a claim about incentives against one-player deviations, not a claim that the outcome is fair, cooperative, or jointly best. We now have a mechanical test for pure equilibria. Next we need a game where that test leaves no cell at all.

Matching Pennies is the smallest game in which the pure-equilibrium search fails completely. Each player secretly chooses Heads or Tails. Row earns one when the faces match, Column earns one when they differ, and the loser receives minus one. Apply the same marking procedure. If Column chooses Heads, Row wants Heads. If Column chooses Tails, Row wants Tails. Row's best responses are the two matching cells, so mark Row's payoff in each of those cells red. Column wants exactly the opposite pattern. If Row chooses Heads, Column wants Tails. If Row chooses Tails, Column wants Heads. Column's two best responses are the mismatching cells, so mark those second payoffs blue. Inspect all four cells. Every cell carries one player's mark, but no cell carries both. At every deterministic outcome, one player is losing and can reverse the result by switching. Therefore Matching Pennies has no pure Nash equilibrium. The incentives form a cycle. Start at Heads, Heads. Row wins there, so Column wants to switch to Tails. That change carries the outcome to Heads, Tails. Now Row is losing, so Row switches to Tails. At Tails, Tails, Column is losing and switches to Heads. Then Row is losing and switches back to Heads. We return to the starting outcome without ever reaching a cell where both players want to stay. This cycle is another way to check the matrix. A pure equilibrium would be a stopping point with no profitable arrow leaving it. Here every one of the four outcomes has an escape for exactly one player. The practical problem is predictability. Suppose Row always chooses Heads, or follows a pattern Column has learned. Column chooses Tails, forces a mismatch, and wins every round. The same vulnerability runs in the other direction. If Column always chooses Heads, Row copies Heads and wins every round. In this game, any deterministic pattern that an opponent can predict reveals the pure reply that defeats it. Randomizing does not mean alternating according to a visible schedule. Heads, Tails, Heads, Tails is deterministic and therefore exploitable once noticed. A mixed strategy assigns probabilities and uses genuine random choice so that past actions do not reveal the next one. The next question is precise. Can we choose probabilities that leave the opponent indifferent between Heads and Tails? If both pure replies give the same expected payoff, the opponent has no profitable way to exploit one of them. That indifference condition is the key to a mixed-strategy equilibrium. We will write each pure action's expected payoff as a function of the opponent's probability, plot the two functions, and find exactly where they cross.

Let q be the probability that Column chooses Heads. We are choosing Column's probability, but the equations we compare are Row's payoffs. Column's mix must remove Row's preference between Row's Heads and Row's Tails. If Row chooses Heads, Row earns one when Column chooses Heads and minus one when Column chooses Tails. Weight those two payoffs by q and one minus q. Simplifying gives two q minus one. The red line on the plot rises with q because Heads becomes more attractive as Column chooses Heads more often. If Row instead chooses Tails, matching Column's Heads loses one and differing from Column's Tails wins one. The weighted payoff is q times minus one plus one minus q times one. That simplifies to one minus two q. The blue line falls as q rises because Tails becomes less attractive when Column chooses Heads more often. At q equals zero, Column always chooses Tails. Row strongly prefers Tails, so the blue payoff is one and the red payoff is minus one. Move q upward and that advantage shrinks. Indifference occurs where the two payoff lines cross. Set two q minus one equal to one minus two q. Adding two q and adding one gives four q equals two, so q equals one half. At the yellow crossing, Row's Heads and Tails both have expected payoff zero. This calculation found Column's probability from Row's payoffs. That cross-player direction is not a trick. A player's mix is chosen to control the opponent's incentives. Now let p be the probability that Row chooses Heads. To find p, compare Column's two pure actions using Column's payoffs. Row's probability must make Column indifferent. If Column chooses Heads, Column loses one when Row also chooses Heads and wins one when Row chooses Tails. The expected payoff is p times minus one plus one minus p times one. That simplifies to one minus two p. If Column chooses Tails, Column wins against Row's Heads and loses against Row's Tails, giving two p minus one. Set the two expressions equal. One minus two p equals two p minus one, so p equals one half. The symmetry of Matching Pennies produced the same half-and-half probability for both players. That symmetry is special. The method, comparing the opponent's pure-action payoffs, is the part that generalizes. The mixed-strategy equilibrium has each player choose Heads with probability one half and Tails with probability one half. Each of the four outcome cells then occurs with probability one quarter. Verify Row first. Against Column's half-and-half mix, Heads wins one half the time and loses one half, so its expected payoff is zero. Tails has the same calculation and also gives zero. Verify Column in the same way. Either pure action wins half the time and loses half, so both give zero. Because neither player has a better pure reply, neither can improve by changing to any other mixture either. Notice what equilibrium does not require. The realized actions can differ from round to round, and after any particular round one player may wish the coin had landed differently. Equilibrium says the probability rule itself cannot be profitably replaced while the opponent keeps the equilibrium mix. We now know the general shape of a mixing calculation. Assign a probability to one player's first action, compute the other player's two pure-action payoffs, set them equal, and solve. The asymmetric example next will show why remembering whose payoffs to use is essential.

Now take a genuinely asymmetric game. A worker chooses Work or Shirk, while an inspector chooses Inspect or Do Not Inspect. The roles differ, the available actions differ, and the first payoff in each cell belongs to the worker. Read the worker's incentives. If inspection occurs, Work gives two while Shirk gives minus two, so Work is better. Without inspection, Shirk gives three while Work gives two, so Shirk is better. Now read the inspector's incentives using the second payoffs. If the worker Works, avoiding the inspection cost gives zero instead of minus one. If the worker Shirks, inspection gives one instead of minus two. No cell carries both marks. Work makes Do Not Inspect attractive, which then makes Shirk attractive. Shirk makes Inspect attractive, which then makes Work attractive. The pure incentives cycle, so there is no pure equilibrium. A natural guess says that if the worker Works most of the time, the inspector must also Inspect most of the time. Perhaps inspection needs probability one half or more. That guess confuses the frequency of an action with the strength of the incentive created by it. Let q be the probability that the inspector inspects. To find q, use the worker's payoffs. The inspector must choose q so that the worker is indifferent between Work and Shirk. Work pays the worker two if inspected and two if not inspected. Its expected payoff is two q plus two times one minus q, which is simply two. The blue line is flat. Shirk pays minus two if inspected and three if not inspected. Its expected payoff is minus two q plus three times one minus q. Simplifying gives three minus five q. The red line slopes downward because more inspection makes shirking less attractive. At an inspection probability of zero, Shirk pays three and beats Work's two. Move q to one tenth and Shirk still pays two point five, so the worker still prefers Shirk. Set the two worker payoffs equal. Two equals three minus five q. Solving gives q equals one fifth, or twenty percent. At exactly twenty percent inspection, Work and Shirk both give the worker an expected payoff of two. Below that crossing, Shirk is better. Above it, Work is better. The mixing point is the boundary between those strict preferences. Now let p be the probability that the worker Works. To find p, switch to the inspector's payoffs. The worker must choose p so that Inspect and Do Not Inspect give the inspector the same expected payoff. Inspect gives the inspector minus one against Work and one against Shirk. Its expected payoff is minus p plus one minus p, which simplifies to one minus two p. Do Not Inspect gives zero against Work and minus two against Shirk. Its expected payoff is zero times p minus two times one minus p, which simplifies to minus two plus two p. Set the two inspector payoffs equal. One minus two p equals minus two plus two p. Rearranging gives three equals four p, so p equals three quarters. Move to p equals zero point seven five. The two payoff lines cross at minus one half. Inspect and Do Not Inspect are equally good, so the inspector can genuinely mix. The equilibrium has the worker Work with probability three quarters and the inspector Inspect with probability one fifth. Frequent work is supported by infrequent inspection because being caught while shirking is costly. Verify the worker. At q equals one fifth, Work gives two. Shirk gives three minus five times one fifth, also two. Verify the inspector. At p equals three quarters, both Inspect and Do Not Inspect give minus one half. Here is the rule worth carrying away. The worker's payoff numbers determined the inspector's mixing probability. The inspector's payoff numbers determined the worker's mixing probability. Your probability is chosen to erase the opponent's strict preference. That is why intuition based only on how often an action appears can be misleading. Mixed equilibrium probabilities are not direct measures of effort, importance, or virtue. They are the probabilities that balance the other player's expected payoffs.

Let's finish by solving a new game from beginning to end. Row chooses Up or Down. Column chooses Left or Right. The question asks for every equilibrium, so we must check pure cells first and then ask whether a mixed equilibrium also exists. Pause at the matrix and begin with Row. Against Column's Left, compare Row's first payoffs three and zero. Up is the unique best response, so mark the three red. Against Column's Right, Row compares zero with one. Down is the best response, so mark the one in the lower-right cell red. Now analyse Column using the second numbers. Against Row's Up, Column compares two from Left with zero from Right. Left is better, so mark the two blue. Against Row's Down, Column compares zero from Left with three from Right. Right is better, so mark the three blue. Inspect complete cells. Up, Left carries both marks, and Down, Right carries both marks. Those are two pure-strategy Nash equilibria. Do not stop merely because pure equilibria exist. Some two-by-two games, including this one, also have a mixed equilibrium. We find it with exactly the same indifference method used before. Let p be the probability that Row chooses Up, and q the probability that Column chooses Left. Remember the cross-player rule. Use Row's payoffs to find q, because q must make Row indifferent. If Row chooses Up, the payoff is three against Left and zero against Right. The expected payoff is therefore three q. If Row chooses Down, the payoff is zero against Left and one against Right. Its expected payoff is one minus q. Set Row's two payoffs equal. Three q equals one minus q, so four q equals one and q equals one quarter. Column chooses Left with probability one quarter to keep Row willing to mix. Now use Column's payoffs to find p. If Column chooses Left, the expected payoff is two p. If Column chooses Right, the expected payoff is three times one minus p. Set those equal. Two p equals three minus three p, so five p equals three and p equals three fifths. Row chooses Up with probability three fifths to keep Column indifferent. Notice that the probabilities are not mirror images of the most attractive payoffs. Column's one-quarter probability came from Row's payoff comparison, while Row's three-fifths probability came from Column's payoff comparison. Always verify a mixed result before accepting it. At q equals one quarter, Row's Up payoff is three times one quarter, or three quarters. Row's Down payoff is one minus one quarter, also three quarters. At p equals three fifths, Column's Left payoff is two times three fifths, or six fifths. Column's Right payoff is three times two fifths, also six fifths. Both indifference conditions hold. A directional check catches sign errors. If q rises above one quarter, Up becomes better for Row; if q falls below one quarter, Down becomes better. The crossing is exactly where Row changes preferred actions. Likewise, if p rises above three fifths, Left becomes better for Column; below three fifths, Right becomes better. At the crossing, each player is willing to use either pure action. This game therefore has three equilibria: the two double-marked pure cells and the interior mixed equilibrium we just verified. Finding one equilibrium is not permission to stop when the question asks for all of them. Here is the complete hand method in six steps. First, label the payoff order. Second, for each column mark every largest Row payoff. Third, for each row mark every largest Column payoff. Fourth, every cell with both marks is a pure equilibrium. Fifth, if mixing is relevant, introduce probabilities and use each probability to make the opponent indifferent. Your mix controls the opponent's incentives. Sixth, verify. Probabilities must lie between zero and one. Pure actions used with positive probability must give equal expected payoffs, and any unused action must not give more. The shortest memory aid is this: your mixing probability is solved from the opponent's payoff comparison. It is chosen to make the opponent indifferent, not to make your own two payoffs equal directly. With that discipline, a two-by-two game becomes a small sequence of comparisons and two linear equations. Read the pairs, mark the best responses, keep every double mark, balance the opponent when mixing, and check the result.