Why Derive a Mixed Equilibrium?
In a strategic-form game, a mixed strategy assigns probabilities to a player’s available actions. In a two-player, two-action game, a fully mixed Nash equilibrium has each player choose each action with positive probability. The probabilities matter: they determine how often each action is used, and they must make the other player willing to randomize rather than switch to one action for sure.
The central calculation is therefore an opponent-indifference condition. To find the probability with which one player mixes, make the other player indifferent between their actions. Then solve the two conditions together and check that the resulting probabilities are feasible. This is a direct application of best responses and Nash equilibrium as mutual best response.
Notation and the Two Indifference Conditions
Represent the players’ actions as Up and Down for Player A, and Left and Right for Player B. Each cell gives the payoffs to A and B, in that order. Let \(p\) be the probability that A chooses Up, and let \(q\) be the probability that B chooses Left.
A’s expected payoff from choosing Up depends on \(q\), the probability that B chooses Left. A’s expected payoff from choosing Down depends on the same probability. Thus A’s indifference condition determines \(q\), not \(p\). In the same way, B’s indifference condition determines \(p\).
Setting A’s two payoffs equal and collecting the terms in \(q\) gives \(q(a-b-c+d)=d-b\). Setting B’s two payoffs equal gives \(p(e-f-g+h)=h-g\). When the denominators are nonzero, these equations give the candidate mixing probabilities.
Why Indifference Is Required
Suppose B chooses Left with a probability strictly between zero and one. If A strictly preferred Up, then choosing Up for sure would give A a higher expected payoff than randomizing; if A strictly preferred Down, A would choose Down for sure. A can rationally assign positive probability to both actions only when the two actions yield equal expected payoffs. The same argument applies to B.
The algebra also shows exactly how the probability is obtained. A’s payoff difference, Up minus Down, is \(qa+(1-q)b-[qc+(1-q)d]\). Rearranging yields \(q(a-b-c+d)+(b-d)\). Setting this difference to zero and solving gives A’s indifference condition. The corresponding payoff difference for B is \(p(e-f-g+h)+(g-h)\); setting it to zero gives B’s condition. This derivation explains why each player’s mixing probability is found by making the opponent indifferent.
Use \(p\) for A’s probability of Up and \(q\) for B’s probability of Left. Keep track of which player’s expected payoff each equation represents.
For each action, take the expected payoff over the opponent’s two actions, using the opponent’s mixing probability.
A’s equality solves for \(q\); B’s equality solves for \(p\).
Both probabilities must be strictly between zero and one. Substitute them into the expected payoffs to verify indifference.
Worked Derivations
Worked Example: A Fully Mixed Equilibrium
Two firms choose either a high or low service level. The payoffs are hypothetical and measured in the same units for both firms. Find a fully mixed equilibrium, if one exists.
Method: impose each opponent-indifference condition. If B chooses Left with probability \(q\), A’s payoff from Up is \(3q+1(1-q)=1+2q\), while A’s payoff from Down is \(0q+4(1-q)=4-4q\). Equating these gives \(1+2q=4-4q\), so \(6q=3\) and \(q=1/2\).
For B, if A chooses Up with probability \(p\), B’s payoff from Left is \(1p+0(1-p)=p\), and the payoff from Right is \(0p+3(1-p)=3-3p\). Equating gives \(p=3-3p\), so \(4p=3\) and \(p=3/4\).
The formula denominators confirm the calculations: \(3-1-0+4=6\), and \(1-0-0+3=4\). Thus \(q=(4-1)/6=1/2\) and \(p=(3-0)/4=3/4\). Both probabilities are strictly between zero and one.
Check. At \(q=1/2\), A receives \(1+2(1/2)=2\) from Up and \(4-4(1/2)=2\) from Down. At \(p=3/4\), B receives \(3/4\) from Left and \(3-3(3/4)=3/4\) from Right. Each player is indifferent, so each player’s mixture is a best response to the other’s. The candidate is a fully mixed Nash equilibrium.
Common mistake avoided: using A’s probability \(p\) in A’s indifference equation. A compares the payoffs from Up and Down across B’s actions, so A’s condition must use \(q\).
Worked Example: A Zero Denominator and No Fully Mixed Equilibrium
Use the following hypothetical payoff matrix to determine whether a fully mixed equilibrium exists, and identify a pure-strategy equilibrium if possible.
A’s indifference condition would require \(q=(3-1)/(4-1-0+3)=2/6=1/3\). But B’s condition has denominator \(0-2-1+3=0\), while its numerator is \(3-1=2\). The equation for B is therefore \(0p=2\), which has no solution. In payoff terms, B gets 2 rather than 0 from Right rather than Left when A chooses Up, and gets 3 rather than 1 from Right rather than Left when A chooses Down. B strictly prefers Right for every value of \(p\). Consequently, B cannot be indifferent and there is no fully mixed equilibrium.
To check for a pure equilibrium, let B choose Right. A’s payoff is 1 from Up and 3 from Down, so Down is A’s best response. When A chooses Down, B gets 3 from Right and 1 from Left, so Right is B’s best response. Thus (Down, Right) is a pure-strategy Nash equilibrium.
Common mistake avoided: a probability calculated from one player’s indifference equation is not by itself an equilibrium. Both players must be able to mix consistently.
Worked Example: Feasible Probabilities with a Negative Denominator
Find the fully mixed equilibrium, if one exists, for the following hypothetical game. Also calculate each player’s expected payoff at equilibrium.
A’s indifference condition gives \(q=(1-6)/(2-6-5+1)=(-5)/(-8)=5/8\). Both numerator and denominator are negative, so the resulting probability is positive. B’s condition gives \(p=(3-0)/(4-1-0+3)=3/6=1/2\). Both probabilities are strictly between zero and one.
At \(q=5/8\), A’s payoff from Up is \(2(5/8)+6(3/8)=10/8+18/8=28/8=7/2\). A’s payoff from Down is \(5(5/8)+1(3/8)=25/8+3/8=28/8=7/2\).
At \(p=1/2\), B’s payoff from Left is \(4(1/2)+0(1/2)=2\). B’s payoff from Right is \(1(1/2)+3(1/2)=2\). Hence both players are indifferent, and the expected payoffs are \(7/2\) for A and \(2\) for B.
Common mistake avoided: rejecting a candidate because one denominator is negative. Feasibility depends on the resulting probability, not on the sign of the denominator alone.
Feasibility, Degenerate Cases, and Common Traps
When both denominators are nonzero, calculate the two ratios and check them separately. Values strictly between zero and one produce a fully mixed candidate. A value below zero or above one cannot be a mixing probability. A value equal to zero or one is a boundary case, not a fully mixed strategy; it calls for checking pure strategies or other boundary equilibria.
A zero denominator needs more care than simply dividing. For A’s condition, write \(\Delta_A=a-b-c+d\), so the equation is \(q\Delta_A=d-b\). If \(\Delta_A=0\) but \(d-b\ne0\), A cannot be made indifferent by any \(q\). If both are zero, A is indifferent for every \(q\), and this equation does not determine B’s probability. Analyze the other player’s condition and check best responses rather than concluding that no equilibrium exists. The same reasoning applies to B’s equation.
- Mixing probability assigned to the wrong player. A’s indifference determines B’s probability \(q\); B’s indifference determines A’s probability \(p\).
- Arithmetic or sign error in a denominator. For example, \(3-1-0+4=2-0+4=6\), not zero. Recompute the denominator from the four relevant payoffs before interpreting the result.
- Stopping after solving the equations. A candidate must satisfy both equations and have feasible probabilities.
- Assuming zero denominator means no equilibrium. It rules out indifference only when the corresponding numerator is nonzero; if both are zero, the player is indifferent for every opponent probability.
- Calling a boundary solution fully mixed. A probability of zero or one means the player uses a pure action, so the equilibrium must be described and checked as a boundary or pure case.
Check Your Understanding
For each game, identify the relevant opponent-indifference probabilities and check whether the stated conclusion follows.
- For the matrix with rows \((6,2),(2,0)\) and \((1,1),(5,4)\), find \(q\) and \(p\), and state whether a fully mixed equilibrium is feasible. (Answer: \(q=3/8,\ p=3/5\); both are feasible.)
- For the matrix with rows \((3,1),(1,0)\) and \((0,0),(4,3)\), find both mixing probabilities and the equilibrium payoffs. (Answer: \(q=1/2,\ p=3/4\); A’s payoff is \(2\) and B’s payoff is \(3/4\).)
- For the matrix with rows \((4,2),(3,0)\) and \((1,1),(2,3)\), calculate A’s indifference probability \(q\). Does it allow a fully mixed equilibrium? (Answer: \(q=(2-3)/(4-3-1+2)=-1/2\); no fully mixed equilibrium is possible.)
- For the matrix with rows \((2,4),(2,0)\) and \((2,1),(2,3)\), A’s payoff is the same across its two actions for either B action. Find \(p\), and describe the fully mixed equilibria. (Answer: \(p=(3-1)/(4-0-1+3)=1/3\); any \(q\) strictly between zero and one, together with \(p=1/3\), is a fully mixed equilibrium.)
- For the matrix with rows \((1,5),(4,0)\) and \((6,2),(2,4)\), find \(q\), \(p\), and each player’s expected payoff. (Answer: \(q=2/7,\ p=2/7\); A’s payoff is \(22/7\) and B’s payoff is \(20/7\).)