Matching Pennies and Zero-Sum Games

Why Matching Pennies Is a Useful Strategic Model

Some strategic choices reward one player precisely when the other player loses. A player choosing a direction for a pass, for example, may want to predict an opponent’s choice while the opponent tries to conceal it. Matching pennies is a compact model of this conflict: one player wants the two choices to match, and the other wants them not to match.

The key lesson is not that either player should choose randomly without a plan. Each player selects a probability distribution that makes the opponent indifferent between the opponent’s available actions. The resulting mixed-strategy equilibrium avoids being predictably exploitable.

Definition: A zero-sum game is a game in which the players’ payoffs sum to zero for every outcome. In a two-player payoff matrix, this means that if the row player receives $x$, the column player receives $-x$. In matching pennies, each player has two actions, conventionally Heads and Tails. The row player prefers a match; the column player prefers a mismatch.

We use the row player’s payoff in each cell and infer the column player’s payoff by changing its sign. The players choose simultaneously, know the payoff matrix, and maximize their expected payoffs. A mixed strategy assigns probabilities to a player’s actions. The randomizations are independent: a player cannot condition a choice on seeing the opponent’s simultaneous choice.

Why There Is No Pure-Strategy Equilibrium

A pure strategy selects one action with certainty. In the standard matching-pennies pattern, each possible pure outcome gives at least one player a reason to switch. If the choices match, the column player prefers to change actions and create a mismatch. If they do not match, the row player prefers to change actions and create a match. Therefore, no outcome is a mutual best response.

This is different from saying the game has no equilibrium. Allowing players to mix can remove the predictable action that the other player would exploit. A mixed-strategy equilibrium occurs when each player’s probability choices make the other player willing to use all actions that receive positive probability.

Worked Example: A Pure-Strategy Cycle

A row player receives $1 for a match and loses $1 for a mismatch. The column player receives the negative of the row player’s payoff. Check each possible pure outcome for a Nash equilibrium.

Step 1: Write the row player’s payoff matrix. Rows are the row player’s actions; columns are the column player’s actions.

\begin{array}{c|cc} & H & T\\ \hline H & 1 & -1\\ T & -1 & 1 \end{array}

Step 2: Check $ (H,H) $. The row player receives 1. If the column player switches from $H$ to $T$, the row player’s payoff becomes $-1$, so the column player improves its own payoff from $-1$ to 1. Thus $ (H,H) $ is not an equilibrium.

Step 3: Check $ (H,T) $. The row player receives $-1$. Switching from $H$ to $T$ would give the row player 1, so the row player wants to switch. This is not an equilibrium.

Step 4: Check $ (T,H) $. The row player receives $-1$. Switching to $H$ would give the row player 1, so the row player wants to switch. This is not an equilibrium.

Step 5: Check $ (T,T) $. The row player receives 1. The column player can switch to $H$, changing the row player’s payoff to $-1$ and raising its own payoff from $-1$ to 1. This is not an equilibrium either.

Conclusion and method. Checking each cell shows that no pure-strategy Nash equilibrium exists. The common mistake is to treat a matching outcome as stable just because the row player prefers it; equilibrium requires both players to have no profitable unilateral deviation.

Deriving the Mixed Equilibrium

Write the row player’s payoffs as $a,b,c,d$, with $H,T$ as the row and column actions in that order. Let $p$ be the probability that the row player chooses $H$, and let $q$ be the probability that the column player chooses $H$. Since the column player’s payoff is the negative of the row player’s, the column player chooses actions to minimize the row player’s expected payoff.

\begin{array}{c|cc} & H & T\\ \hline H & a & b\\ T & c & d \end{array}

For the row player to mix between $H$ and $T$, the column player’s probability $q$ must make the row player indifferent between those actions. If the row player chooses $H$, the expected payoff is $qa+(1-q)b$. If the row player chooses $T$, it is $qc+(1-q)d$. Equating these payoffs gives the probability $q$.

$$ qa+(1-q)b=qc+(1-q)d $$

Rearranging gives $q(a-b-c+d)=d-b$. For the column player to mix, the row player’s probability $p$ must make the column player indifferent between choosing $H$ and $T$. Equivalently, those two column choices must give the same expected payoff to the row player:

$$ pa+(1-p)c=pb+(1-p)d $$

Rearranging this equation gives $p(a-b-c+d)=d-c$. When the resulting probabilities are strictly between 0 and 1, the equilibrium is interior: each player uses both actions with positive probability.

Mixed-equilibrium result: For a two-action zero-sum game with row payoffs $a,b,c,d$, define $D=a-b-c+d$. If $D\ne0$ and the probabilities below lie between 0 and 1, the equilibrium mixing probabilities are

q=\frac{d-b}{D}, \qquad p=\frac{d-c}{D}, \qquad D=a-b-c+d.

The row player’s equilibrium expected payoff, often called the value of the game to the row player, can be found by evaluating either row action against $q$. They give the same result because $q$ makes the row player indifferent. The column player’s expected payoff is the negative of this value. The equations describe an interior mixed equilibrium; if the calculated probabilities are outside the interval from 0 to 1, the interior formula does not describe an equilibrium with both actions used.

Set up the row-payoff matrix.
Record one player’s payoff in each cell. In a zero-sum game, the other player’s payoff is its negative.

Find the opponent’s mixing probability.
Set the row player’s expected payoffs from its two actions equal and solve for $q$.

Find the row player’s mixing probability.
Set the row player’s expected payoff under the column player’s two actions equal and solve for $p$.

Verify the equilibrium and value.
Check that both probabilities are valid, confirm indifference, and calculate the expected payoffs with the other player’s equilibrium mixture.

Worked Equilibrium Calculations

Worked Example: Unequal Rewards for Matching and Mismatching

Suppose the row player earns $6 for matching and loses $4 for a mismatch. The column player’s payoff is the negative of the row player’s. Find both equilibrium probabilities and each player’s expected payoff.

Step 1: Write the matrix and identify its entries. Here $a=6$, $b=-4$, $c=-4$, and $d=6$.

\begin{array}{c|cc} & H & T\\ \hline H & 6 & -4\\ T & -4 & 6 \end{array}

Step 2: Make the row player indifferent. If the row player chooses $H$, its expected payoff is $6q-4(1-q)=10q-4$. If it chooses $T$, its expected payoff is $-4q+6(1-q)=6-10q$. Set these equal:

10q-4=6-10q,\qquad 20q=10,\qquad q=0.5.

Step 3: Make the column player indifferent. Against column action $H$, the row player’s expected payoff is $6p-4(1-p)=10p-4$. Against column action $T$, it is $-4p+6(1-p)=6-10p$. Equating them gives $10p-4=6-10p$, so $20p=10$, and $p=0.5$.

Step 4: Calculate and check the value. Against $q=0.5$, the row player’s payoff from either action is $10(0.5)-4=1$, or $6-10(0.5)=1$. Thus the row player’s expected payoff is $1 and the column player’s is \(-$1. Both probabilities are valid and both players are indifferent between their actions.

Method and mistake avoided. Equate the row player’s payoffs to solve for the column player’s probability, then equate payoffs across the column player’s choices to solve for the row player’s probability. Do not swap these roles: $q$ makes the row player indifferent, while $p$ makes the column player indifferent.

Worked Example: A Matching Game with Asymmetric Cell Payoffs

In a hypothetical tactical game, the row player’s payoff is $8$ when both players choose $H$, $-2$ for $H,T$, $-4$ for $T,H$, and $6$ for $T,T$. The column player receives the negative of each payoff. Find the equilibrium and check the value.

Step 1: Record the entries and denominator. Here $a=8$, $b=-2$, $c=-4$, and $d=6$. Therefore,

$$ D=a-b-c+d=8-(-2)-(-4)+6=20. $$

Step 2: Find $q$. Indifference for the row player requires $q(8)+(1-q)(-2)=q(-4)+(1-q)(6)$. The left side is $10q-2$; the right side is $6-10q$. Hence $20q=8$, so $q=0.4$.

Step 3: Find $p$. Against column action $H$, the row player’s expected payoff is $8p-4(1-p)=12p-4$. Against column action $T$, it is $-2p+6(1-p)=6-8p$. Thus $12p-4=6-8p$, so $20p=10$ and $p=0.5$.

Step 4: Calculate both expected payoffs. Against $q=0.4$, choosing row action $H$ gives $8(0.4)+(-2)(0.6)=3.2-1.2=2$. Choosing $T$ gives $(-4)(0.4)+6(0.6)=-1.6+3.6=2$. The row player’s equilibrium payoff is 2; the column player’s is $-2$.

Check. Against $p=0.5$, column action $H$ gives the row player $8(0.5)+(-4)(0.5)=2$, and column action $T$ gives $(-2)(0.5)+6(0.5)=2$. The column player is therefore indifferent as well. The common mistake is to assume the equilibrium probabilities must each be one-half just because each player has two actions; here $q=0.4$.

Worked Example: Solving and Verifying a Dollar-Payoff Game

Two hypothetical contestants choose $H$ or $T$ simultaneously. The row contestant receives $9 for $H,H$, loses $3 for $H,T$, loses $5 for $T,H$, and receives $7 for $T,T$. The column contestant receives the negative of the row contestant’s payoff. Find the equilibrium probabilities and expected payoffs.

Step 1: Compute the denominator. The entries are $a=9$, $b=-3$, $c=-5$, and $d=7$, so $D=9-(-3)-(-5)+7=24$.

Step 2: Solve for $q$. The row player’s payoff from $H$ is $9q-3(1-q)=12q-3$. Its payoff from $T$ is $-5q+7(1-q)=7-12q$. Indifference requires $12q-3=7-12q$, so $24q=10$, and $q=10/24=5/12$.

Step 3: Solve for $p$. Against column action $H$, the row player’s expected payoff is $9p-5(1-p)=14p-5$. Against column action $T$, it is $-3p+7(1-p)=7-10p$. Equating gives $14p-5=7-10p$, so $24p=12$, and $p=1/2$.

Step 4: Calculate and verify the value. Against $q=5/12$, row action $H$ yields $9(5/12)-3(7/12)=(45-21)/12=2$. Row action $T$ yields $-5(5/12)+7(7/12)=(-25+49)/12=2$. Thus the row player’s expected payoff is $2 and the column player’s is $-$2. As a second check, against \(p=1/2$, column action $H$ gives $9(1/2)-5(1/2)=2$, and column action $T$ gives $-3(1/2)+7(1/2)=2$.

Method and mistake avoided. Compute each indifference condition separately and verify the value from both pure actions. A frequent error is to calculate the row player’s expected payoff using $p$; the row player’s payoff against a chosen row action depends on the column player’s probability $q$.

Interpreting the Equilibrium and Avoiding Traps

An equilibrium mixture does not mean that each player is indifferent before the opponent’s strategy is specified. Indifference is created by the opponent’s equilibrium probabilities. For example, the row player’s probability $p$ is chosen to make the column player indifferent; the column player’s probability $q$ is chosen to make the row player indifferent. This is why calculating one probability does not require using that same player’s own indifference equation.

The expected payoff is not necessarily zero for each player. Zero-sum means the two players’ payoffs add to zero, not that the row player’s equilibrium value must be zero. Unequal rewards and losses can give one player a positive expected payoff at equilibrium, as in the examples above. That value is the payoff each player can secure against an opponent who responds optimally, with the column player receiving its negative.

A practical check is to evaluate both of a player’s pure actions against the opponent’s proposed mixture. They should give equal expected payoffs when that player mixes between them. Also confirm that both probabilities lie between 0 and 1. If a proposed probability is invalid, do not interpret it as a probability or force it into the interval; the assumed interior equilibrium does not apply, and the game requires checking possible boundary strategies.

Key takeaway: In a matching-pennies game, no pure outcome is stable because one player can always benefit by switching. In an interior mixed equilibrium, each player chooses probabilities that make the opponent indifferent, and the expected payoffs sum to zero.

Check Your Understanding

For each zero-sum game, the displayed entries are the row player’s payoffs; the column player’s payoffs are their negatives. Find the equilibrium probabilities and row player’s expected payoff where requested.

The row player earns 5 for a match and loses 3 for a mismatch. What are $p$, $q$, and the row player’s equilibrium expected payoff? (Answer: $p=0.5,\ q=0.5$; row payoff $=1$, column payoff $=-1$.)
For the matrix $\begin{pmatrix}7&-1\\-5&3\end{pmatrix}$, find $p$, $q$, and the equilibrium payoffs. (Answer: $p=0.5,\ q=0.25$; row payoff $=1$, column payoff $=-1$.)
For the matrix $\begin{pmatrix}10&-6\\-2&4\end{pmatrix}$, find both mixing probabilities and the row player’s value as a fraction. (Answer: $p=3/11,\ q=5/11$; row payoff $=14/11$, column payoff $=-14/11$.)
For the matrix $\begin{pmatrix}4&-8\\-2&10\end{pmatrix}$, calculate $D$, $p$, $q$, and both expected payoffs. (Answer: $D=24,\ p=0.5,\ q=0.75$; row payoff $=1$, column payoff $=-1$.)
For the matrix $\begin{pmatrix}11&1\\-3&7\end{pmatrix}$, determine the equilibrium probabilities and value, then state why no pure-strategy equilibrium exists. (Answer: $p=0.5,\ q=0.3$; row payoff $=4$, column payoff $=-4$; no pure equilibrium because each pure outcome gives at least one player a profitable switch.)

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