Mixed Strategies: Expected Payoffs 🎲

students, imagine a game where you do not always pick the same move. Instead, you sometimes choose one action and sometimes another, on purpose. That is the idea behind a mixed strategy. In this lesson, you will learn how to compute expected payoffs, which means the average result you can expect when strategies are chosen with probabilities.

Objectives

Calculate expected payoffs in a mixed game.
Use probabilities to evaluate outcomes.
Set up payoff expressions correctly.

Why this matters

In real life, randomization can be useful. A soccer goalie may dive left or right unpredictably. A business may choose between two pricing plans to stay competitive. In game theory, mixed strategies help players avoid being too predictable. To work with them, we need a way to measure the value of each choice. That is where expected payoff comes in. 💡

What an Expected Payoff Means

An expected payoff is the weighted average of all possible outcomes. The weights are the probabilities of those outcomes happening. If an action leads to payoff $8$ with probability $\tfrac{1}{4}$ and payoff $2$ with probability $\tfrac{3}{4}$, then the expected payoff is

$\frac{1}{4}(8)+\frac{3}{4}(2).$

This does not mean you always get that exact number in one play. It means that if you repeated the game many times, the long-run average would approach that value.

For mixed strategies, each player assigns probabilities to their available actions. The expected payoff depends on both players’ probabilities. Because of that, we often compute payoffs by looking at every possible combination of moves.

Building the Probability Model

Suppose Player 1 chooses between actions $A$ and $B$. Player 2 chooses between actions $C$ and $D$. If Player 1 plays $A$ with probability $p$ and $B$ with probability $1-p$, then Player 1 is using a mixed strategy. If Player 2 plays $C$ with probability $q$ and $D$ with probability $1-q$, then Player 2 is also mixing.

To find expected payoffs, students, follow these steps:

List every outcome pair, such as $(A,C)$, $(A,D)$, $(B,C)$, and $(B,D)$.
Find the probability of each pair by multiplying the players’ probabilities, assuming their choices are independent.
Multiply each outcome’s payoff by its probability.
Add everything together.

If the payoff to Player 1 is shown in the matrix below:

$(A,C)$ gives $6$
$(A,D)$ gives $1$
$(B,C)$ gives $4$
$(B,D)$ gives $3$

and Player 1 uses probability $p$ on $A$, while Player 2 uses probability $q$ on $C$, then Player 1’s expected payoff is

E_1 = pq(6) + p(1-q)(1) + (1-p)q(4) + (1-p)(1-q)(3).

That formula may look long, but it is just the sum of “probability times payoff” for every possible result. 🧠

A Step-by-Step Example

Let’s work through a complete example.

Suppose the payoff matrix for Player 1 is:

$\begin{array}{c|cc}$

& C & D \\

$\hline$

A & 10 & 0 \\

B & 4 & 6

$\end{array}$

Player 1 chooses $A$ with probability $p=\tfrac{1}{2}$ and $B$ with probability $1-p=\tfrac{1}{2}$. Player 2 chooses $C$ with probability $q=\tfrac{1}{3}$ and $D$ with probability $1-q=\tfrac{2}{3}$.

Now compute the expected payoff for Player 1.

Step 1: Find each outcome probability

$(A,C)$ has probability $\tfrac{1}{2}\cdot \tfrac{1}{3}=\tfrac{1}{6}$
$(A,D)$ has probability $\tfrac{1}{2}\cdot \tfrac{2}{3}=\tfrac{1}{3}$
$(B,C)$ has probability $\tfrac{1}{2}\cdot \tfrac{1}{3}=\tfrac{1}{6}$
$(B,D)$ has probability $\tfrac{1}{2}\cdot \tfrac{2}{3}=\tfrac{1}{3}$

Step 2: Multiply by payoffs

$E_1 = \frac{1}{6}(10)+\frac{1}{3}(0)+\frac{1}{6}(4)+\frac{1}{3}(6).$

Step 3: Add them up

E_1 = $\frac{10}{6}$+0+$\frac{4}{6}$+2 = $\frac{14}{6}$+2 = $\frac{7}{3}$+2 = $\frac{13}{3}$.

So the expected payoff for Player 1 is $\tfrac{13}{3}$, or about $4.33$.

This value is useful because it tells Player 1 what average result to expect if both players keep using these mixed strategies over time. ✅

Writing Expected Payoff Expressions Correctly

A common mistake is to mix up the probability of a row action with the probability of a column action. students, always match each payoff with the correct outcome.

For a $2\times 2$ game, if Player 1’s payoff matrix is

$\begin{array}{c|cc}$

& C & D \\

$\hline$

A & a & b \\

B & c & d

$\end{array}$

and Player 1 uses $p$ on $A$, while Player 2 uses $q$ on $C$, then Player 1’s expected payoff is

E_1 = pq(a) + p(1-q)(b) + (1-p)q(c) + (1-p)(1-q)(d).

If you want to simplify, you can expand it:

E_1 = apq + bp - bpq + cq - cpq + d - dp - dq + dpq.

Sometimes, simplification helps reveal patterns. For example, if you are checking for a mixed-strategy equilibrium, you may need to compare expected payoffs from different actions. The player often chooses probabilities so that the other player is indifferent between certain moves.

Using Expected Payoff to Compare Choices

Expected payoff helps you decide which strategy is better on average. Suppose Player 1 can choose between two pure strategies, and Player 2 is mixing. Then Player 1 can compute the expected payoff from each pure strategy and compare them.

For example, if Player 2 plays $C$ with probability $q$ and $D$ with probability $1-q$, then:

If Player 1 chooses $A$, the expected payoff is

$E(A)=qa+(1-q)b.$

If Player 1 chooses $B$, the expected payoff is

$E(B)=qc+(1-q)d.$

If $E(A)>E(B)$, then $A$ gives a higher average payoff. If $E(A)<E(B)$, then $B$ is better. If they are equal, Player 1 is indifferent between $A$ and $B$.

That indifference idea is very important in mixed equilibrium problems. In equilibrium, a player’s randomization often makes the other player’s choices tie in expected payoff. This prevents the other player from improving by switching to one pure strategy. 🎯

Another Example with Real-World Meaning

Imagine two companies competing on advertising. Company 1 can either run a bold ad campaign $A$ or a quiet campaign $B$. Company 2 can either launch early $C$ or launch late $D$. The payoff numbers represent profit in thousands of dollars.

$\begin{array}{c|cc}$

& C & D \\

$\hline$

A & 5 & 2 \\

B & 3 & 4

$\end{array}$

If Company 2 launches early with probability $q$, then Company 1’s expected payoff from choosing $A$ is

$E(A)=5q+2(1-q).$

Simplify:

$E(A)=5q+2-2q=3q+2.$

If Company 1 chooses $B$, its expected payoff is

$E(B)=3q+4(1-q).$

Simplify:

$E(B)=3q+4-4q=4-q.$

Now students, compare the two expressions. If Company 1 wants to know which action is better for a given $q$, it just checks whether $3q+2$ is bigger than $4-q$. Solving

$3q+2=4-q$

gives

$4q=2$

$q=\frac{1}{2}.$

This means if Company 2 launches early half the time, Company 1 is indifferent between bold and quiet ads. That kind of balance is a key idea in mixed-strategy analysis.

Common Mistakes to Avoid

Here are a few mistakes students often make:

Forgetting to multiply by both players’ probabilities
Matching the wrong payoff to the wrong outcome
Adding probabilities instead of multiplying them for a single outcome
Assuming expected payoff is guaranteed in one round
Forgetting that probabilities like $p$ and $q$ must satisfy $0\le p\le 1$ and $0\le q\le 1$

A good habit is to write every outcome as a pair first, then attach its probability, then attach its payoff. This keeps your work organized and accurate.

Conclusion

Expected payoff is one of the most important tools in mixed strategies. It turns a game with randomness into a number you can analyze. By multiplying each outcome’s payoff by its probability and then adding the results, you can compare strategies, solve for best responses, and understand mixed equilibria.

students, the key idea is simple: probability times payoff, summed across all outcomes. Once you can set up those expressions correctly, you are ready for more advanced mixed-strategy problems. 🌟

Study Notes

A mixed strategy uses probabilities to choose among actions.
The expected payoff is the weighted average of possible payoffs.
For each outcome, use $\text{probability} \times \text{payoff}$.
In a $2\times 2$ game, the four outcome probabilities are products like $pq$, $p(1-q)$, $(1-p)q$, and $(1-p)(1-q)$.
Player 1’s expected payoff in a payoff matrix $$\begin{array}{c|cc} & C & D \\ \hline A & a & b \\ B & c & d \end{array}$$ is $E_1 = pq(a) + p(1-q)(b) + (1-p)q(c) + (1-p)(1-q)(d)$.
To compare pure strategies against a mixed opponent, compute the expected payoff for each pure strategy.
If two strategies give the same expected payoff, the player is indifferent between them.
Expected payoff describes the long-run average, not necessarily the result of a single game.