Expected Value

Discovering Game Theory

This course uses an online textbook: Game Theory, a Discovery Approach, by Jennifer Nordstrom.

In probability theory, we find the general formula for expected value. Consider a numeric data set \(X\) of sample size \(n\) with outcomes \(X=\left\{x_1, x_2, \dots , x_n\right\}\) and associated probabilities of \(\left\{p(x_1), p(x_2), \dots , p(x_n)\right\}\). A t-chart often helps to visualize the scenario as is shown in the example below:

\[\begin{split}\begin{array}{r|c}x&p(x)\\ \hline 2&\frac{1}{5}\\0&\frac{1}{7}\\-1&\frac{1}{4} \\ \vdots&\vdots \\3&\frac{1}{3} \end{array}\end{split}\]

Formula

Expected Value

Given a set \(X\) of \(n\in\mathbb{N}\) outcomes and \(n\) associate probabilities, the expected value of \(X\) is given as follows:

\[E(X) =\sum x_ip(x_i)\]

In expanded form, we have the formula as follows:

\[E(X) =x_1p(x_1)+x_2p(x_2)+\dots+x_np(x_n)\]

We convert to summation notation to see how the traditional formula shown above arrives:

\[E(X) =\sum_{i=1}^{n}x_ip(x_i)\]

Game Theory Application

In a matrix game, each row and column have an expected value only if we know (or can guess) the opponent’s strategy mixture.

Example

Given that Colin will play the strategy mixture \(\vec c = \left(\frac{3}{4},\frac{1}{4}\right)\), calculate the expected values for Rose’s strategies.

\[\begin{split}\begin{array}{cc}&\text{Colin}\\\text{Rose}&\begin{array}{r|rr}&A&B\\\hline A&12&-2\\B&4&3\\C&9&11\end{array}\end{array}\end{split}\]

Solution. We first notice that \(\textbf{Rose C}\geq\textbf{Rose B}\) because

• \(9\geq 4\), and

• \(11\geq 3\)

Since Rose will never play a dominated strategy, we can see that \(E\left(\text{Rose B}\right) = 0\), and we are now able to \textbf{reduce by dominance} to the following \(2\times 2\) game: $\(\begin{array}{cc}&\text{Colin}\\\text{Rose}&\begin{array}{r|rr}&A&B\\\hline A&12&-2\\C&9&11\end{array}\end{array}\)$

With Colin playing the strategy set \(\vec c = \left[\begin{array}{r}\frac{3}{4}\\\frac{1}{4}\end{array}\right]\), the expected value calculations for Rose’s strategies \(A\) and \(C\) are as follows:

\begin{align} E\left(\text{Rose }A\right) &= 12c_1-2c_2=12\left(\frac{3}{4}\right) - 2\left(\frac{1}{4}\right)\ E\left(\text{Rose }C\right) &= 9c_1+11c_2= 9\left(\frac{3}{4}\right)+11\left(\frac{1}{4}\right)\[3mm] E\left(\text{Rose }A\right) &= \frac{36}{4}-\frac{2}{4}=\frac{34}{4}\ E\left(\text{Rose }C\right) &= \frac{27}{4}-\frac{11}{4}=\frac{26}{4} \hspace{5mm} \implies \[3mm] E\left(\text{Rose }A\right) &= \frac{17}{2}=8.5\ E\left(\text{Rose }C\right) &= \frac{13}{2}=6.5 \end{align}

Gathering the results, we see that

\[E\left(\text{Rose C}\right) = 9.5 > 8.5 = E\left(\text{Rose A}\right)\]

and that Rose, given Colin’s plays \(\vec c = \left(\frac{3}{4},\frac{1}{4}\right)\), will be better off playing strategy \(C\) than any other option. Given the rational play principle, we know that Rose will play \(C\).