Question

Matching pennies. In this simple two-player game, the players (call them $\mathbf{R}$and $\mathbf{C}$) each choose an outcome, heads or tails. If both outcomes are equal, $\mathbf{C}$gives a dollar to $\mathbf{R}$; if the outcomes are different, $\mathbf{R}$gives a dollar to $\mathbf{C}$.

(a) Represent the payoffs by a$\mathbf{2}\mathbf{}\mathbf{\times }\mathbf{}\mathbf{2}$ matrix.

(b) What is the value of this game, and what are the optimal strategies for the two players?

Short answer

The value of the game is$0$ and the optimal strategy of both the player will be equal i.e., $\frac{1}{2}$.

Step-by-step solution

Represent the payoffs by a matrix.

(a)It is given that for$R$ to win, the two coins must have same outcome, i.e., either both heads or both tails.

The above condition can be represented as:

	$H$	$T$
$H$	$+1$	$-1$
$T$	$-1$	$+1$

The matrix represents the money$R$ got by game.

$\mathrm{H}=\mathrm{Head},\mathrm{T}=\mathrm{Tail}$

$+1$indicates that$R$ got a dollar whereas$-1$ shows that$C$ got a dollar.

Calculate the value of this game and the optimal strategies for the two players

(b)

Let, Probability of $R$to get head and tail be given as${\mathrm{X}}_1$and $X_2$.

And, Probability of $C$to get head and tail be given as $y_1$and $y_2$.

Max:$z$: Min:$w$

$\begin{array}{c}\mathrm{z}\le {\mathrm{x}}_1-{\mathrm{x}}_2\\ \mathrm{z}\le -{\mathrm{x}}_1+{\mathrm{x}}_2\\ {\mathrm{x}}_1+{\mathrm{x}}_2=1\\ {\mathrm{x}}_1,{\mathrm{x}}_2>0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}$

$\begin{array}{c}w\ge y_1-y_2\\ w\ge -y_1+y_2\\ y_1+y_2=1\\ y_1,y_2>0\end{array}$

The value of this game is$0$ .

Therefore, the optimal strategy of both the player is equal to i.e., $\frac{1}{2}$.