Re-implementing Matching Pennies

Question

I'm a little confused on a problem from my game theory course. I am reviewing the standard ``matching pennies'' game where player $1$ wins $1$ and player $2$ loses $1$ if the their two pennies match and player $1$ loses $1$ and player $2$ wins $1$ if the their two pennies don't match. We can represent the game as this, where player $1$ has the rows and player $2$ has the columns: $$\begin{array}{|c|c|c|} \hline &\text{H}&\text{T} \\ \hline \text{H}&1,-1&-1,1 \\ \hline \text{T}&-1,1&1,-1 \\ \hline \end{array}$$ Unfortunately, we know that this game does not have a pure strategy Nash Equilibrium. However, we want to show that this game has a Nash Equilibrium in the case of the sequential game where player $1$ moves first and player $2$ moves second. However I am having issues proving this in standard simultaneous form. Does anyone have suggestions on how I might degenerate this sequential form game to a simultaneous form?

Answer 1

In the sequential game, player 1's pure strategies are $\{H,T\}$ and player 2's pure strategies are $\{HH,HT,TH,TT\}$. So the strategic form of the game is

$$ \begin{array}{|r|c|c|c|c|} \hline &HH&HT&TH&TT \\\hline H&\mathbf{1},-1&\mathbf{1},-1&\mathbf{-1},\mathbf{1}&-1,\mathbf{1}\\\hline T&-1,\mathbf{1}&\mathbf{1},-1&\mathbf{-1},\mathbf{1}&\mathbf{1},-1\\ \hline \end{array} $$ The best responses are boldfaced. We can therefore see the two pure NEs: $(H,TH)$ and $(T,TH)$. Player 1 gets $-1$ in both equilibria.