2.5b Expected Payoffs

Expected Payoff

The expected payoff to player $i\in N$ playing a mixed strategy ${\sigma }_i(s_i^k)$ for $s_i^k\in S_i$, with other players playing pure strategies $s_{-i}$, is denoted as
$U_i(s_i^k)={\sum }_{k=1}^{|S_i|}{\sigma }_i(s_i^k)\cdot u_i(s_i^k,s_{-i}).$

The expected payoff to player $i\in N$ playing a pure strategy $s_i^{\prime }\in S_i$, with other players mixing such that ${\sigma }_j(s_j^k)$ for $s_j^k\in S_j$, is denoted as
$TODO:U_i(s_i^{\prime })={\sum }_{s=}^{}\left({\prod }_{j=1}^{|N|-1}{\sigma }_j(s_j^l)\right)u_i(s_i,s_{-i})$

• If the $|N|-1$ other players are also mixing, then the expected payoff to player $i$ for a strategy $s_i$ is
  $TODO:U_i(s_i)={\sum }_{s=}^{}\left({\prod }_{j=1}^{|N|-1}{\sigma }_j(s_j^l)\right)u_i(s_i,s_{-i})$

Properties

(Mixing Criterion) Player $i$ must be indifferent over at least two strategies of all other players in order to mix.

Check example #1.

Examples

Matching Pennies

Consider a game of matching pennies, where player 1 wins if both players flip the same side, and the contrary for player 2, in the following game matrix:

\begin{document}
 
\begin{tikzpicture}
   % Draw the vertical lines
   \draw (0, 0) -- (0, -3);
   \draw (1.5, 0) -- (1.5, -3);
   \draw (3, 0) -- (3, -3);
 
   % Draw the horizontal lines
   \draw (0, 0) -- (3, 0);
   \draw (0, -1.5) -- (3, -1.5);
   \draw (0, -3) -- (3, -3);
 
   % Add labels for players
   \node at (1.5, 1) {Player 2};
   \node[rotate=90] at (-1, -1.5) {Player 1};
 
   % Add labels for Player 2's choices
   \node at (0.75, 0.25) {H};
   \node at (2.25, 0.25) {T};
 
   % Add labels for Player 1's choices
   \node at (-0.25, -0.75) {H};
   \node at (-0.25, -2.25) {T};
 
   % Add payoffs for each cell
   \node at (0.75, -0.75) {(1, -1)};
   \node at (2.25, -0.75) {(-1, 1)};
 
   \node at (0.75, -2.25) {(-1, 1)};
   \node at (2.25, -2.25) {(1, -1)};
\end{tikzpicture}
 
\end{document}

Let’s take a deeper look at $P_1$‘s expected utility:

\begin{align} U_1(H) &= \sigma_2(H)u_1(H, H)+\sigma_2(T)u_1(H, T) \\ &=\sigma_2(H)(1)+\sigma_2(T)(-1) \\ \\ U_1(T)&=\sigma_2(H)u_1(T, H)+\sigma_2(T)u_1(T, T) \\ &=\sigma_2(h)(-1)+\sigma_2(T)(1).

\end{align}

We can clearly see that in the game, if $U_1(H)>U_1(T)$, then substituting, $2{\sigma }_2(H)>2{\sigma }_2(T)$. As such, we can conclude that if ${\sigma }_2(H)>{\sigma }_2(T)$ (i.e. the probability of $P_2$ playing heads is higher than the probability of $P_2$ playing tails), then $P_1$ gets higher expected utility from $H$ (i.e. 1) than $T$ (i.e. -1).

We then get an interesting result: if $U_1(H)>U_1(T)⟺{\sigma }_2(H)>{\sigma }_2(T)$, $P_1$ will play pure heads (since they benefit from matching pennies). Conversely, if $U_1(H)<U_1(T)⟺{\sigma }_2(H)<{\sigma }_2(T)$, $P_1$ will play pure tails. So when does $P_1$ mix?

$P_1$ mixes only when $U_1(H)=U_1(T)⟺{\sigma }_2(H)={\sigma }_2(T)$. Since $P_1$ gets the same utility from both strategies, they are indifferent to these strategies and can only then begin mixing. Per the rational choice postulate, any other case will be irrational since all players are utility maximizers.

Mixed Strategy Nash Equilibrium ${\sigma }_2(H)=p$ and ${\sigma }_2(T)=1-p$ (since probabilities have to sum up to 1). Then, since in Nash equilibrium $P_1$ has to be indifferent between two strategies to mix, $U_1(H)=U_1(T)$, such that $p(1)+(1-p)(-1)=p(-1)+(1-p)(1)⟺p=\frac{1}{2}.$ Therefore, in the matching pennies game, mixed Nash equilibrium occurs at $({\sigma }_1,{\sigma }_2)$ when ${\sigma }_1=(\frac{1}{2},\frac{1}{2})$ and ${\sigma }_2=(\frac{1}{2},\frac{1}{2})$.

So what is the Nash equilibrium in this game? Since there are two pure strategies here, we can denote