= Solution
Represent satisfied and unsatisfied goals by <payoffs> one and zero. The second player's goal is an <exclusive or>: it is true precisely when $q\ne r$, independently of $p$. The first goal is false only when all three <Boolean variables> are false. The resulting <payoff matrix> is below; each boxed entry marks that player's <best response> to the opposing <pure strategy>, including ties:
$$
\begin{array}{c|cccc}
&pq&p\bar q&\bar p q&\bar p\bar q\\\hline
r&(\boxed{1},0)&(\boxed{1},\boxed{1})&(\boxed{1},0)&(\boxed{1},\boxed{1})\\
\bar r&(\boxed{1},\boxed{1})&(\boxed{1},0)&(\boxed{1},\boxed{1})&(0,0)
\end{array}
$$
For player 1, choosing $r$ satisfies the goal against every opposing <pure strategy>. Choosing $\bar r$ does equally well except against $\bar p\bar q$, where it does worse. Thus \b[$r$ is player 1's unique <dominant strategy> in the weak sense], and there is no <strictly dominant strategy>. The ties are why most first-player entries in both rows are boxed.
For player 2, either <pure strategy> with $q=0$ is a <best response> against $r$, while either with $q=1$ is a <best response> against $\bar r$. No one choice of $q$ works against both rows, so \b[player 2 has no <dominant strategy>], and \b[there is no <dominant-strategy equilibrium>]. The value of $p$ does not affect player 2's own <payoff>.
The cells with both entries boxed give exactly four <pure strategy> <Nash equilibria>:
$$
\boxed{(r,p\bar q),\quad(r,\bar p\bar q),\quad(\bar r,pq),\quad(\bar r,\bar p q).}
$$
All four give <payoffs> $(1,1)$. In particular, the <weakly dominated> choice $\bar r$ can occur in a <Nash equilibrium> because player 1 is already satisfied there and cannot strictly improve.
Back to article page
