= Asymmetric coordination example
{title2=$(p^*,q^*)=(3/5,2/5)$}
Two players independently choose venues $A$ or $B$. Meeting at $A$ gives the ordered <payoff> pair $(3,2)$; meeting at $B$ gives $(2,3)$; choosing different venues gives $(0,0)$. As throughout this article, the row player's <payoff> comes first:
|| Row choice / column choice
|| $A$
|| $B$
| $A$
| $(3,2)$
| $(0,0)$
| $B$
| $(0,0)$
| $(2,3)$
Against a column player choosing $A$, the row player gets $3$ from $A$ and $0$ from $B$, so $A$ is the unique <best response>. Against $B$, the row player gets $0$ from $A$ and $2$ from $B$, so $B$ is the unique <best response>. The column player likewise prefers to match the row choice. Thus $(A,A)$ and $(B,B)$ are <pure strategy> <Nash equilibria>, while each mismatched <strategy profile> has a profitable <unilateral deviation>.
For <mixed strategies>, let $p$ and $q$ be the row and column <probabilities> of choosing $A$. The pure-choice <expected payoffs> are
$$
U_1(A,q)=3q,\quad U_1(B,q)=2(1-q),\qquad
U_2(p,A)=2p,\quad U_2(p,B)=3(1-p).
$$
The row player is indifferent when $3q=2(1-q)$, giving $q=2/5$. The column player is indifferent when $2p=3(1-p)$, giving $p=3/5$. The <best responses> are therefore
$$
\operatorname{BR}_1(q)=\begin{cases}\{0\}&q<2/5,\\ {[0,1]}&q=2/5,\\\{1\}&q>2/5,\end{cases}
\qquad
\operatorname{BR}_2(p)=\begin{cases}\{0\}&p<3/5,\\ {[0,1]}&p=3/5,\\\{1\}&p>3/5.\end{cases}
$$
These give exactly three <Nash equilibria>: $(p,q)=(0,0),(1,1),(3/5,2/5)$. If either player uses both choices, their indifference condition places the other player's <probability> strictly between zero and one, so both conditions must hold; otherwise both choices are pure and must match. Notice whose equation determines whose <probability>: the row player's indifference fixes the column player's $q$, and the column player's indifference fixes the row player's $p$.
At the mixed <Nash equilibrium>, <independence> gives
$$
\Pr(A,A)=\frac6{25},\quad \Pr(A,B)=\frac9{25},\quad
\Pr(B,A)=\frac4{25},\quad \Pr(B,B)=\frac6{25}.
$$
Thus they meet with <probability> $12/25$, and each receives <expected payoff> $3(6/25)+2(6/25)=6/5$. Holding $q=2/5$ fixed, every row deviation $p'$ gives $p'(6/5)+(1-p')(6/5)=6/5$; holding $p=3/5$ fixed, every column deviation $q'$ gives $q'(6/5)+(1-q')(6/5)=6/5$. This directly verifies the <Nash equilibrium> inequalities.
Both coordinated <pure strategy> outcomes strictly improve both players' <expected payoffs> over $(6/5,6/5)$: $(3,2)$ and $(2,3)$ each do so. The mixed <Nash equilibrium> therefore fails <Pareto efficiency>, even though no <unilateral deviation> helps. Switching together can help when switching alone cannot. Neither coordinated outcome improves both players over the other, and both are <Pareto efficient>, even when a <lottery over joint action profiles> is allowed: any positive weight on another outcome reduces the <payoff> of the player receiving $3$.
\Image[/coordination-game-equilibria.png]
{title=Asymmetric coordination: three best-response intersections and the cost of independent mixing}
Back to article page
