 Solution

For a bimatrix game with row and column payoff matrices $A,B$, an equilibrium pair is a pair of mixed strategies $(p,q)$ for which neither player gains by a unilateral change:This is a Nash equilibrium, or mutual best response pair. Every finite game with nonempty finite pure-strategy sets has at least one mixed-strategy equilibrium by Nash's theorem. Equivalently, the strategy domains are compact convex probability simplexes and the expected payoffs are continuous and linear in each player's own strategy. The existence assertion need not hold if only pure strategies are allowed.

Write the probability vectors in the order Bach, Mozart, Schubert. Against $q=(q_B,q_M,q_S)$ the row player's three pure-action payoffs areAgainst $p=(p_B,p_M,p_S)$ the column player's payoffs areA supported action must attain the corresponding maximum, and an omitted action must give no larger payoff.

We first justify that both players have the same support of a mixed strategy, rather than assume it. Each player's maximum payoff is strictly positive for every opponent probability vector. Thus $p_M>0$ implies $q_M>0$, $q_B>0$ implies $p_B>0$, and $p_S>0$ is equivalent to $q_S>0$. If $p_B>0$ but $q_B=0$, then either $q_M>0$, making Mozart's payoff $2q_M$ strictly exceed Bach's $q_M$, or $q_S=1$, making Bach's payoff zero and Schubert's positive. Both contradict best response. Therefore $p_B>0$ also implies $q_B>0$. Similarly, if $q_M>0$ but $p_M=0$, then $p_B>0$ makes the column payoff for Bach $2p_B$ exceed Mozart's $p_B$, or $p_S=1$ makes Mozart's payoff zero and Schubert's positive. Thus $q_M>0$ implies $p_M>0$. The supports coincide, and there are only seven nonempty possibilities.

For singleton supports the three matching pure pairs are equilibria. For each two-element support, solve the two players' indifference equations and normalize their probabilities. For example, on Bach–Mozart, the equations are $4q_B=q_M$ and $p_B=4p_M$, giving $q=(1/5,4/5,0)$ and $p=(4/5,1/5,0)$. On Bach–Schubert they are $4q_B=3q_S$ and $2p_B=3p_S$. On Mozart–Schubert they are $2q_M=3q_S$ and $4p_M=3p_S$. In each case the omitted action gives strictly less than the common supported payoff.

For full support, the row indifference equations give $q_M=4q_B$ and $3q_S=2q_M$; the column equations give $p_B=4p_M$ and $3p_S=2p_B$. Normalization determines the remaining pair. The full list, with row strategy first, isAll displayed supported probabilities are positive. The supported payoff pairs in the four nonpure cases are respectively $(8/5,8/5)$, $(12/7,6/5)$, $(6/5,12/7)$ and $(24/23,24/23)$; these values also verify the omitted-action inequalities directly. The common-support argument excludes every other support pair, and each indifference system has a unique solution. Hence there are exactly seven equilibrium pairs, three pure and four genuinely mixed. This is a complete support enumeration for a bimatrix game, not just a list of selected equilibria.