Solution
A Unique Mixed Equilibrium
Show the problem again
Row chooses a row and Column chooses a column, simultaneously and once. Row's payoffs: Row 1: 3, 1, 5 Row 2: 4, 4, 0 Row 3: 2, 1, 6 Column's payoffs: Row 1: 4, 6, 0 Row 2: 2, 3, 7 Row 3: 7, 6, 4 (The entry in position (i, j) is the payoff to that player when Row plays i and Column plays j.) The game has exactly one Nash equilibrium. What is Row's expected payoff in it? Give a reduced fraction.
Worked solution
The answer is 14/5. No cell is a mutual best reply, so there is no pure equilibrium, and checking the support pairs shows that the only equilibrium has both players using all three strategies. Column's mix (y1, y2, y3) must make Row indifferent among her rows: 3y1 + y2 + 5y3 = 4y1 + 4y2 = 2y1 + y2 + 6y3, with y1 + y2 + y3 = 1. Equating the first and third expressions gives y1 = y3; equating the first and second then gives 4y1 = 3y2. Normalising yields y = (3/10, 2/5, 3/10), and Row's common payoff is 4(3/10) + 4(2/5) = 14/5. Symmetrically, Row's mix (1/13, 5/13, 7/13) makes Column indifferent at 63/13, and no deviation helps either player.
Source: Standard bimatrix-game exercise (support enumeration); payoffs original to AxiomIQ.