Solution
Dominated by a Mixture
Show the problem again
Row chooses a row and Column chooses a column, simultaneously and once. Row's payoffs: Row 1: 3, 3, 4 Row 2: 4, 1, 2 Row 3: 3, 2, 0 Column's payoffs: Row 1: 4, 2, 1 Row 2: 3, 0, 1 Row 3: 0, 2, 4 (The entry in position (i, j) is that player's payoff when Row plays i and Column plays j.) Rational players repeatedly delete every strategy that is strictly dominated by some pure or mixed strategy among those remaining. No pure strategy is strictly dominated by another pure strategy at the start. Exactly one outcome survives the full process. What is the product of the two players' payoffs at that outcome? Give an integer.
Worked solution
The answer is 12. Row 3 is never strictly dominated by a single row, but the mixture of 3/4 Row 1 and 1/4 Row 2 pays (3.25, 2.5, 3.5) against Row 3's (3, 2, 0), strictly more in every column, so Row 3 goes. With Row 3 gone, Column 2 pays Column (2, 0) in Rows 1 and 2, strictly less than Column 1's (4, 3), so Column 2 goes. Column 3 pays (1, 1), again strictly less than Column 1, so it goes too. Row then faces only Column 1, where Row 2 pays 4 and Row 1 pays 3, so Row 1 goes. The survivor is (Row 2, Column 1) with payoffs (4, 3), whose product is 12. This is also the unique Nash equilibrium.
Source: Dominance by mixed strategies is standard textbook material (e.g. Osborne and Rubinstein); numbers original to AxiomIQ.