Solution
A Three-by-Three Value
Show the problem again
In a zero-sum game, the row player chooses a row and the column player chooses a column, simultaneously. The entries give the row player's payoff (the column player pays that amount): Row 1: 2, −1, 5 Row 2: −4, 5, −1 Row 3: 6, −1, −1 Both players use optimal mixed strategies. What is the value of the game to the row player? Give a reduced fraction.
Worked solution
The answer is 11/10. No entry is simultaneously the smallest in its row and largest in its column, and no row or column is dominated, so both players mix over all three options. Let the column player's optimal mix be (a, b, c). Making the row player indifferent among the three rows gives 2a − b + 5c = −4a + 5b − c = 6a − b − c = v together with a + b + c = 1. Solving gives (a, b, c) = (3/10, 1/2, 1/5) and v = 11/10. The row player's optimal mix (7/20, 7/20, 3/10) guarantees at least 11/10 in every column, for example 2(7/20) − 4(7/20) + 6(3/10) = 11/10, matching the column player's guarantee, so 11/10 is the value.
Source: Standard zero-sum matrix game exercise; matrix original to AxiomIQ.