Solution
Four Strategies Each, Two Needed
Show the problem again
In a zero-sum game, the row player picks a row and the column player picks a column, simultaneously. The entries are the row player's winnings (the column player pays them): Row 1: −1, −4, 4, 6 Row 2: 3, 3, 0, 4 Row 3: 6, 2, −1, 6 Row 4: 8, −1, 7, 8 Both players play optimally. What is the value of the game to the row player? Give a reduced fraction.
Worked solution
The answer is 21/11. Row 4 beats Row 1 in every column, so delete Row 1. Against the remaining rows, the column player (who wants small numbers) strictly prefers Column 3, which gives (0, −1, 7), to Column 1, which gives (3, 6, 8), so delete Column 1. Column 4 gives (4, 6, 8), which is strictly worse for the column player than Column 3, so delete it too. Now Row 2 gives (3, 0) against Row 3's (2, −1) in the two remaining columns, so delete Row 3. This leaves Rows 2 and 4 against Columns 2 and 3: [[3, 0], [−1, 7]]. There is no saddle point, so the value is (3·7 − 0·(−1)) / (3 + 7 − 0 − (−1)) = 21/11. The row player mixes Rows 2 and 4 with probabilities 8/11 and 3/11, and the column player mixes Columns 2 and 3 with probabilities 7/11 and 4/11.
Source: Standard dominance-then-mix exercise for zero-sum games; matrix original to AxiomIQ.