Solution
Peeling Away Dominated Strategies
Show the problem again
Row chooses U, M or D and Column chooses L, C or R, simultaneously and once. Each cell below is (Row's payoff, Column's payoff). Row plays U: vs L gives (4, 0), vs C gives (0, 2), vs R gives (1, 2) Row plays M: vs L gives (3, 2), vs C gives (2, 1), vs R gives (4, 3) Row plays D: vs L gives (2, 3), vs C gives (1, 3), vs R gives (5, 4) Both players are rational and this is common knowledge, so each repeatedly deletes strategies that are strictly dominated by another remaining pure strategy. Exactly one outcome survives. What is the sum of the two players' payoffs at that outcome? Give an integer.
Worked solution
The answer is 9. Row has nothing to delete at the start (no row is beaten in every column), but Column's L pays (0, 2, 3) against R's (2, 3, 4) in rows U, M, D, so R strictly dominates L and L goes. With L gone, Row's U yields (0, 1) against M's (2, 4) in columns C and R, so U goes. Among the remaining rows M and D, Column compares C, which pays (1, 3), with R, which pays (3, 4), so C goes. Finally Row faces only column R, where D pays 5 and M pays 4, so M goes. The surviving outcome is (D, R) with payoffs (5, 4), summing to 9. Because the deletions are strict, the order in which they are made does not matter, and (D, R) is also the unique pure Nash equilibrium.
Source: Textbook-style iterated strict dominance exercise; no single originator. Original numbers written for AxiomIQ.