Why Elimination Is Useful—and Where It Stops
A payoff matrix can contain actions that a player would never choose: one action may yield a lower payoff than another no matter what the opponent does. Removing such an action simplifies the game and can expose additional dominated strategies. The key discipline is to repeat the comparison using the current game, not the original matrix.
For strict dominance, the order of valid eliminations does not change the final set of surviving strategies in a finite game. That guarantee does not extend to weak dominance. A weakly dominated strategy can tie another action in some situations, and deleting it can remove a strategically relevant tie or change what can be deleted next. The distinction matters when using elimination as a reasoning tool.
All comparisons are made in the game as it currently stands. A strategy that is dominated in the original game remains dominated when opponents' strategies are removed, because there are fewer opponent profiles to check. Conversely, removing an opponent strategy can make a new dominance comparison possible. In this tutorial, a deletion sequence is maximal if it continues until no remaining strategy is dominated under the chosen dominance rule.
How to Apply Iterated Strict-Dominance Elimination
For each candidate strategy, compare its payoff with the payoffs from another available pure strategy or mixed strategy, holding fixed every currently available opponent profile. For a row player, compare row payoffs across columns; for a column player, compare column payoffs down rows. Delete a strategy only after the required strict inequality has been verified everywhere.
Crossed-out strategies are no longer available and must not be used as dominators.
For a proposed strict domination, check that the candidate replacement pays strictly more against every remaining opponent profile.
After a valid deletion, retest all players. An elimination may reveal a new strict dominance relation.
Report the surviving strategies and distinguish “not eliminated” from “best” or “optimal.”
Worked Example: Two Valid Elimination Orders
A supplier and a buyer choose strategies simultaneously. Each cell lists (supplier payoff, buyer payoff). Eliminate strictly dominated strategies and check whether two possible orders yield the same survivors.
| Buyer L | Buyer R | |
|---|---|---|
| Supplier A | (6, 4) | (4, 2) |
| Supplier B | (5, 3) | (3, 2) |
| Supplier C | (1, 1) | (0, 0) |
Check the supplier's strategies. Against L, A pays \(6\), B pays \(5\), and C pays \(1\). Against R, A pays \(4\), B pays \(3\), and C pays \(0\). Thus A strictly dominates B and C: \(6>5,\ 4>3,\ 6>1,\) and \(4>0\).
Check the buyer's strategies. The buyer receives \(4,3,1\) from L against A, B, C, respectively, and \(2,2,0\) from R. Therefore L strictly dominates R because \(4>2,\ 3>2,\) and \(1>0\).
One sequence removes B, then C, then R, leaving A and L. Another removes R first; with only L remaining, A still pays more than B and C, so removing them also leaves A and L. The method is direct payoff comparison, and the trap to avoid is comparing a buyer's payoffs using the supplier's payoff entries.
Strict-Dominance Elimination Is Order-Independent
In a finite strategic-form game, if each deletion removes a strategy strictly dominated by an available pure or mixed strategy, every maximal deletion sequence leaves the same strategies for each player. This statement concerns the final survivors, not the number of steps or the order of intermediate deletions. Finiteness matters: with finitely many strategies, each deletion reduces the number remaining, so a process that continues must stop.
The reason is that strict domination is preserved when opponent strategies are removed. If a mixture gives more than a strategy against every profile in a set of opponent strategies, it continues to give more against any subset. Also, if a strategy used in a dominating mixture is itself eliminated, it can be replaced by a mixture that strictly dominates it. Repeating such replacements transfers the original strict improvement to strategies that remain.
Here is the replacement argument more formally. Suppose a strategy \(s_i\) is deleted in one sequence because an available mixture \(\sigma_i\) gives a strictly higher payoff than \(s_i\) against every opponent profile then remaining. Compare that deletion with another maximal sequence. If some strategy in the support of \(\sigma_i\) was deleted in the other sequence, replace it by the mixture that strictly dominated it when it was removed. This replacement strictly improves the payoff wherever that component had positive probability. Continue replacing deleted components; the process terminates because the game is finite. The result is a mixture of strategies surviving the other sequence that still strictly beats \(s_i\) against all opponents who survive there. Therefore, if \(s_i\) were still present at the end of that sequence, it would be strictly dominated in its terminal game, contradicting maximality. The same reasoning in the opposite direction shows that the two terminal sets coincide.
This result does not say that every strategy that survives is part of a particular outcome, nor that elimination identifies every strategically relevant feature. It says only that strict-dominance deletion has a stable endpoint. A strategy may survive because no available pure or mixed strategy strictly dominates it, even when the game requires further analysis.
Worked Example: A Deletion Does Not Justify Further Deletions
A row player has strategies U, V, W; a column player has strategies L, R. Each cell lists (row payoff, column payoff). Apply strict-dominance elimination to the matrix and state what remains.
| Column L | Column R | |
|---|---|---|
| Row U | (5, 2) | (1, 0) |
| Row V | (4, 0) | (4, 1) |
| Row W | (2, 0) | (2, 3) |
For the row player, V pays \(4\) against L and \(4\) against R, while W pays \(2\) against each column. Since \(4>2\) in both cases, W is strictly dominated by V. No other row strictly dominates another initially: U pays more than V against L (\(5>4\)), but less against R (\(1<4\)); and neither U nor V strictly dominates W because U pays less than W against R (\(1<2\)).
For the column player, the payoffs from L across U, V, W are \(2,0,0\); those from R are \(0,1,3\). L does not strictly dominate R because \(0<1\) against V, and R does not strictly dominate L because \(0<2\) against U. After deleting W, the same comparisons for the column player still fail. The row strategies U and V also still trade off: U pays more against L, while V pays more against R. The survivors are U, V and L, R. The method is to retest every comparison after deletion; the mistake to avoid is claiming an additional elimination without verifying every remaining opponent action.
Why Weak-Dominance Elimination Has Different Limits
Weak dominance allows equality at some opponent profiles. That equality is not a minor technical detail: it can make the order of deletion matter. If one strategy is deleted first, a comparison that was tied may no longer permit any deletion. A weakly dominated strategy may also be a best response to an opponent action, because “at least as good everywhere” permits equal payoffs.
Worked Example: Weak-Dominance Order Changes the Survivors
Consider this game, again listing (row payoff, column payoff). Compare two valid deletion orders when weakly dominated strategies may be removed.
| Column L | Column R | |
|---|---|---|
| Row A | (3, 0) | (2, 1) |
| Row B | (1, 0) | (2, 1) |
The row player’s A weakly dominates B: against L, \(3>1\), and against R, \(2=2\). For the column player, R strictly dominates L: against A, \(1>0\), and against B, \(1>0\).
Order one: remove B first. The row strategies then reduce to A. R strictly dominates L, so the terminal strategies are A and R.
Order two: remove L first. Only R remains. The row payoffs are now \(2\) for A and \(2\) for B. Neither weakly dominates the other under the definition requiring at least one strict inequality, so both survive. The terminal strategies are A, B and R. Thus weak-dominance elimination can produce different survivors depending on the order.
The tie has a strategic interpretation: against R, B is as good as A for the row player. Deleting B may therefore discard an action that is a best response to R. The common mistake is to treat weak dominance as if it had the order-independence guarantee of strict dominance; this example shows why that is unsafe.
Practical Checks and Exam Traps
A reliable solution records the payoff comparisons that justify each deletion. In a matrix with several rows and columns, a single failed inequality invalidates a claim of strict dominance. When mixed strategies are allowed, calculate the expected payoff of the proposed mixture against each remaining opponent action; a mixture that improves against most, but not all, actions does not strictly dominate.
The treatment here is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus. For a clear answer, state whether dominance is strict or weak, identify the dominating action or mixture, show the comparisons against all currently available opponent strategies, and list the terminal strategy sets. If the question specifies pure-strategy domination, do not quietly use a mixed strategy; if it permits mixed domination, do not overlook a useful probability mixture.
Check Your Understanding
For each problem, apply the stated dominance rule to the current game and check every comparison against the remaining opponent strategies.
- A row player chooses M or N and a column player chooses X or Y. Payoffs are M,X: \((7,6)\), M,Y: \((3,2)\), N,X: \((5,5)\), and N,Y: \((1,1)\). Under strict dominance, identify the dominated strategies and the survivors. (Answer: N is strictly dominated by M, and Y is strictly dominated by X; the survivors are M and X.)
- Rows P, Q, R and columns C, D have payoffs P,C \((5,1)\), P,D \((2,0)\), Q,C \((4,0)\), Q,D \((4,2)\), R,C \((1,0)\), R,D \((1,0)\). Under strict dominance, delete any dominated strategies and state the survivors. (Answer: R is strictly dominated by P; afterward neither P nor Q dominates the other, and neither column dominates the other; the survivors are P, Q, C and D.)
- In the U, V, W by L, R game in the worked example, why is it invalid to claim that L strictly dominates R after W is deleted? (Answer: Against V, the column player receives 0 from L and 1 from R, so L does not strictly dominate R; the surviving game has no strictly dominated strategy.)
- In the A, B by L, R game in the weak-dominance example, compare deleting B first with deleting L first. Give the terminal strategy sets in both cases. (Answer: Deleting B first leaves A and R; deleting L first leaves A, B and R.)
- A row player has strategies A, B, C, and the opponent has L, R. The row payoffs are A: \((4,0)\), B: \((0,4)\), and C: \((1,1)\). Does the equal mixture of A and B strictly dominate C? Show its expected payoffs. (Answer: Yes. Against L its expected payoff is \((4+0)/2=2>1\), and against R it is \((0+4)/2=2>1\); the mixture strictly dominates C.)