Iterated Elimination of Strictly Dominated Strategies

Watch dominated strategies get deleted one by one. Push the round slider to eliminate them in order.

If a strategy always does worse than another one, no rational player uses it. Delete it and the table shrinks. In the smaller table, another bad strategy shows up. In this scene C, M, and D vanish in turn until only U remains.

For example, say a 🏪 shop sells three kinds of 🍬 candy. One kind is always smaller and less tasty than another kind. Nobody ever picks it, so it can leave the shelf. Once it is gone, another always-worse candy stands out. Deleting step by step leaves the final choice.

si′ strictly dominates si  ⟺  ui(si′,s−i)>ui(si,s−i)    ∀s−is_i' \text{ strictly dominates } s_i \iff u_i(s_i', s_{-i}) > u_i(s_i, s_{-i}) \;\; \forall s_{-i}

Part of these learning paths

Examples & Exercises

Run iterated elimination of strictly dominated strategies for three rounds in the 3×33 \times 3 game, identifying what is removed and why each round.

LCRU(4,3)(5,1)(6,2)M(2,1)(8,4)(3,6)D(3,0)(9,6)(2,8)\begin{matrix} & L & C & R \\ U & (4,3) & (5,1) & (6,2) \\ M & (2,1) & (8,4) & (3,6) \\ D & (3,0) & (9,6) & (2,8) \end{matrix}
After 3 rounds — only UU survives for P1; P2 picks LL.
Show proof

  1. Round 1: comparing player 2's payoffs, column CC (1,4,6)(1,4,6) vs column RR (2,6,8)(2,6,8) — RR is larger in every row, so RR strictly dominates CC. Remove CC.

  2. Round 2 (against {L,R}\{L,R\}): player 1's row MM (2,3)(2,3) vs row UU (4,6)(4,6) — UU strictly dominates. Remove MM.

    Round 3: row DD (3,2)(3,2) vs row UU (4,6)(4,6) — UU strictly dominates again. Remove DD.

  3. Only UU survives for player 1.

    Against UU, player 2 prefers LL (payoff 33) over RR (payoff 22).

    Final outcome (U,L)=(4,3)(U, L) = (4, 3) — also a Nash equilibrium (unilateral deviation loses).

Prove that iterated elimination of strictly dominated strategies never eliminates a Nash equilibrium.

Elimination complete — the equilibrium (U,L)(U, L) survived.
Show proof

  1. That si′s_i' strictly dominates sis_i means ui(si′,s−i)>ui(si,s−i)u_i(s_i', s_{-i}) > u_i(s_i, s_{-i}) for every s−is_{-i}. If a Nash equilibrium s∗s^* used a strictly dominated si∗s_i^*, deviating to si′s_i' would strictly raise the payoff against s−i∗s_{-i}^* — contradicting equilibrium.

  2. So no first-round elimination can remove part of a Nash equilibrium.

    The same argument applies inside the reduced game, so by induction every round preserves all Nash equilibria: the IEDS survivor set always contains every Nash equilibrium.

8. Game Theory