Solve a Game by Crossing Out Bad Moves (Iterated Elimination)

Why this matters

Eliminating one dominated strategy sometimes exposes another strategy as now-dominated in the smaller game. Repeat. Sometimes the game collapses to a single strategy profile — and that profile is often the 'solution' to the game without ever mentioning Nash equilibrium. Iterated elimination is the first real solution concept you'll meet, and every time you use it you're quietly invoking common knowledge of rationality from the previous lesson.

Demo

Start from the game $G$. Remove strictly dominated strategies for every player simultaneously, producing $G_1$. Do it again on $G_1$ to get $G_2$. Keep going until nothing can be removed; the fixed point $G^\infty$ is the set of iteratively un-dominated strategies:

# A tiny demo: a 3x3 game that reduces to a single profile.
# Rows: U, M, D.  Columns: L, C, R.  Payoffs are (Row, Col).
payoffs = {
    ("U","L"): (3,0), ("U","C"): (0,2), ("U","R"): (0,3),
    ("M","L"): (2,1), ("M","C"

Try it yourself

1. On paper, work through a 3x3 game where the first round of elimination removes one strategy per player, and the second round collapses to a single outcome.
2. Try the order-sensitivity trap: pick a game where weak dominance is present and demonstrate that different elimination orders give different surviving sets.
3. What does it mean if iterated elimination leaves more than one profile? (Preview: those are the candidates for Nash equilibrium — Module 3.)
4. Explain in one sentence: which level of common knowledge justifies three rounds of elimination?

References

• Bernheim — Rationalizable Strategic Behavior (Econometrica, 1984)
• Pearce — Rationalizable Strategic Behavior and the Problem of Perfection (Econometrica, 1984)