# Can a dominance solvable game have a mixed strategy equilibrium?

Prelude: we get this question about specific finite games. Let us answer it generally. (examples 1, 2)

Suppose there is a two-player finite game (a "matrix game") where both players maximize their expected payoff - e.g., if player $$1$$ attributes probability $$q_i$$ to any strategy $$s_{2i}$$ of player $$2$$, then given her payoff function $$f_1$$ and a strategy $$s_{1k}$$ her payoff is $$\sum_i q_i \cdot f_1(s_k,s_{2i}).$$

The game is dominance solvable; that is, after the iterated elimination of strictly dominated strategies only one pure strategy profile remains.

Is it possible for this game to have a mixed strategy Nash-equilibrium?

• A game is dominance solvable means the process of iterated deletion of strictly dominated strategies leads to a unique outcome, which in matrix games amounts to a profile of sure strategies. We also know that the process does not eliminate any NEs. Therefore dominance solvable games must have unique pure strategy NE. Jul 15, 2022 at 20:54
• @HerrK. "We also know that the process does not eliminate any NEs." Do the people who asked the questions in examples 1 and 2 know? Jul 15, 2022 at 20:57
• They should know the result, if not the proof. I don't think the result itself is difficult to comprehend. Jul 15, 2022 at 21:05

No, that is not possible.

Assume that player $$2$$ has a strategy $$s_{2j'}$$ that is strictly dominated by $$s_{2j}$$. This means that for all pure strategies $$s_1$$ of player $$1$$, we have $$f_2(s_1,s_{2j}) > f_2(s_1,s_{2j'}).$$

Lemma 1. This above inequality is also true for any mixed strategies $$s_1$$ of player $$1$$.

Proof. Assume player $$1$$ mixes with a probability vector $$p$$. Thus the expected payoffs of player $$2$$ when playing strategies $$s_{2j}$$ and $$s_{2j'}$$ are respectively $$\sum_i p_i \cdot f_2(s_{1i},s_{2j}), \hskip 10pt \text{and} \hskip 10pt \sum_i p_i \cdot f_2(s_{1i},s_{2j'}).$$ For positive probabilities $$p_i$$ any member $$p_i \cdot f_2(s_{1i},s_{2j})$$ of the left hand sum will be larger than the corresponding $$p_i \cdot f_2(s_{1i},s_{2j'})$$ member of the right hand sum. When $$p_i = 0$$ the members too will be equal. $$\sum_i p_i = 1$$, so at least one $$p_i$$ is positive, thus the left hand sum is itself larger than the right hand sum; hence $$s_{2j}$$ dominated $$s_{2j'}$$ even against mixed strategies. $$QED$$

Lemma 2. If a mixed strategy $$s_2'$$ of player $$2$$ places positive weight $$q_{j'}$$ on her pure strategy $$s_{2j'}$$, then player $$2$$ has a mixed strategy $$s_2$$ that dominates $$s_2'$$.

Proof. Player $$2$$ can increase her payoff by removing the probability from $$s_{2j'}$$, because regardless of the strategy $$s_1$$ player $$1$$ plays we have \begin{align*} f_2(s_{1},s_{2j'}) & < f_2(s_{1},s_{2j}) \\ q_{j'} \cdot f_2(s_{1},s_{2j'}) & < q_{j'} \cdot f_2(s_{1},s_{2j}) \\ q_{j'} \cdot f_2(s_{1},s_{2j'}) + \sum_{i\neq j'} q_i \cdot f_2(s_{1},s_{2i}) & < q_{j'} \cdot f_2(s_{1},s_{2j}) + \sum_{i\neq j'} q_i \cdot f_2(s_{1},s_{2i}). \end{align*} The left hand side of the final inequality is the expected playoff of player $$2$$ while playing $$s_2'$$, while the larger right hand side is her expected payoff while playing another strategy. This other strategy - which we will denote by $$s_2$$ - yields a larger payoff than $$s_2'$$ irrespective of $$s_1$$, thus $$s_2$$ strictly dominated $$s_2'$$. $$QED$$

One can now perform the iteration of strictly dominated pure strategies again. In the end a player should never put positive weight on a strictly dominated pure strategy, even when facing mixed strategies. In equilibrium players act optimally, they will place 0 positive weight on all eliminated strategies. Thus in a dominance solvable game in equilibrium players will play pure strategies.

• But why shouldn't it be enough to say: (a) In a mixed NE at least one player uses at least two pure strategies. (b) Players don't use interatively strictly dominated pure strategies. From (a) and (b), a dominance solvable game cannot have a mixed NE. Oct 21, 2022 at 11:48
• I think the question is essentially about how to prove your (b) lemma. Oct 21, 2022 at 13:20