Okay consider a game $G$ if a strategy $s_i$ has the following property we call $s_i$ the strictly dominant strategy

$$u_i(s_i,s_{-i})>u_i(s_i',s_{-i}) \\ \forall s_{-i} \ \forall s_i' \epsilon S'_i$$

Where $s_{i}$ indicates the strategies of players other then $i$ in the game and $S'_i$ is the set for strategies of player $i$ except the specific strategy $s_i$

Now let's look at the definition of a weakly dominant strategy

if a strategy $s_i$ has the following property we call $s_i$ the weakly dominant strategy

$$u_i(s_i,s_{-i})≥u_i(s_i',s_{-i}) \\ \forall s_{-i} \ \forall s_i' \epsilon S'_i \ and \\ \exists s_i' \epsilon S'_i \ such \ that: \ u_i(s_i,s_{-i})>u_i(s_i',s_{-i})$$

Okay I believe from these two definitions we can derive that any strictly dominant strategy $s_i$ is also a weakly dominant strategy

Definition of strict dominance solvable is as follows :

A strict dominance solvable game is a game where the equilibrium outcome is strict dominance equilibrium. A weakly dominance solvable games is a game where the equilibrium outcome is weakly dominance equilibrium.

So It seems like then a strictly dominance solvable game is always a weakly dominance solvable game. Am I wrong?


1 Answer 1


You're right. Let $s^*=(s_1^*,\dots,s_N^*)$ be the equilibrium of a strictly dominance solvable game. Then by definition, $$ u_i(s_i^*,s_{-i})>u_i(s_i,s_{-i}) $$ for all $i$, all $s_i\ne s_i^*$ and all $s_{-i}$. This implies that $$ u_i(s_i^*,s_{-i})\ge u_i(s_i,s_{-i}) $$ for all $i$, all $s_i\ne s_i^*$, all $s_{-i}$ and with strict inequality for at least some $s_i$ (in fact, for all $s_i\ne s_i^*$). This makes $s^*$ an equilibrium satisfying the weak dominance solvability criterion.

Your quoted definition of strict/weak dominance solvability makes me a little uncomfortable though. I'd say

A game is strict (or weak) dominance solvable if the process of iteratively removing strictly (or weakly) dominated strategies leads to a unique outcome (i.e. only one strategy for each player survives).

  • 1
    $\begingroup$ You are right to be uneasy because iteratively removing weakly dominated strategies can lead to different unique outcomes depending on the order of the removal of the strategies. I would say that these are in some sense strange solutions. $\endgroup$
    – Giskard
    Commented Feb 21, 2017 at 20:15
  • 1
    $\begingroup$ An example: Both players have to choose between strategy $A$ and $B$. If your opponent choses $A$ your payoff is 0, otherwise it is 1. In these games both $A$ and $B$ are weakly dominant strategies. You can eliminate either one and have whatever you want as a solution: All strategy profiles are weakly dominant equilibria. $\endgroup$
    – Giskard
    Commented Feb 21, 2017 at 20:20
  • $\begingroup$ @denesp: Iteratively deleting weakly dominated strategies has long been known to be problematic. Aside from the point you mentioned, the fact that the procedure can remove Nash equilibria from a game makes it an undesirable method for analyzing games. $\endgroup$
    – Herr K.
    Commented Feb 21, 2017 at 21:07
  • $\begingroup$ I think that is included in my argument (you can remove weakly dominant equilibria and all w.d.equilibra are NE) but yes, we are in agreement. That is why the phrase 'solvable' is somewhat problematic in this context in my opinion. $\endgroup$
    – Giskard
    Commented Feb 21, 2017 at 22:56
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    $\begingroup$ @denesp in your example, for the player with payoffs specified, A and B are payoff equivalent and thus neither weakly dominates the other, no? A proper example of order dependence for iterated elimination of weakly dominated strategies necessarily needs more than one step. $\endgroup$ Commented Feb 22, 2017 at 23:41

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