<p>No, that is not possible.</p>
<p>Assume that player <span class="math-container">$2$</span> has a strategy <span class="math-container">$s_{2j'}$</span> that is strictly dominated by <span class="math-container">$s_{2j}$</span>. This means that for all pure strategies <span class="math-container">$s_1$</span> of player <span class="math-container">$1$</span>, we have
<span class="math-container">$$
f_2(s_1,s_{2j}) &gt; f_2(s_1,s_{2j'}). 
$$</span></p>
<p><strong>Lemma 1.</strong>
This above inequality is also true for any mixed strategies <span class="math-container">$s_1$</span> of player <span class="math-container">$1$</span>.</p>
<p><strong>Proof.</strong> Assume player <span class="math-container">$1$</span> mixes with a probability vector <span class="math-container">$p$</span>. Thus the expected payoffs of player <span class="math-container">$2$</span> when playing strategies <span class="math-container">$s_{2j}$</span> and <span class="math-container">$s_{2j'}$</span> are respectively
<span class="math-container">$$
\sum_i p_i \cdot f_2(s_{1i},s_{2j}), \hskip 10pt \text{and} \hskip 10pt \sum_i p_i \cdot f_2(s_{1i},s_{2j'}). 
$$</span>
For positive probabilities <span class="math-container">$p_i$</span> any member <span class="math-container">$p_i \cdot f_2(s_{1i},s_{2j})$</span> of the left hand sum will be larger than the corresponding <span class="math-container">$p_i \cdot f_2(s_{1i},s_{2j'})$</span> member of the right hand sum. When <span class="math-container">$p_i = 0$</span> the members too will be equal. <span class="math-container">$\sum_i p_i = 1$</span>, so at least one <span class="math-container">$p_i$</span> is positive, thus the left hand sum is itself larger than the right hand sum; hence <span class="math-container">$s_{2j}$</span> dominated <span class="math-container">$s_{2j'}$</span> even against mixed strategies. <span class="math-container">$QED$</span></p>
<hr />
<p><strong>Lemma 2.</strong>
If a mixed strategy <span class="math-container">$s_2'$</span> of player <span class="math-container">$2$</span> places positive weight <span class="math-container">$q_{j'}$</span> on her pure strategy <span class="math-container">$s_{2j'}$</span>, then player <span class="math-container">$2$</span> has a mixed strategy <span class="math-container">$s_2$</span> that dominates <span class="math-container">$s_2'$</span>.</p>
<p><strong>Proof.</strong>  Player <span class="math-container">$2$</span> can increase her payoff by removing the probability from <span class="math-container">$s_{2j'}$</span>, because regardless of the strategy <span class="math-container">$s_1$</span> player <span class="math-container">$1$</span> plays we have
<span class="math-container">$$
\begin{align*}
f_2(s_{1},s_{2j'}) &amp; &lt; f_2(s_{1},s_{2j}) \\
q_{j'} \cdot f_2(s_{1},s_{2j'}) &amp; &lt; q_{j'} \cdot f_2(s_{1},s_{2j}) \\
q_{j'} \cdot f_2(s_{1},s_{2j'}) + \sum_{i\neq j'} q_i \cdot f_2(s_{1},s_{2i}) &amp; &lt; q_{j'} \cdot f_2(s_{1},s_{2j}) + \sum_{i\neq j'} q_i \cdot f_2(s_{1},s_{2i}).
\end{align*}
$$</span>
The left hand side of the final inequality is the expected playoff of player <span class="math-container">$2$</span> while playing <span class="math-container">$s_2'$</span>, while the larger right hand side is her expected payoff while playing another strategy. This other strategy - which we will denote by <span class="math-container">$s_2$</span> - yields a larger payoff than <span class="math-container">$s_2'$</span> irrespective of <span class="math-container">$s_1$</span>, thus <span class="math-container">$s_2$</span> strictly dominated <span class="math-container">$s_2'$</span>. <span class="math-container">$QED$</span></p>
<hr />
<p>One can now perform the iteration of strictly dominated <em>pure</em> 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.</p>

CC BY-SA 4.0
2022-10-21T11:48:46.393Z
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.
VARulle
CC BY-SA 4.0
2022-10-21T13:20:17.877Z
I think the question is essentially about how to prove your (b) lemma.
Giskard
CC BY-SA 4.0
2022-07-15T20:49:54.343Z
2022-07-15T20:49:54.343Z
Giskard
<p>Prelude: we get this question about specific finite games. Let us answer it generally. (examples <a href="https://economics.stackexchange.com/questions/51947/can-a-dominance-solvable-game-have-mixed-strategies">1</a>, <a href="https://economics.stackexchange.com/questions/52081/is-there-a-mixed-strategy-in-dominance-solvable-game">2</a>)</p>
<hr />
<p>Suppose there is a two-player finite game (a &quot;matrix game&quot;) where both players maximize their expected payoff - e.g., if player <span class="math-container">$1$</span> attributes probability <span class="math-container">$q_i$</span> to any strategy <span class="math-container">$s_{2i}$</span> of player <span class="math-container">$2$</span>, then given her payoff function <span class="math-container">$f_1$</span> and a strategy <span class="math-container">$s_{1k}$</span> her payoff is
<span class="math-container">$$
\sum_i q_i \cdot f_1(s_k,s_{2i}). 
$$</span></p>
<p>The game is dominance solvable; that is, after the iterated elimination of strictly dominated strategies only one pure strategy profile remains.</p>
<p>Is it possible for this game to have a mixed strategy Nash-equilibrium?</p>

CC BY-SA 4.0
2022-07-15T20:54:37.997Z
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.
Herr K.
CC BY-SA 4.0
2022-07-15T20:57:23.687Z
@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?
Giskard
CC BY-SA 4.0
2022-07-15T21:05:55.473Z
They *should* know the result, if not the proof. I don't think the result itself is difficult to comprehend.
Herr K.
<game-theory><nash-equilibrium>
Can a dominance solvable game have a mixed strategy equilibrium?
Giskard
# Can a dominance solvable game have a mixed strategy equilibrium?

(Asked by: username_0 on 2022-07-15)

Prelude: we get this question about specific finite games. Let us answer it generally. (examples [1](https://economics.stackexchange.com/questions/51947/can-a-dominance-solvable-game-have-mixed-strategies), [2](https://economics.stackexchange.com/questions/52081/is-there-a-mixed-strategy-in-dominance-solvable-game))

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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?

**Tags:** game-theory, nash-equilibrium

> **Comments:**
> **username_1**: 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.  
> **username_0**: @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?  
> **username_1**: They *should* know the result, if not the proof. I don't think the result itself is difficult to comprehend.  

---

## 1 answers

### Answer by username_0 on 2022-07-15. Score: 4

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 $s\_2'$. $QED$

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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.

> **Comments:**
> **username_2**: 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.  
> **username_0**: I think the question is essentially about how to prove your (b) lemma.  

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