<blockquote>
  <p>First, because action <span class="math-container">$M$</span> is weakly dominated for Player 3, does everyone assume that <span class="math-container">$r_2=0$</span>? </p>
</blockquote>

<p>Without seeing the game you're considering, it's hard to tell. But the general answer is No. Weakly dominated strategies <em>can</em> be played in a Nash equilibrium (either mixed or pure). In the following game, both <span class="math-container">$B$</span> and <span class="math-container">$C$</span> are weakly dominated strategies for both players. But each player randomizing between them in any proportion is still a NE. </p>

<p><span class="math-container">\begin{array}{|c|c|c|c|}
\hline &amp; A &amp; B&amp;C\\\hline 
A&amp;1,1&amp;0,0&amp;0,0\\\hline 
B&amp;0,0&amp;0,0&amp;0,0\\\hline 
C&amp;0,0&amp;0,0&amp;0,0\\\hline
\end{array}</span></p>

<blockquote>
  <p>Does Player 3 need to be indifferent between <span class="math-container">$L$</span>, <span class="math-container">$M$</span>, and <span class="math-container">$R$</span>, or just between <span class="math-container">$L$</span> and <span class="math-container">$R$</span>, since <span class="math-container">$M$</span> is weakly dominated?</p>
</blockquote>

<p>The indifference condition for MSNE applies to pure strategies <em>in the support of the equilibrium mixed strategy</em>. If <span class="math-container">$M$</span> is played with positive probability in equilibrium, then you'd need indifference across all three. If <span class="math-container">$M$</span> is played with zero probability, then you'd need <span class="math-container">$u_3(\sigma_1,\sigma_2,L)=u_3(\sigma_1,\sigma_2,R)\ge u_3(\sigma_1,\sigma_2,M)$</span></p>

<hr>

<p>More generally, the <strong>necessary and sufficient conditions</strong> for a Nash equilibrium (either pure or mixed) are summarized in the following theorem: </p>

<blockquote>
  <p>A strategy profile <span class="math-container">$(\sigma_1,\dots,\sigma_n)$</span> is a Nash equilibrium if and only if for all <span class="math-container">$i=1,\dots,n$</span>, 
  <span class="math-container">\begin{equation}
u_i(s_i,\sigma_{-i})= u_i(s_i',\sigma_{-i}), \qquad \forall s_i,s_i'\in\mathrm{supp}(\sigma_i)
\end{equation}</span>
  and 
  <span class="math-container">\begin{equation}
u_i(s_i,\sigma_{-i})\ge u_i(s_i',\sigma_{-i}), \qquad \forall s_i\in\mathrm{supp}(\sigma_i)\text{ and }\forall s_i'\notin\mathrm{supp}(\sigma_i).
\end{equation}</span></p>
</blockquote>

<p>Hence, we see that the indifference condition is only required for pure strategies in the support of the equilibrium mixed strategy but not necessarily outside it. </p>

CC BY-SA 4.0
2019-03-11T21:25:48.883Z
2019-03-11T22:00:51.873Z
2019-03-11T22:00:51.873Z
Herr K.
<p>Suppose we are playing a game where the Action set for Player 1 is <span class="math-container">$(a,b)$</span>, for Player 2 is <span class="math-container">$(c,d)$</span>, and for Player 3 is <span class="math-container">$(L,M,R)$</span>. Assume that for Player 3, the action <span class="math-container">$M$</span> is weakly dominated by some mix between <span class="math-container">$L$</span> and <span class="math-container">$R$</span>. I am attempting to find all of the possible mixed equilibria. </p>

<p>Let <span class="math-container">$p$</span> be the probability that Player 1 chooses <span class="math-container">$a$</span>, let <span class="math-container">$q$</span> be the probability that Player 2 chooses <span class="math-container">$c$</span>, and let <span class="math-container">$r_1, r_2, r_3$</span> be the probabilities that Player 3 choose <span class="math-container">$L, M, R$</span> respectively. </p>

<p>I have two questions:</p>

<p>First, because action <span class="math-container">$M$</span> is weakly dominated for Player 3, does everyone assume that <span class="math-container">$r_2=0$</span>? </p>

<p>Second, the probabilities <span class="math-container">$p$</span> and <span class="math-container">$q$</span> are subject to the constraint that Player 3 be indifferent between his/her own actions. Does Player 3 need to be indifferent between <span class="math-container">$L$</span>, <span class="math-container">$M$</span>, and <span class="math-container">$R$</span>, or just between <span class="math-container">$L$</span> and <span class="math-container">$R$</span>, since <span class="math-container">$M$</span> is weakly dominated?</p>

<game-theory><nash-equilibrium>
Necessary indifference conditions in mixed equilibrium
David
# Necessary indifference conditions in mixed equilibrium

(Asked by: username_0 on 2019-03-11)

Suppose we are playing a game where the Action set for Player 1 is $(a,b)$, for Player 2 is $(c,d)$, and for Player 3 is $(L,M,R)$. Assume that for Player 3, the action $M$ is weakly dominated by some mix between $L$ and $R$. I am attempting to find all of the possible mixed equilibria.

Let $p$ be the probability that Player 1 chooses $a$, let $q$ be the probability that Player 2 chooses $c$, and let $r\_1, r\_2, r\_3$ be the probabilities that Player 3 choose $L, M, R$ respectively.

I have two questions:

First, because action $M$ is weakly dominated for Player 3, does everyone assume that $r\_2=0$?

Second, the probabilities $p$ and $q$ are subject to the constraint that Player 3 be indifferent between his/her own actions. Does Player 3 need to be indifferent between $L$, $M$, and $R$, or just between $L$ and $R$, since $M$ is weakly dominated?

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


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## 1 answers

### Answer by username_1 on 2019-03-11. Score: 3 | Accepted answer

> First, because action $M$ is weakly dominated for Player 3, does everyone assume that $r\_2=0$?

Without seeing the game you're considering, it's hard to tell. But the general answer is No. Weakly dominated strategies *can* be played in a Nash equilibrium (either mixed or pure). In the following game, both $B$ and $C$ are weakly dominated strategies for both players. But each player randomizing between them in any proportion is still a NE.

\begin{array}{|c|c|c|c|}
\hline & A & B&C\\\hline
A&1,1&0,0&0,0\\\hline
B&0,0&0,0&0,0\\\hline
C&0,0&0,0&0,0\\\hline
\end{array}

> Does Player 3 need to be indifferent between $L$, $M$, and $R$, or just between $L$ and $R$, since $M$ is weakly dominated?

The indifference condition for MSNE applies to pure strategies *in the support of the equilibrium mixed strategy*. If $M$ is played with positive probability in equilibrium, then you'd need indifference across all three. If $M$ is played with zero probability, then you'd need $u\_3(\sigma\_1,\sigma\_2,L)=u\_3(\sigma\_1,\sigma\_2,R)\ge u\_3(\sigma\_1,\sigma\_2,M)$

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More generally, the **necessary and sufficient conditions** for a Nash equilibrium (either pure or mixed) are summarized in the following theorem:

> A strategy profile $(\sigma\_1,\dots,\sigma\_n)$ is a Nash equilibrium if and only if for all $i=1,\dots,n$,
> \begin{equation}
> u\_i(s\_i,\sigma\_{-i})= u\_i(s\_i',\sigma\_{-i}), \qquad \forall s\_i,s\_i'\in\mathrm{supp}(\sigma\_i)
> \end{equation}
> and
> \begin{equation}
> u\_i(s\_i,\sigma\_{-i})\ge u\_i(s\_i',\sigma\_{-i}), \qquad \forall s\_i\in\mathrm{supp}(\sigma\_i)\text{ and }\forall s\_i'\notin\mathrm{supp}(\sigma\_i).
> \end{equation}

Hence, we see that the indifference condition is only required for pure strategies in the support of the equilibrium mixed strategy but not necessarily outside it.


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