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Given game also has one pure-strategy Nash equilibrium besides the two you mentioned: Row player plays <span class="math-container">$C$</span>, Column player (left) plays <span class="math-container">$C$</span> and right plays <span class="math-container">$H$</span> i.e. <span class="math-container">$C,H$</span>.</p>
<p>Also, when player 1 chooses a mixed strategy in order to make player 2 indifferent, that does not necessarily mean that player 1 is choosing optimally. It is only when player 2 makes player 1 indifferent, then the mixed strategy chosen by 1 becomes optimal. In fact as <span class="math-container">$p$</span> gets close to <span class="math-container">$1$</span> (not necessarily equals 1), given that player 2 (left) chooses <span class="math-container">$H$</span> and player 2 (right) chooses <span class="math-container">$(\frac{2}{3}, \frac{1}{3})$</span>, player 1 will not choose <span class="math-container">$(\frac{1}{3},\frac{2}{3})$</span>, since player 1 is no longer indifferent between <span class="math-container">$C$</span> and <span class="math-container">$H$</span>. Playing <span class="math-container">$C$</span> will give row player a payoff of <span class="math-container">$\frac{20}{3}(1-p)$</span>, while playing <span class="math-container">$H$</span> gives him <span class="math-container">$\frac{10p}{3}+\frac{5}{3}$</span> and these are only equal when <span class="math-container">$p=\frac{1}{2}$</span>.
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CC BY-SA 4.0
2023-02-08T17:47:41.237Z
2023-02-08T18:06:34.027Z
2023-02-08T18:06:34.027Z
Amit
<p>Suppose player 1 chooses a mixed strategy such that the left type of player 2 is indifferent between C and H. There is a unique mixed strategy that does this. The same mixed strategy, however, does NOT make the right type of player 2 indifferent. Loosely speaking, this is simply because the preferences of the left type are exactly those of the right type switched around. Thus it is impossible for player 1 to make both types indifferent simultaneously. Therefore, the probability <span class="math-container">$p$</span> plays no role when choosing a strategy such that at least one of them is indifferent.</p>
<p>This also indicates why there are exactly two equilibria.</p>

CC BY-SA 4.0
2020-09-21T09:27:43.217Z
2020-09-21T09:27:43.217Z
induction_is_a_laddah
<p>I have a question about calculating mixed strategies in a Bayes Nash Equilibrium in a simple 2-player bimatrix game. To demonstrate the issue, consider ``Bayesian Battle of the Sexes.'' Suppose P1 faces a P2 whose type is unknown with probability <span class="math-container">$p$</span>. An example is below in which <span class="math-container">$p=.5$</span>.</p>
<p><a href="https://i.sstatic.net/oqb6T.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/oqb6T.jpg" alt="enter image description here" /></a></p>
<p>Here's what's surprising to me. If you calculate the mixing strategy for P1 by making sure that P2 is indifferent, you wind up with P1 mixing strategies <strong>that don't depend at all on <span class="math-container">$p$</span></strong>. This seems strange to me.</p>
<p><strong>If P2's type is more clear, shouldn't the optimal mixed strategy shift to anticipate this?</strong></p>
<p>End of question. Addendum: The game above apparently has two mixed strategy EQMs:</p>
<ol>
<li>P1 mixes <span class="math-container">$(\frac{1}{3},\frac{2}{3})$</span>, P2 left mixes (0,1), P2 right (<span class="math-container">$\frac{2}{3},\frac{1}{3}$</span>).</li>
<li>P1 mixes <span class="math-container">$(\frac{2}{3},\frac{1}{3})$</span>, P2 left mixes (<span class="math-container">$\frac{2}{3},\frac{1}{3}$</span>), P2 right (0,1).</li>
</ol>
<p>Like I said, this doesn't seem to depend on <span class="math-container">$p$</span> (the probability of facing the first type).</p>
<p><strong>Also:</strong> Shouldn't this game have an odd # of EQM? I thought this was a rule of thumb. I am contemplating the idea that there's a third equilibrium that mixes the two above.</p>

<game-theory><nash-equilibrium><bayesian-game>
Mixed Strategies in Bayes Nash Equilibrium (Bayesian Battle of the Sexes). Shouldn't it depend on $p$?
ABC
# Mixed Strategies in Bayes Nash Equilibrium (Bayesian Battle of the Sexes). Shouldn't it depend on $p$?

(Asked by: username_0 on 2020-09-09)

I have a question about calculating mixed strategies in a Bayes Nash Equilibrium in a simple 2-player bimatrix game. To demonstrate the issue, consider ``Bayesian Battle of the Sexes.'' Suppose P1 faces a P2 whose type is unknown with probability $p$. An example is below in which $p=.5$.

[![enter image description here](https://i.sstatic.net/oqb6T.jpg)](https://i.sstatic.net/oqb6T.jpg)

Here's what's surprising to me. If you calculate the mixing strategy for P1 by making sure that P2 is indifferent, you wind up with P1 mixing strategies **that don't depend at all on $p$**. This seems strange to me.

**If P2's type is more clear, shouldn't the optimal mixed strategy shift to anticipate this?**

End of question. Addendum: The game above apparently has two mixed strategy EQMs:

1. P1 mixes $(\frac{1}{3},\frac{2}{3})$, P2 left mixes (0,1), P2 right ($\frac{2}{3},\frac{1}{3}$).
2. P1 mixes $(\frac{2}{3},\frac{1}{3})$, P2 left mixes ($\frac{2}{3},\frac{1}{3}$), P2 right (0,1).

Like I said, this doesn't seem to depend on $p$ (the probability of facing the first type).

**Also:** Shouldn't this game have an odd # of EQM? I thought this was a rule of thumb. I am contemplating the idea that there's a third equilibrium that mixes the two above.

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


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## 2 answers

### Answer by username_1 on 2023-02-08. Score: 4

[![enter image description here](https://i.sstatic.net/bEc7p.png)](https://i.sstatic.net/bEc7p.png)
Given game also has one pure-strategy Nash equilibrium besides the two you mentioned: Row player plays $C$, Column player (left) plays $C$ and right plays $H$ i.e. $C,H$.

Also, when player 1 chooses a mixed strategy in order to make player 2 indifferent, that does not necessarily mean that player 1 is choosing optimally. It is only when player 2 makes player 1 indifferent, then the mixed strategy chosen by 1 becomes optimal. In fact as $p$ gets close to $1$ (not necessarily equals 1), given that player 2 (left) chooses $H$ and player 2 (right) chooses $(\frac{2}{3}, \frac{1}{3})$, player 1 will not choose $(\frac{1}{3},\frac{2}{3})$, since player 1 is no longer indifferent between $C$ and $H$. Playing $C$ will give row player a payoff of $\frac{20}{3}(1-p)$, while playing $H$ gives him $\frac{10p}{3}+\frac{5}{3}$ and these are only equal when $p=\frac{1}{2}$.
[![enter image description here](https://i.sstatic.net/tLDQe.png)](https://i.sstatic.net/tLDQe.png)


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### Answer by username_2 on 2020-09-21. Score: 0

Suppose player 1 chooses a mixed strategy such that the left type of player 2 is indifferent between C and H. There is a unique mixed strategy that does this. The same mixed strategy, however, does NOT make the right type of player 2 indifferent. Loosely speaking, this is simply because the preferences of the left type are exactly those of the right type switched around. Thus it is impossible for player 1 to make both types indifferent simultaneously. Therefore, the probability $p$ plays no role when choosing a strategy such that at least one of them is indifferent.

This also indicates why there are exactly two equilibria.


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