<blockquote>
<p>you would get a higher payoff by assigning a probability of 0 to the pure strategy that is dominated</p>
</blockquote>
<p>This is not relevant, as no one is claiming that <span class="math-container">$\sigma_1$</span> is optimal in any way.</p>
<blockquote>
<p>I can't find any way to formally argue for this or give a concise example.</p>
</blockquote>
<p>Study the following game with two players: <br>
Player 1 chooses A or B, while player 2 chooses between A, B and &quot;pass&quot;. We are not trying to find an equilibrium, we will be focusing on player 2, laying out their payoff only. <br>
Should player 2 choose the same strategy (A or B) as player 1, they will get a payoff of 5. Should player 2 choose &quot;pass&quot; they will get a payoff of 4. In all other cases player 2 gets a payoff of 0.</p>
<p>Show the following:</p>
<ol>
<li>Player 2 has no dominated pure strategies in this game.</li>
<li>Player 2 has a mixed strategy <span class="math-container">$\sigma_1$</span> (likely your first guess) in this game that is dominated by a pure strategy.</li>
</ol>

CC BY-SA 4.0
2023-04-11T05:12:24.090Z
2023-04-11T05:12:24.090Z
Giskard
<p>I would like help with the following:</p>
<p>Assume a game with two individuals. Suppose that for an individual a mixed strategy <span class="math-container">$\sigma_1$</span> is dominated by an strategy <span class="math-container">$\sigma_2, $</span> it is necessary for <span class="math-container">$\sigma_1$</span> to put a positive probability <span class="math-container">$p$</span> to a dominated pure strategy?</p>
<p>My guess is that it isn't necessary, since you would get a higher payoff by assigning a probability of 0 to the pure strategy that is dominated, but I can't find any way to formally argue for this or give a concise example.</p>

<game-theory><dominant-strategy>
Understanding mixed strategies
Haus
# Understanding mixed strategies

(Asked by: username_0 on 2023-04-11)

I would like help with the following:

Assume a game with two individuals. Suppose that for an individual a mixed strategy $\sigma\_1$ is dominated by an strategy $\sigma\_2, $ it is necessary for $\sigma\_1$ to put a positive probability $p$ to a dominated pure strategy?

My guess is that it isn't necessary, since you would get a higher payoff by assigning a probability of 0 to the pure strategy that is dominated, but I can't find any way to formally argue for this or give a concise example.

**Tags:** game-theory, dominant-strategy

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## 1 answers

### Answer by username_1 on 2023-04-11. Score: 1 | Accepted answer

> you would get a higher payoff by assigning a probability of 0 to the pure strategy that is dominated

This is not relevant, as no one is claiming that $\sigma\_1$ is optimal in any way.

> I can't find any way to formally argue for this or give a concise example.

Study the following game with two players:   
Player 1 chooses A or B, while player 2 chooses between A, B and "pass". We are not trying to find an equilibrium, we will be focusing on player 2, laying out their payoff only.   
Should player 2 choose the same strategy (A or B) as player 1, they will get a payoff of 5. Should player 2 choose "pass" they will get a payoff of 4. In all other cases player 2 gets a payoff of 0.

Show the following:

1. Player 2 has no dominated pure strategies in this game.
2. Player 2 has a mixed strategy $\sigma\_1$ (likely your first guess) in this game that is dominated by a pure strategy.


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