<p>It is not possible for a mixed (non-pure) strategy to be strictly dominant. A mixed strategy can be weakly dominant, but only if all pure strategies in its support are weakly dominant.</p>
<p>This essentially follows from the fact that it is impossible for the (weighted) average of several numbers to be larger than all the original numbers. In this setting, fix any (possibly mixed) strategy for player 2. Consider a mixed strategy for player 1 that plays each pure strategy <span class="math-container">$s_i$</span> with probability <span class="math-container">$p_i$</span>. Let <span class="math-container">$u(s_i)$</span> be the expected utility that player <span class="math-container">$1$</span> receives by playing <span class="math-container">$s_i$</span> against the strategy of player <span class="math-container">$2$</span>. Then their overall expected utility is <span class="math-container">$\sum_{i} p_i u(s_i)$</span>. But since <span class="math-container">$\sum_{i} p_i = 1$</span>, this in turn is at most <span class="math-container">$\max_{i} u(s_i)$</span>, which is the utility they would have gotten from playing their best pure strategy, meaning the mixed strategy could not have been strictly dominant.</p>

CC BY-SA 4.0
2024-05-15T05:36:29.353Z
2024-05-15T05:36:29.353Z
jschnei
<p>Consider a simultaneous-move game with two players, 1 and 2. Player 1 does not have any pure strategy that is a strictly dominant strategy. Is it possible for player 1 to have a mixed strategy that is a strictly dominant strategy?</p>
<p>A strictly dominant strategy can be defined as a strategy, let's say <span class="math-container">$s_d$</span>, with a payoff that is larger than the payoff of any other strategy <span class="math-container">$s_i$</span>, irrespective of the strategy played by player 2. Following this, is a strictly dominant mixed strategy defined as the combination of <span class="math-container">$n$</span> strategies, let's say <span class="math-container">$S_m$</span> with associated probabilities <span class="math-container">$p_m$</span> such that the expected payoff of the combination of strategies <span class="math-container">$S_m$</span> is larger than all other possible expected payoffs of mixed strategies? Should mixed strategies with the same number of strategies, n, be looked at, or should all others also be included into the comparison? If it is defined like that, how should the question above be approached? Is there any way to prove the answer mathematically?</p>

<game-theory><dominant-strategy>
Can a mixed strategy that is strictly dominant exist if there is no strictly dominant pure strategy?
Jovan Jezdic
# Can a mixed strategy that is strictly dominant exist if there is no strictly dominant pure strategy?

(Asked by: username_0 on 2024-05-14)

Consider a simultaneous-move game with two players, 1 and 2. Player 1 does not have any pure strategy that is a strictly dominant strategy. Is it possible for player 1 to have a mixed strategy that is a strictly dominant strategy?

A strictly dominant strategy can be defined as a strategy, let's say $s\_d$, with a payoff that is larger than the payoff of any other strategy $s\_i$, irrespective of the strategy played by player 2. Following this, is a strictly dominant mixed strategy defined as the combination of $n$ strategies, let's say $S\_m$ with associated probabilities $p\_m$ such that the expected payoff of the combination of strategies $S\_m$ is larger than all other possible expected payoffs of mixed strategies? Should mixed strategies with the same number of strategies, n, be looked at, or should all others also be included into the comparison? If it is defined like that, how should the question above be approached? Is there any way to prove the answer mathematically?

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

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## 1 answers

### Answer by username_1 on 2024-05-15. Score: 2 | Accepted answer

It is not possible for a mixed (non-pure) strategy to be strictly dominant. A mixed strategy can be weakly dominant, but only if all pure strategies in its support are weakly dominant.

This essentially follows from the fact that it is impossible for the (weighted) average of several numbers to be larger than all the original numbers. In this setting, fix any (possibly mixed) strategy for player 2. Consider a mixed strategy for player 1 that plays each pure strategy $s\_i$ with probability $p\_i$. Let $u(s\_i)$ be the expected utility that player $1$ receives by playing $s\_i$ against the strategy of player $2$. Then their overall expected utility is $\sum\_{i} p\_i u(s\_i)$. But since $\sum\_{i} p\_i = 1$, this in turn is at most $\max\_{i} u(s\_i)$, which is the utility they would have gotten from playing their best pure strategy, meaning the mixed strategy could not have been strictly dominant.


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