Explaining mixed strategies for one-shot games

Question

In the classic introduction to non-cooperative game theory, the mixed strategy for a player is taught as a distribution over strategy space for the player. The distribution essentially gives us the probabilities (say, discrete strategy set) with which a player should play the strategies in a Nash equilibrium.

However probabilities carry the notion of being frequencies and these essentially mean the long-run fraction of games in which the player should play the strategy. However the setting is a one-shot game and this is a contradiction.

How do we resolve the contradiction when explaining what a mixed strategy is?

Answer 1

Ariel Rubinstein tends to be insightful regarding these kinds of questions.

He addresses the interpretation of mixed strategies in section 3 of this paper.

A few possible interpretations aside from deliberate randomization:

1. Purification: A mixed strategy is a plan of action based on information not specified in the model.
2. A fictitious long run story.
3. Population averages, so imagine the player's being pulled from some population distribution where different types play different pure strategies. The population distribution is the mixed strategy distribution.

An interesting quote regarding player $i$'s mixed strategy reflecting uncertainty among $-i$'s regarding what $i$ will do:

Mixed strategy can alternatively be viewed as the belief held by all other players concerning a player's actions. A mixed strategy equilibrium is then an n-tuple of common knowledge expectations, which has the property that all the actions to which a strictly positive probability is assigned are optimal, given the beliefs. A player's behavior may be perceived by all the other players as the outcome of a random device even though this is not the case. Adopting this interpretation requires the reassessment of much of applied game theory. In particular, it implies that an equilibrium does not lead to a prediction (statistical or otherwise) of the players' behavior. Any player i's action which is a best response given his expectation about the other players' behavior (the other n - 1 strategies) is consistent as a prediction for i's action (this might include actions which are outside the support of the mixed strategy). This renders meaningless any comparative statics or welfare analysis of the mixed strategy equilibrium and brings into question the enormous economic literature which utilizes mixed strategy equilibrium.

Answer 2

Let $s_i = \{p_A^i, p_B^i\}$ denote a strategy that attaches probabilities to playing $A,B$, and let $s = \{s_i, s_i\}_i$ be the set of such strategies that result in an equilibrium in a two-player symmetric game.

As you say, we think about $s_i$ to be probabilities with which a specific action is played. Whenever $s$ is not a singleton, we have multiple-equilibria, something which most branches of economics dislike, because it makes solving models quite difficult, and non-uniqueness is difficult to work with: How should we simulate the model? Which one of the equilibria is actually being played?

At least, with mixed-strategy equilibria, we know the likelihood of each of the equilibria happening. You dislike probabilities to the extent that they carry frequencies, which you say are contradicted by the notion of the game being one-shot.

Simultaneously However, the game being one-shot does not mean the game is only played once. In a world with many individuals, everyone can find a partner and play one of the strategies in $s$, to the extent that we (at the same time!) find $p_A$ of them in the equilibrium $\{A, A\}$, and the fraction $p_B$ of individuals playing the next equilibrium, etc.

Non-Simulatenously As an alternative, you could argue that in a world with a lot of anonymity, people forget the partners that they played with before. We have many people playing strategies in $s$ at time $t$, then we de-couple them, give everyone new partners and let them play again. Even if there is the possibility of meeting the same guy again: Since that possibility goes to zero, you could model this as a repeated game with a discount factor $\delta\rightarrow 0$.

Lack of Commitment Finally, think about situations which are actually repeated games, such as interactions between the government and the consumers. While this could be modeled as a repeated game, we might think that the government is not able to commit to a strategy sequence. Therefore, instead of modeling this as a repeated game, we model it as repetitions of the one-shot equilibrium: Given a time horizon $T$, we will see that $T\cdot p_A$ of the times, the government and the consumers play equilibrium $\{A, A\}$, etc.

Answer 3

This is a supplement of Pburg's quote:

One view in Aumann and Brandenburger(1995) is that mixed strategy is only in the eyes of opponents. In a $N$-player game, the set of states of world $\mathbf {S} : = \times_{i \in N} S_i$. For a state $s \in \mathbf S$, it satisfies the following specification:

1. For a player $i$, let $\pi_i : \mathbf {S} \to S_i$ be the projection on a state´s $i$th component. When a state is realized, player $i$ is absolutly sure about her own type $s_i$, but unsure about the exact state. In other word, she doesn't know which state in $\pi_i^{-1}(s_i)$ is obtained. Instead, she held a belief on $\pi_i^{-1}(s_i)$, which is specified by $s_i$.
2. Let $A_i$ be player $i$'s action space. Her action is a random variable $a_i : \mathbf{S} \to A_i$, while its restriction $\left.a_i\right|_{\pi_i^{-1}(s_i)}$ is constant.
3. Player $i$´s utility function $g_i$ is defined in the same fashion as $a_i$, which means $g(s) : \mathbf{A} \to \mathbb{R}$ is referring to the same utility function, for all $s \in \pi_i^{-1}(s_i)$ for all $s_i$.

Answer 4

Well, here is my shot at answering, following this paper in Physics http://bayes.wustl.edu/etj/articles/prob.in.qm.pdf . I think, that propensity is a nice interpretation of mixed strategies, but more formally we should say it captures the ignorance of the modeller. We say, anything goes, in fact all strategies could be taken (if the support is everywhere positive) but the solution concept says certain are more likely. Probabilities here measure the ignorance of the modeller and are a result of the lack of information of the game theorist about the game. To clarify this think of an enhanced dataset where we know additional information about the game, say we speak with one of the players an he assures us that he is going to take one strategy no matter what, then we can make a sharper prediction in the form of a pure strategy. Frequencies arise when we think of the game as a typical game, and we see several independent situations where players with the same preferences play this game several times then mixed strategies will correspond indeed in the limit (asymptotic argument) to the frequency of strategies observed if our model is indeed correct.

Answer 5

It doesn't apply to all games, but there are also situations in which (at least some of) the players actually use randomization devices in games which could be viewed as one-shot. Here, probability distributions are not frequencies, they are the distributions that the randomization device uses. Any mixed-strategy equilibrium is then an equilibriun in an ex-ante sense (although players might very well draw from the randomization device a single time, and there might not be any sense in which the ex-post situation is an equilibrium).

Examples include:

• https://www.geek.com/apps/tsa-paid-1-4-million-for-randomizer-app-that-chooses-left-or-right-1651337/
• https://www.prnewswire.com/news-releases/new-startup-formed-to-commercialize-patented-usc-technology-that-uses-game-theory-to-intelligently-randomize-security-patrols-for-safer-airports-borders-municipalities-and-waterways-198028371.html, http://teamcore.usc.edu/news-content.htm, https://magazine.viterbi.usc.edu/spring-2015-2/features/a-safer-world/