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A pure strategy is a definite, complete plan for a player's choices. In a game with one simultaneous choice, it selects one available action with certainty: heads or tails in matching pennies, for example. In a sequential game it specifies an action at every situation where that player could be called upon to move, including situations never reached when the plan is followed. The plan can use only information available to the player: indistinguishable decision situations must prescribe the same action. An action at one such situation is only part of the plan.

If $S_i$ is player $i$'s set of pure strategies, the payoff $u_i(s_i,s_{-i})$ depends on the whole strategy profile. Thus the same pure strategy can be a best response against one opposing choice and perform poorly against another. A dominant strategy is the stronger case in which it is a best response against every opposing strategy profile.

Two strategies for the same player are realization-equivalent when they induce the same probability distribution on terminal histories against every fixed opposing strategy profile. The requirement is stronger than having equal expected payoff against one particular opponent. For a finite extensive-form game with perfect recall, Kuhn's theorem gives this equivalence between appropriate mixed strategies and behavioural strategies.

A strategy profile specifies one strategy for every player. A pure strategy strategy profile lies in the Cartesian product $S_1\times \cdots \times S_n$ of the players' strategy sets. A mixed strategy strategy profile specifies one probability distribution ${\sigma }_i$ over each $S_i$. Write $s_{-i}$ or ${\sigma }_{-i}$ for the choices of all players except player $i$.

A strategy profile that is a Nash equilibrium in every subgame, including those outside the realized play path. In finite sequential games, backward induction constructs such profiles by solving continuation games before earlier decisions. When a stage is an all-pay contest, its solution can involve mixed strategies.

A pure strategy is very weakly dominated by another strategy when its payoff is no greater against every opposing pure strategy profile, with no strict advantage required anywhere. The definition permits payoff-equivalent strategies and even domination by the original strategy itself. A weakly dominated strategy instead has a strict inequality for at least one opposing strategy profile.

In a binary game, suppose ${\sigma }_i$ very weakly dominates $s_i$ and assigns positive probability to a different pure action. At every opposing strategy profile where $s_i$ pays one, all actions in the support of a mixed strategy must also pay one. Where $s_i$ pays zero, their payoffs are automatically no smaller. Therefore every distinct supported action very weakly dominates $s_i$. If the mixed domination is weak with a strict improvement somewhere, choose a supported action paying one at a profile where $s_i$ pays zero; that action also supplies a strict improvement. The requirement of nontrivial support is indispensable for the very-weak version.

A strategy giving at least as large a payoff as every alternative against every opponent strategy profile. Under the convention distinguishing weak dominance from complete payoff equivalence, require a strict advantage somewhere. In a two-action finite game, it suffices to check the opponent's pure strategies: expected payoff differences against an opponent's mixed strategy are weighted averages of those pure-action comparisons.

In a zero-sum game, the players' payoffs sum to zero at every pure strategy strategy profile. For two players this means $u_2=-u_1$: maximizing the column player's payoff is minimizing the row player's. A single payoff matrix $A$ therefore specifies a finite two-player matrix game, and independent mixed strategies $x,y$ give expected payoffs $x^{\mathsf{T}}Ay$ and $-x^{\mathsf{T}}Ay$.

Matching pennies and the usual rock paper scissors are two-player zero-sum games. The asymmetric coordination example is not: its total payoff is $5$ when choices match and $0$ otherwise. Applying a two-player minimax theorem to general bimatrix games would discard this possibility of shared gains. For a two-player zero-sum game, however, optimal mixed strategies guarantee the value of a zero-sum game, and pairing any two opposing optimal strategies gives a Nash equilibrium.