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A binary game is a finite normal-form game in which each player's payoff at every pure strategy profile is either zero or one. This describes the numerical payoffs, not the number of players or available actions. A mixed strategy can have an intermediate expected payoff. If a weighted average of binary payoffs is at least one, every positive-weight term must equal one; this observation makes pure-by-mixed domination unusually simple in binary games.

Write column $j$ of a two-row binary game as $((a_j,c_j),(b_j,d_j){)}^T$, with the row player's payoff first. If there is no pure Nash equilibrium, the column player's maximum payoff in each row must be one. Otherwise every column would be a best response in a zero-payoff row, and excluding an equilibrium there would force the opposite row to dominate in every column, giving an equilibrium in that opposite row.

A column maximizing the column player's top-row payoff must then have $(a_j,b_j)=(0,1)$, or its top cell would be an equilibrium. Its bottom column-player payoff must be zero, or its bottom cell would be one. Thus a column $((0,1),(1,0){)}^T$ occurs. The same argument for the bottom row produces $((1,0),(0,1){)}^T$. No column can give the column player payoffs $(1,1)$, since a row-player best response within that column would be a pure equilibrium.

Consequently, if neither pure equilibrium is found, mix equally over the two rows and equally over those two special columns. The row player's two expected payoffs are $1/2$; every column gives the column player at most $1/2$, and the supported columns achieve $1/2$. This is a Nash equilibrium. Computing the two row maxima, scanning for pure equilibria and locating the two special columns takes $O(n)$ running time for $n$ columns.

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.