Past exam of the computer science course of the University of Oxford / 2019 / Computational game theory / 2 / c / Solution

As literally stated, the claim is false. Give player one two pure actions, with payoffs one and zero against every opposing action. A mixed strategy assigning probability one to the first action has exactly that action's payoffs, so very weakly dominates it under the printed non-strict inequality. The other pure action nevertheless fails to dominate it.

The intended statement becomes true if the mixture gives positive probability to some distinct action, equivalently $p_i<1$. Fix any opposing pure strategy profile at which $s_i$ pays one. Since all payoffs are in $\{0,1\}$,Equality forces every positive-probability supported action to pay one there. At a profile where $s_i$ pays zero, every other action already pays at least zero. Therefore every action in the support of a mixed strategy is no worse than $s_i$ against every opposing profile. Choosing a distinct supported action proves the corrected conclusion: nontrivial mixed very weak dominance implies very weak domination by a distinct pure action. This proof does not insert the missing support condition silently.